2024届福建省福州市高三下学期4月末质量检测(三模)数学答案

2024-04-29 · 9页 · 196.7 K

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-1…nn-1当n2时,a=(a-+a)(a-a)+L+()a-+aa,…nnn-1nn--12211所以an=+2nn2-2+L++42,·························································3分nn(2+2)所以an=,2,所以an=+2nn,2,··································4分n2…n…又因为a1=2,···············································································5分2*所以an=n+Înn,N.······································································6分2*(2)由(1)可知an=n+n=n(nn+Î1),N,·············································7分1111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,µσ==0,0.2,···························································1分所以零件为合格品的概率为P(-0.6===.mn51´5参考答案第3页共9页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,H1G所以GQ∥AB,GQ=AB,····························2分2BEQ∥,C又因为ABEF,AB=EFP∥所以GQHE,GQ=HE,·································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AF因为GH,分别为线段AP,EF的中点,DIH所以GI∥BP,HI∥EB,GB又因为GIËÌ面面BCE,BPBCE,ECP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

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