福建省福州市2024届高三下学期4月末三模试题 数学 --简略答案4月9日

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参考答案 第 1 页 共9页
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．D 2．C 3．A 4．C 5．D 6．B 7．B 8．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．ABC 10．BC 11．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．
2
−
13．
8
14．6，22
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 1
2,2
nn
aann
−
=+
…
，所以 1
2
nn
aan
−
−=
，··································· 1分
当
n
…
时，
112211
()()()
nnnnn
aaaaaaaa
−−−
=−+−++−+
L，
所以
22242
n
ann
=+−+++
L，························································· 3分
所以 (22)
,2
2
n
nn
an
+
=
…
，所以 2
,2
n
annn=+
…
，·································· 4分
又因为 1
2
a
=
，··············································································· 5分
所以
2*
,
n
annn=+∈
N
.······································································ 6分
（2）由（1）可知
2*
(1),
n
annnnn=+=+∈
N
，············································· 7分
所以
( )
1111
11
n
annnn
==−
++
，···························································· 9分
所以 1111
1223(1)(1)
n
Snnnn
=++++
××−+
L
1111111
1
22311
nnnn
=−+−++−+−
−+
L，····································· 11 分
所以
1
1
1
n
Sn
=−
+
，········································································· 12 分
又因为
1
n
…
，所以
1
n
S
<
.································································································· 13 分
参考答案 第 2 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，
0,0.2
µσ
==
，··························································· 1分
所以零件为合格品的概率为
(0.60.6)(33)0.9973
PXPX
µσµσ
−<<=−<<+=
，
··································································································· 2分
零件为优等品的概率为
(0.20.2)()0.6827
PXPX
µσµσ
−<<=−<<+=
，····· 3分
所以零件为合格品但非优等品的概率为
0.99730.68270.3146
P
=−=
，··········· 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
1000.314631
×≈
.············································································· 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
DABC
=+
，且
A
与
BC
互斥，······················································· 9分
所以
()()()
PDPAPBC
=+
································································· 10 分
()()(|)
PAPBPCB
=+
························································· 11 分
221
22
0.68270.68270.31460.6827
CC=×+×××
2
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
()
(|)
()
PAD
PAD
PD
=··········································································· 13 分
2
2
0.6827
1.62920.6827
=×
1
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 3 页 共9页
因为
,
GH
分别为线段
,
APEF
中点，
所以
HFHE
=
，
所以 Rt
AFH
△≌Rt
KEH
△，
所以
AHKH
=
，····························· 4分
所以
GHPK
∥.································ 5分
又因为 ,
GHBCEPKBCE
⊄⊂面面
，
所以
GHBCE
∥面.··········································································· 7分
（2）依题意得，
ABBCE
⊥面，又因为
BPBCE
⊂面，所以
ABBP
⊥
.
又因为
BPAE
⊥
,
ABAEA
=
I,,
ABAEABEF
⊂面，
所以
BPABEF
⊥面，········································································ 8分
又
BEABEF
⊂面，所以
BPBE
⊥
，····················································· 9分
所以
,,
BPBEBA
两两垂直.
以
B
为原点，
,,
BPBEBA
所在直线分别为
,,
xyz
轴建立空
间直角坐标系，如图所示.
································································ 10 分
不妨设
1
AB
=
,
则
31
(1,0,0),(,,1)
22
PD−，
()
31
=1,0,0,,,1
22
BPBD

=−



uuuruuur ，························································· 11 分
设平面
BPD
的法向量为
(
)
,,
xyz
=m，则
0,
0,
BP
BD
⋅=
⋅=
uuur
uuur m
m
即
0,
31
0,
22
x
xyz
=
−+=
取
2
y
=
，得
0,1
xz
==
，
所以平面
BPD
的一个法向量是
(
)
0,2,1
=m，········································· 13 分
又平面
BPA
的一个法向量为
(
)
0,1,0
=n.················································ 14 分
设平面
BPD
与平面
BPA
的夹角为
θ
，则
225
coscos,
5
51
θ⋅
=<>===
×
mn
mn mn .
K
F
E
P
C
D
H
A
G
B
z
y
x
B
G
A
H
D
C
P
E
F
福建省福州市2024届高三下学期4月末三模试题 数学 --简略答案4月9日.pdf

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