福建省福州市2023-2024学年高三下学期4月末质量检测数学试卷＋答案

{#{QQABJYCUogigQIBAARhCQQGgCgEQkBEAACoGhAAIIAAByANABAA=}#} {#{QQABJYCUogigQIBAARhCQQGgCgEQkBEAACoGhAAIIAAByANABAA=}#} {#{QQABJYCUogigQIBAARhCQQGgCgEQkBEAACoGhAAIIAAByANABAA=}#} {#{QQABJYCUogigQIBAARhCQQGgCgEQkBEAACoGhAAIIAAByANABAA=}#} 2023～2024学年福州市高三年级4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．D6．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．ABC10．BC11．ACD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．-213．814．6，22四、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.【考查意图】本小题主要考查递推数列与数列求和等基础知识，考查运算求解能力、推理论证能力等；考查分类与整合、化归与转化等思想方法；考查数学运算、逻辑推理等核心素养；体现基础性和综合性.满分13分.解：（1）因为a=+a2nn,2，所以a-=an2，···································1分nn-1nn-1当n2时，a=(a-+a)(a-a)+L+()a-+aa，nnn-1nn--12211所以a=+2nn2-2+L++42，·························································3分nnn(2+2)2所以an=,2，所以an=+nn,2，··································4分nn2又因为a=2，···············································································5分12*所以a=n+Înn,N.······································································6分n2*（2）由（1）可知a=n+n=n(nn+Î1),N，·············································7分n1111所以==-，····························································9分ann(n++11)nn1111所以S=++L++n1´22´3(n-+1)nnn(1)1111111=1-+-+L+-+-，·····································11分223n-+11nnn1所以S=-1，·········································································12分nn+1又因为n1，所以S<1.·································································································13分n参考答案第1页共9页{#{QQABJYCUogigQIBAARhCQQGgCgEQkBEAACoGhAAIIAAByANABAA=}#} 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、逻辑思维能力和运算求解能力等，考查分类与整合思想、概率与统计思想等，考查数学建模、数据分析、数学运算等核心素养，体现基础性、综合性和应用性．满分15分.解：（1）依题意得，ms==0,0.2，···························································1分所以零件为合格品的概率为P(-0.6<X<0.6)=PX(m-3s<<ms+=3)0.9973，···································································································2分零件为优等品的概率为P(-0.2<X<0.2)=PX(m-s<<ms+=)0.6827，·····3分所以零件为合格品但非优等品的概率为P=0.9973-=0.68270.3146，···········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´»0.314631.·············································································6分（2）设从这批零件中任取2个作检测，2个零件中有2个优等品为事件A，恰有1个优等品，1个为合格品但非优等品为事件B，从这批零件中任取1个检测是优等品为事件C，这批产品通过检测为事件D，····························································8分则D=+ABC，且A与BC互斥，·······················································9分所以P(D)=+P(A)P()BC·································································10分=+P(A)PB()P(CB|)·························································11分221=CC´0.6827+´´´0.68270.31460.6827222=´1.62920.6827，····························································12分所以这批零件通过检测时，检测了2个零件的概率为P()ADP(AD|)=···········································································13分PD()20.6827=21.6292´0.68271=1.6292»0.61.·············································································15分答：这批零件通过检测时，检测了2个零件的概率约为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、平面与平面的夹角、空间向量、三角函数的概念等基础知识，考査直观想象能力、逻辑推理能力、运算求解能力等,考查数形结合思想、化归与转化思想等，考査直观想象、逻辑推理、数学运算等核心素养，体现基础性、综合性.满分15分.解法一：（1）在正方形ABEF中，连接AH并延长，交BE的延长线于点K，连接PK.···································································································2分参考答案第2页共9页{#{QQABJYCUogigQIBAARhCQQGgCgEQkBEAACoGhAAIIAAByANABAA=}#} 因为GH,分别为线段AP,EF中点，AFD所以HF=HE，H所以Rt△AFH≌Rt△KEH，GBE所以AH=KH，·····························4分KCP所以GH∥PK.