文档内容
{{##{{QQQQAABBJIYYAC8U5oggCigwQwIIBTAAACRJ5hqCRQUQGG0gCCgggEQQskIABhELAKAoCEoRGRhCAHAqIIAARALBiJyNAINBAIAB=A}A#}=}#}{{##{{QQQQAABBJIYYAC8U5oggCigwQwIIBTAAACRJ5hqCRQUQGG0gCCgggEQQskIABhELAKAoCEoRGRhCAHAqIIAARALBiJyNAINBAIAB=A}A#}=}#}{{##{{QQQQAABBJIYYAC8U5oggCigwQwIIBTAAACRJ5hqCRQUQGG0gCCgggEQQskIABhELAKAoCEoRGRhCAHAqIIAARALBiJyNAINBAIAB=A}A#}=}#}{{##{{QQQQAABBJIYYAC8U5oggCigwQwIIBTAAACRJ5hqCRQUQGG0gCCgggEQQskIABhELAKAoCEoRGRhCAHAqIIAARALBiJyNAINBAIAB=A}A#}=}#}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aan=n +2, 2 ，所以aa- =n 2 ，···································1分
n n- 1 … n n- 1
当n 2时，aaa=a-(a)a()a( +-++a- ) + L ，
… nnnn n -- 11221 - 1
所以an=22n2+4-++2 + L ，·························································3分
n
n(22+)n
所以a = ,n 2，所以a =nn2 +n , 2， ··································4分
n 2 … n …
又因为a =2，···············································································5分
1
所以ann=+n2 Î , N*.······································································6分
n
（2）由（1）可知ann=n+n=2+ nÎ (1), N*，·············································7分
n
111 1
所以 == - ，····························································9分
annn ( n+1 ) +1
n
111 1
所以S =+++ +L
n 12´´- 23(1)(1) + nnn n
111111 1
=-1 +-++- + - L ，·····································11分
2231 nn- n1 n +
1
所以S =1- ，·········································································12分
n n+1
又因为n 1，所以S <1.·································································································13分
… n
参考答案 第 1 页 共9页
{{##{{QQQQAABBJIYYAC8U5oggCigwQwIIBTAAACRJ5hqCRQUQGG0gCCgggEQQskIABhELAKAoCEoRGRhCAHAqIIAARALBiJyNAINBAIAB=A}A#}=}#}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，m=0,0s.2= ，···························································1分
所以零件为合格品的概率为PX(0-==, m n = .
m n 5´ 1 5
参考答案 第 3 页 共9页
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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GQE, Q . ·····1分 A F
D
因为G,H 分别为线段APE, F 的中点，
H
G
1
所以GQ∥AB ，GQA=B ，····························2分
2 B E
又因为AB∥EFAB,EF = ， C Q
P
GQ∥HEGQ,HE=
所以 ，·································3分
所以四边形GQEH 是平行四边形，······················································4分
所以GHQ∥E ，··············································································5分
又因为GHBË C面EQEBC,EÌ 面 ，
所以GHB∥C面E .············································································7分
（2）同解法一.····················································································15分
解法三：（1）证明：取AB的中点I ，连接GIH, I .········1分
A F
D
因为G,H 分别为线段APE, F 的中点，
I
H
所以GIB∥PHIE, B∥ ， G
又因为GIBËC面EBPBC,E Ì 面 ， B E
C
所以GIB∥C面E .············································3分 P
因为HIBË C面EBEBC,E Ì 面 ，
所以HIB∥C面E .············································································5分
又因为GIHIIIGIG=ÌIH,HIGIH面 Ì , 面 ，
所以面GIHB∥C面E ，·····································································6分
又因为GHGÌ IH面 ，
所以GHB∥C面E .············································································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.
