2023～2024 学年福州市高三年级4月份质量检测--简略答案4月9日

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