································5分又因为GHËÌ面面BCE,PKBCE，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,ABIAEA=,AB,AEÌ面ABEF，所以BP^面ABEF，········································································8分又BEÌ面ABEF，所以BP^BE，·····················································9分所以BP,,BEBA两两垂直.z以B为原点，BP,,BEBA所在直线分别为x,,yz轴建立空AFD间直角坐标系，如图所示.HG································································10分BEy不妨设AB=1,CP31则PD(1,0,0),(,-,1)，x22uuuruuuræö31BP=(1,0,0),BD=-ç÷ç÷,,1，·························································11分èø22uuurìïBP×=m0,设平面BPD的法向量为m=(xyz,,)，则íuuurïîBD×=m0,ìx=0,ï即í31取y=2，得xz==0,1，ïx-yz+=0,î22所以平面BPD的一个法向量是m=(0,2,1)，·········································13分又平面BPA的一个法向量为n=(0,1,0).················································14分mn×225设平面BPD与平面BPA的夹角为q，则cosq=cos,<mn>===.mn51´5参考答案第3页共9页{#{QQABJYCUogigQIBAARhCQQGgCgEQkBEAACoGhAAIIAAByANABAA=}#} 25所以平面DBP与平面BPA夹角的余弦值为.····································15分5解法二：（1）证明：取BP的中点Q，连接GQ,EQ.·····1分AFD因为GH,分别为线段AP,EF的中点，HG1所以GQ∥AB，GQ=AB，····························2分2BEQ又因为AB∥EF,AB=EF，CPGQ∥HE,GQ=HE所以，·································3分所以四边形GQEH是平行四边形，······················································4分所以GH∥QE，··············································································5分又因为GHËÌ面BCE,QE面BCE，所以GH∥面BCE.············································································7分（2）同解法一.····················································································15分解法三：（1）证明：取AB的中点I，连接GI,HI.········1分AFD因为GH,分别为线段AP,EF的中点，IH所以GI∥BP,HI∥EB，GBE又因为GIËÌ面面BCE,BPBCE，CP所以GI∥面BCE.············································3分因为HIËÌ面面BCE,BEBCE，所以HI∥面BCE.············································································5分又因为GIIHI=I,,GIÌÌ面面GIHHIGIH，所以面GIH∥面BCE，·····································································6分又因为GHÌ面GIH，所以GH∥面BCE.············································································7分（2）同解法一.····················································································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,考查数形结合思想、化归与转化思想、分类与整合思想等，考査直观想象、逻辑推理、数学运算等核心素养，体现基础性、综合性与创新性.满分17分.222解法一：（1）依题意，b+=ca．··························································1分参考答案第4页共9页{#{QQABJYCUogigQIBAARhCQQGgCgEQkBEAACoGhAAIIAAByANABAA=}#} 22xy00设P(xy,)，则+=1，-a<<xa，00220ab222222x0b222所以|PF|=(x-c)+y=(x-c)+b(1-)=(1-)2x-cx++bc，20002200aa2cc22所以|PF|=x-2cx+a=-||xa，············································3分22000aa22caca又ac>，所以ax>，>x，所以||PF=-ax，dx=-00200acaccax-||PFa0c||PFc2==2所以，即为定值，且这个定值为．··················4分2daada-x0cxy00（2）（ⅰ）依题意，G(,)，33x0设直线IG与x轴交于点C，因为IG⊥x轴，所以C(,0)，·······················5分3xx200所以|FC|-|FC|=(+-c)()cx-=，··········································6分120333因为△PF1F2的内切圆与x轴切于点C，2所以|PF|-|PF|=|FC|-=||FCx，·················································7分121203又因为|PF|+=|PFa|2，12xy0解得||PFa=-，··········································8分23Pc由（1）得||PF=-ax，20GaIxFOCFcx120所以a-xa=-，0a3c1所以椭圆E的离心率e==．·························10分a3c1222（ⅱ）由26a=，得a=3，又=，所以c=1，b=ac-=8，a322xy所以椭圆E的方程为+=1．······················································11分98根据椭圆对称性，不妨设点P在第一象限或y轴正半轴上，即03x0<，0<y022，又F(-1,0)，F(1,0)，12y0所以直线PF1的方程为yx=+(1)，x+10xyx(+3)000设直线IG与PF1交于点D，因为xD=，所以yD=，33(x+1)0△F1CD的面积S1与△PF1F2的面积S之比为1xyx(+3)000(+´1)2S233(xx++1)(3)100==，················································13分Sx118(+1)´´2y0022(x+3)(xx+3)(-1)令fx()=（03x<），则fx¢()=，218(x+1)18(1x+)参考答案第5页共9页{#{QQABJYCUogigQIBAARhCQQGgCgEQkBEAACoGhAAIIAAByANABAA=}#} 当xÎ[0,1)，fx¢()0<，当xÎ(1,3)，fx¢()0>，所以函数fx()在[0,1)单调递减，在(1,3)单调递增.141又因为f(0)=，f(1)=，f(3)=，29241所以fx()的值域是[,]，9241S1所以，··········································································15分92S4S1所以1，·······································································16分5SS-1根据对称性，45△PF1F2被直线IG分成两个部分的图形面积之比的取值范围是[,]．