文档内容
数学试题 第1页（共14 页）
2024-2025 学年福州市高三年级第一次质量检测
数学答案
一、单项选择题：本题共8 小题，每小题5 分，共40 分．在每小题给出的四个选项中，
只有一项是符合题目要求的。
题目
1
2
3
4
5
6
7
8
答案
D
C
C
B
A
D
B
C
二、多项选择题：本题共3 小题，每小题6 分，共18 分。在每小题给出的四个选项中，
有多项符合题目要求。全部选对的得6 分，部分选对的得部分分，有选错的得0 分。
题目
9
10
11
答案
AD
ABD
BCD
三、填空题：本大题共3 小题，每小题5 分，共15 分。
题目
12
13
14
答案
3
5
(
)
24 25
，
四、解答题：本题共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. （13 分）
已知数列
na
满足
1
2
a =
，
1
3
2
n
n
a
a
+ =
+
．
（1）证明：数列
1
na +
是等比数列；
（2）求
na
的前n 项和
n
S ．
【解法一】（1）证明：因为
1
3
2
n
n
a
a
+ =
+
，且
1
2
a =
，
所以
1
0
na +
， ··················································································· 1 分
所以
1
1
3
2
1
1
1
n
n
n
n
a
a
a
a
+ +
+
+
=
+
+
········································································ 3 分
3(
1)
3
1
n
n
a
a
+
=
=
+
， ···································································· 5 分
又
1
1
3
a + =
，
所以数列
1
na +
是以3为首项，3为公比的等比数列． ······························· 6 分
（2）由（1）得
1
3n
na + =
，所以
3
1
n
na =
−， ············································· 8 分
所以
(
) (
)
(
)
2
3 1
3
1
3
1
n
nS =
−
+
−
+
+
−
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
数学试题 第2页（共14 页）
(
)
2
3
3
3
3
3n
n
=
+
+
+
+
−
···························································· 10 分
1
3
3
1
3
n
n
+
−
=
−
−
············································································ 12 分
1
3
3
.
2
n
n
+ −
=
−
············································································ 13 分
【解法二】（1）证明：因为
1
3
2
n
n
a
a
+ =
+
，
所以
(
)
1
1
3
3
3
1
n
n
n
a
a
a
+ + =
+
=
+
， ····························································· 2 分
因为
1
2
a =
，所以
1
1
3
0
a + =
，所以
1
0
na +
， ········································ 4 分
所以
1
1
3
1
n
n
a
a
+ +
=
+
,
所以数列
1
na +
是以3为首项，3为公比的等比数列． ······························· 6 分
（2）略，同解法一．
16. （15 分）
已知
ABC
△
的内角, ,
A B C 的对边分别为, ,
a b c ，且2 cos
3 cos
3
cos
a
C
b
C
c
B
=
+
．
（1）求角C ；
（2）若
4
a =
，
3
b =
，D 为AB 中点，求CD 的长．
【解法一】（1）因为2 cos
3 cos
3
cos
a
C
b
C
c
B
=
+
，
由正弦定理，
得2sin
cos
3sin
cos
3 cos
sin
A
C
B
C
B
C
=
+
·············································· 2 分
(
)
3sin B
C
=
+
·································································· 4 分
(
)
3sin π
A
=
−
3sin A
=
， ······································································ 6 分
因为0
π
A
，则sin
0
A
，所以
3
cos
2
C =
， ·········································· 7 分
由于0
π
C
，则
π
6
C =
； ···································································· 8 分
（2）因为D 为AB 中点，故
(
)
1
2
CD
CA
CB
=
+
， ······································ 10 分
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
数学试题 第3页（共14 页）
所以
(
)
2
2
1
4
CD
CA
CB
=
+
······································································ 11 分
2
2
1
1
1
π
cos
4
4
2
6
CA
CB
CA CB
=
+
+
············································ 13 分
1
1
1
3
3
16
3
4
4
4
2
2
=
+
+
31
4
=
，················································································· 14 分
所以CD 的长为
31
2
． ······································································ 15 分
【解法二】（1）因为2 cos
3 cos
3
cos
a
C
b
C
c
B
=
+
，
由余弦定理，得
2
2
2
2
2
2
2 cos
3
3
2
2
a
b
c
a
c
b
a
C
b
c
ab
ac
+
−
+
−
=
+
··························· 2 分
=
3a ， ···························································· 4 分
所以
3
cos
2
C =
， ················································································ 6 分
由于0
π
C
，则
π
6
C =
； ···································································· 8 分
（2）由（1）知，
π
6
ACB
=
，
在
ABC
△
中，由余弦定理，得
2
2
2
2
cos
c
a
b
ab
ACB
=
+
−
··································································· 10 分
2
2
3
4
( 3)
2
4
3
2
=
+
−
7
=
， ··························································································· 11 分