解法一：（1）依题意，bc22+ a2= ．··························································1分
参考答案 第 4 页 共9页
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设Px(, y) ，则 0 + 0 =1，-c，所以a> x ， >x ，所以|PFa|= -x ，d = - x
a 0 c 0 2 a 0 c 0
c
a- x
所以 |PF 2 | = a 0 = c ，即 |PF 2 | 为定值，且这个定值为 c ． ··················4分
d a2 a d a
- x
c 0
x y
（2）（ⅰ）依题意，G(, 0 ) 0 ，
3 3
x
设直线IG与x轴交于点C，因为IG⊥x轴，所以C(,00) ，·······················5分
3
x x 2
所以|||F|(C)(FC-=c+c )-- =0 x 0 ，··········································6分
12 33 3 0
因为△PF F 的内切圆与x轴切于点C，
1 2
2
所以|||P|||F|PF-=F- |CFC= x ，·················································7分
1212 0 3
又因为|||P|F2PF+ a= ，
1 2
x y
解得|PF |=a- 0 ，·········································· 8分
2 3
P
c
由（1）得|PFa|= -x ，
2 a 0 G
I
所以ax-= c -a x 0 ， F1 O C F2 x
a 0 3
c 1
所以椭圆E的离心率e= = ．·························10分
a 3
c 1
（ⅱ）由2a=6，得a=3，又 = ，所以c=1，ba22=- c2= 8，
a 3
x2 y2
所以椭圆E的方程为 + =1．······················································11分
9 8
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0 x <3，02< y2 ，
„ 0 0„
又F(1- ,0) ，F (1,0)，
1 2
y
所以直线PF 的方程为y= 0 (1x)+ ，
1 x +1
0
x y (3x) +
设直线IG与PF 交于点D，因为x = 0 ，所以y = 0 0 ，
1 D 3 D 3(1x) +
0
△F CD的面积S 与△PF F 的面积S之比为
1 1 1 2
1 xy x (3) +
(1)00+ 0 ´
S 233(1)(3) x + x + 2
1 = 0 = 0 ，················································13分
S 1 ´ 2´ y 18(1x) 0 +
2 0
(3x)+ 2 (3x)+(1) x-
令 f(x)= （0 x<3），则 f ¢ (x)= ，
18(1x)+ „ 18(x+1)2
参考答案 第 5 页 共9页
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所以函数 f(x)在[0,1)单调递减，在(1,3)单调递增.
1 4 1
又因为 f(0)= ， f(1)= ， f(3)= ，
2 9 2
4 1
所以 f(x)的值域是[, ] ，
9 2
4 S 1
所以 1 ，··········································································15分
„ „
9 S 2
4 S
所以 1 1，·······································································16分
5„ S - S „
1
根据对称性，
4 5
△PF F 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
1 2
5 4
解法二：（1）同解法一···········································································4分
x y
（2）（ⅰ）依题意，G(, 0 ) 0 ，
3 3
x
设直线IG与x轴交于点C，因为IG⊥x轴，所以C(,00) ，·······················5分
3
x x 2
所以|||F|(C)(FC-=c+c )-- =0 x 0 ，··········································6分
12 33 3 0
因为△PF F 的内切圆与x轴切于点C，
1 2
2
所以|||P|||F|PF-=F- |CFC= x ，·················································7分
1212 0 3
ì x
ï ||PF =a+, 0
ï 1 3
又因为|||P|F2PF+ a= ，得í ···········································8分
1 2 ï x
||PF =a.- 0
ïî 2 3
ì x
ï ()xcy++=2a+ 2 , 0
所以 ï í 0 0 3 两式平方后取差，得4cxax= 4 对任意x 成立，
ï x 0 3 0 0
()xcy-+= 2-a 2 , 0
ïî 0 0 3
c 1
所以椭圆E的离心率e= = ．························································10分
a 3
（ⅱ）同解法一···················································································17分
解法三：（1）同解法一···········································································4分
x y x
（2）（ⅰ）依题意，G(, 0 ) 0 ，因为IG⊥x轴，设点I坐标为(, 0) t .··········5分
3 3 3
y
可求直线PF方程为yx= c 0 ( + )，
1 x +c
0
x
yc(tx)(0 +c- ) +
0 3 0
则点I到直线PF的距离 =|t|，·································6分
1 yx2 +(c + )2
0 0
æ x ö 2 ( )
即ç yc(tx)(c0)t(y+x-+=+c) + ÷ 22 2 ，化简得
è 0003 0 ø
x x
yttc2x+c+y2+()-+()0() c= 0 0 2 ，①
00 30 3
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
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yttc2x+c----y2()(=)0() c 0 0 2 ，②············································7分
00 30 3
8 4 y
将式①- 式②，得2tc×x=ycx × ，则t = 0 .·····································8分
3 00 03 4
y
将t = 0 代入式①，得
4
yx2 1 x
00+++(-+)()() c=xc0 0 c 2 ，
1623 0 3
x2 y2
化简得 0 + 0 =1，
9c2 8c2
得9c2 =a2，
c 1
所以椭圆E的离心率e= = ．························································10分
a 3
（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lx( )ÎL ，
0(), fx x Î R
所以"Î-+xxxkRx, + 2 1，且 x R，-+=xx2kx + 1，····················1分
„ 0 00 0
令f()x(x1k=)-+x- 12 - ，D=- ( k- 1 )2 4，
则f()x 0 ，且f(x )=0，
„ 0
ì D 0,
所以í „ 所以D= 0， ···································································3分
î D 0,
…
即 ( k-- 1=) 42 0 ，
解得k =3或- 1. ··············································································4分
（2）（ⅰ）先证必要性.