········17分54解法二：（1）同解法一···········································································4分xy00（2）（ⅰ）依题意，G(,)，33x0设直线IG与x轴交于点C，因为IG⊥x轴，所以C(,0)，·······················5分3xx200所以|FC|-|FC|=(+-c)()cx-=，··········································6分120333因为△PF1F2的内切圆与x轴切于点C，2所以|PF|-|PF|=|FC|-=||FCx，·················································7分121203ìx0|PFa|,=+ï1ï3又因为|PF12|+=|PFa|2，得í···········································8分xï|PFa|.=-02ïî3ì22x0(x+c),+ya=+ï004ï3所以í两式平方后取差，得4cx00=ax对任意x0成立，ï(x-c),22+ya=-x0300ïî3c1所以椭圆E的离心率e==．························································10分a3（ⅱ）同解法一···················································································17分解法三：（1）同解法一···········································································4分xyx000（2）（ⅰ）依题意，G(,)，因为IG⊥x轴，设点I坐标为(,)t.··········5分333y0可求直线PF方程为y=+()xc，1xc+0x0y(+c)-+t()xc003则点I到直线PF的距离=||t，·································6分122y++()xc002æöx0222即ç÷y(+c)-t(x+c)=t(y++()xc)，化简得0000èø322xx00yt+2t(+c)(x+c)-yc(+=)0，①00033同理，由点I到直线PF的距离等于||t，可得2参考答案第6页共9页{#{QQABJYCUogigQIBAARhCQQGgCgEQkBEAACoGhAAIIAAByANABAA=}#} 22xx00yt+2t(-c)(x-c)-yc(-=)0，②············································7分0003384y0将式①-式②，得2t×cx=×ycx，则t=.·····································8分000334y0将t=代入式①，得42y01xx002+(+c)(x+cc)-(+=)0，01623322xy00化简得+=1，2298cc22得9ca=，c1所以椭圆E的离心率e==．························································10分a3（ⅱ）同解法一···················································································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、直观想象能力、运算求解能力和创新能力等，考查函数与方程思想、化归与转化思想、分类与整合思想、数形结合思想等，考查数学抽象、逻辑推理、直观想象、数学运算等核心素养，体现基础性、综合性与创新性.满分17分.解：（1）依题意，因为l()xLÎ，0f(xx),ÎR22所以"xÎR,1-x++xkx，且xR，-x+x=+kx1，····················1分000022令f(x)=-x+(1--kx)1，D=(k--14)，则f(x)0，且f()0x=，0ìD0,所以í所以D=0，···································································3分îD0,2即(k-1)-=40，解得k=3或-1.··············································································4分（2）（ⅰ）先证必要性.x,e1x0+，若直线y=lx()是曲线y=gx()的切线，设切点为()0xy-exx00+1=-e()xx，因为gx¢()e=，所以切线方程为()0xx00即l(x)=exx+(1-+)e1（*）.························································5分0一方面，g(x)=lx()，所以$xÎ=R,g(x)lx()，································6分00000参考答案第7页共9页{#{QQABJYCUogigQIBAARhCQQGgCgEQkBEAACoGhAAIIAAByANABAA=}#} xxx00另一方面，令G(x)=g(x)-l(x)=e-exx--(1)e，则Gx()=0，00xx0因为Gx¢()=-ee，所以当xx<时，Gx¢()<0，Gx()在(-¥,x)单调递减，00当xx>时，Gx¢()>0，Gx()在(x,+¥)单调递增，00所以G(x)Gx()=0，所以g(x)lx().··············································7分0即"ÎxR,g(x)lx()，所以l(xT)Î，即lx()是函数gx()在R上的“最佳下界线”.···············8分g(xx),ÎR再证充分性.若lx()是函数gx()在R上的“最佳下界线”，不妨设l()x=+kxb，由“最佳下界线”的定义，"ÎxR,g(x)lx()，且$xÎ=R,g(x)lx()，····9分000x令H(x)=g(x)-l(x)=e1+--kxb，则Hx()0且Hx()=0，所以Hx()=0.·········································10分0minx因为H¢(xk)=-e，①若k0，则Hx¢()0，所以Hx()在R上单调递增，所以$<xx，使得H(x)<=Hx()0，故k0不符合题意.·····················11分1010②若k>0，令Hx¢()=0，得xk=ln，当xkÎ(-¥,ln)时，Hx¢()<0，得Hx()在(-¥,lnk)单调递减，当xkÎ(ln,+¥)时，Hx()>0，得Hx()在(ln,k+¥)单调递增，所以，当且仅当xk=ln时，Hx()取得最小值Hk(ln).····························12分又由Hx()在x处取得最小值，Hx()=0，0minìxk=ln,0所以í···········································································13分îHk(ln)=0,ïì=x0ke,即解得k=ex0，bx=(1-+)e1x0，····························14分íx0ïîe0+1-kxb-=0,0xx00所以l(x)=exx+(1-+)e1，0参考答案第8页共9页{#{QQABJYCUogigQIBAARhCQQGgCgEQkBEAACoGhAAIIAAByANABAA=}#} x,e1x0+处的切线.············15分由（*）式知直线y=lx()是曲线y=gx()在点()0综上所述，直线y=lx()是曲线y=gx()的一条切线的充要条件是直线y=lx()是函数gx()在R上的“最佳下界线”.（ⅱ）集合LTI元素个数为2个.············································17分h(x),xÎ(1,+¥Î)g(xx),R参考答案第9页共9页{#{QQABJYCUogigQIBAARhCQQGgCgEQkBEAACoGhAAIIAAByANABAA=}#}