( )
若直线yl=x ( ) 是曲线yg= x ( ) 的切线，设切点为 x ,ex0 +1 ，
0
( )
因为g¢()x =e x，所以切线方程为yx-+=ex1x0e( - )x0 ，
0
即lx ( x )=+e-(x01x)e +1 x0 （*）.························································5分
0
一方面，gx ( l x )= ( ) ，所以$Î xgxlRx,(=)( ) ，································6分
0 0 00 0
参考答案 第 7 页 共9页
{{##{{QQQQAABBJIYYAC8U5oggCigwQwIIBTAAACRJ5hqCRQUQGG0gCCgggEQQskIABhELAKAoCEoRGRhCAHAqIIAARALBiJyNAINBAIAB=A}A#}=}#}另一方面，令Gx ( gx ) l=xx-=-- ( - ) x ( ) eex(1)ex0 x0，则G ( x )=0，
0 0
因为G ¢ ( x )=ex - ex0，
所以当x< x 时，G ¢ ( x )<0，G ( x ) 在 (¥- ,x ) 单调递减，
0 0
当x>x 时，G ¢ ( x )>0，G ( x ) 在 ( x ,+¥ ) 单调递增，
0 0
所以Gx
(
G
)
x
( )=0，所以gx (
l
)
x
( )
.··············································7分
… 0 …
即" ÎxgxlRx,()( ) ，
…
所以lx
( )ÎT
，即l(x)是函数g(x)在R上的“最佳下界线”.···············8分
gx(),x Î R
再证充分性.
若l(x)是函数g(x)在R上的“最佳下界线”，不妨设lx(kx)=b + ，
由“最佳下界线”的定义，" ÎxgxlR,x ( ) ( ) ，且$Î xgxlRx, =( ) ( ) ，····9分
… 00 0
令Hx
(
gx
) lx=k-=x+(-
b
)- ( )
ex 1 ，
则H ( x ) 0且H ( x )=0，所以H ( x ) =0. ·········································10分
… 0 min
因为Hx ¢ ( )=exk- ，
①若k 0，则H ¢ ( x ) 0，所以H ( x ) 在R上单调递增，
„ …
所以$ x< x ，使得Hx ( H ) x< ( )=0，故k 0不符合题意.·····················11分
1 0 1 0 „
②若k >0，令H ¢ ( x )=0，得x=lnk，
当x¥Î- ( ,lnk ) 时，H ¢ ( x )<0，得H ( x ) 在 (¥- ,lnk ) 单调递减，
当xÎ+¥( lnk, ) 时，H ( x )>0，得H ( x ) 在 ( lnk,+¥ ) 单调递增，
所以，当且仅当x=lnk时，H ( x ) 取得最小值H ( lnk ) .····························12分
又由H ( x ) 在x 处取得最小值，H ( x ) =0，
0 min
ì x =lnk,
所以í 0 ···········································································13分
î H(ln)k0,=
即 ï í ì k =ex0, 解得k =ex 0 ，b=-( 1e x )+1x0 ，····························14分
ïî e1x 00+,-- =kx b 0
0
所以lx ( x )=+e-(x01x)e +1 x0 ，
0
参考答案 第 8 页 共9页
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由（*）式知直线yl=x( )是曲线yg=x( )在点 x ,ex0 +1 处的切线. ············15分
0
综上所述，直线yl=x( )是曲线yg=x( )的一条切线的充要条件是直线yl=x( )是函数
g(x)在R上的“最佳下界线”.
（ⅱ）集合L T 元素个数为2个. ············································17分
hx()x,g(1x,)Î+(x¥), I Î R
参考答案 第 9 页 共9页
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