故
7
c =
， ······················································································· 12 分
因为D 为AB 中点，
所以cos
cos
0
ADC
BDC
+
=
，
故
2
2
2
2
2
2
0
2
2
AD
CD
AC
BD
CD
BC
AD
CD
BD
CD
+
−
+
−
+
=
， ·········································· 13 分
所以
(
)
2
2
2
2
2
2
7
7
3
4
2
2
0
7
7
2
2
2
2
CD
CD
CD
CD
+
−
+
−
+
=
，
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
数学试题 第4页（共14 页）
解得
2
31
4
=
CD
， ················································································ 14 分
故CD 的长为
31
2
． ··········································································· 15 分
【解法三】（1）略，同解法一或解法二；
（2）由（1）知，
π
6
ACB
=
，
在
ABC
△
中，由余弦定理，得
2
2
2
2
cos
c
a
b
ab
ACB
=
+
−
··································································· 10 分
(
)
2
2
3
4
3
2
4
3
2
=
+
−
7
=
， ··························································································· 11 分
故
7
c =
， ······················································································· 12 分
所以
2
2
2
cos
2
b
c
a
A
bc
+
−
=
(
)
(
)
2
2
2
3
7
4
2
3
7
+
−
=
3
7
= −
， ············································································· 13 分
在
ACD
△
中，由余弦定理，
得
2
2
2
2
cos
CD
AC
AD
AC AD
A
=
+
−
(
)
2
2
7
7
3
3
2
3
2
2
7
=
+
−
−
31
4
=
， ······················································································· 14 分
故CD 的长为
31
2
． ··········································································· 15 分
17. （15 分）
如图，在四棱锥S
ABCD
−
中，BC ⊥平面SAB ，
∥
AD
BC ，
1
SA
BC
=
= ，
2
SB =
，
o
45
SBA
=
．
（1）求证：SA ⊥平面ABCD ；
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
数学试题 第5页（共14 页）
（2）若
1
2
AD =
，求平面SCD 与平面SAB 的夹角的余弦值．
【解法一】（1）在△SAB 中，
因为
1
SA = ，
o
45
SBA
=
，
2
SB =
，
由正弦定理，得
sin
sin
SA
SB
SBA
SAB
=
， ········································································· 1 分
所以
1
2
sin45
sin
SAB
=
， ······································································ 2 分
所以sin
1
SAB
= ，
因为0
180
SAB
，所以
90
SAB
=
，
所以SA
AB
⊥
． ··················································································· 4 分
因为BC ⊥平面SAB ，SA 平面SAB ，
所以BC
SA
⊥
， ··················································································· 5 分
又BC
AB
B
=
，
所以SA ⊥平面ABCD ； ········································································· 6 分
（2）解：由（1）知SA ⊥平面ABCD ，
又
,
AB AD
平面ABCD ，所以SA
AB
⊥
，SA
AD
⊥
，
因为BC ⊥平面SAB ， ··········································································· 7 分
AB
平面SAB ，所以BC
AB
⊥
，
因为
∥
AD
BC ，所以AD
AB
⊥
，
所以
,
,
SA AD AB 两两垂直． ··································································· 8 分
以点A 为原点，分别以AD ，AB ，AS 所在直线为x 轴，y 轴，z 轴建立如图所示的
空间直角坐标系， ················································································ 9 分
则
1
(1,1,0),
,0,0 ,
2
(0,0,1),
D
S
C
所以
(
)
1,1, 1
SC =
−
，
1 ,0, 1
2
SD
=
−
，
设平面SCD 的法向量为
1
( , , )
x y z
=
n
，
则
1
1
,
,
SC
SD
⊥
⊥
n
n
即
1
1
0,
1
0,
2
SC
x
y
z
SD
x
z
=
+
−
=
=
−
=
n
n
取
2
x =
，则
(
)
1
2, 1,1
=
−
n
， ·································································· 11 分
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
数学试题 第6页（共14 页）
显然平面SAB 的一个法向量
(
)
2
1,0,0
=
n
， ················································ 12 分
所以cos
=
1
2
1
2
1
2
n
n
n ,n
n
n
····································································· 13 分
(
)
2
2
2
2
2
1
1
=
+ −
+
6
3
=
， ········································································· 14 分
所以平面SCD 与平面SAB 的夹角的余弦值为
6
3 ． ··································· 15 分
【解法二】（1）证明：设AB
x
=
，在△SAB 中，
因为
1
SA = ，
o
45
SBA
=
，
2
SB =
，
由余弦定理，得
2
2
2
2
cos
SA
SB
AB
SB
S
AB
BA
=
+
−
， ······················································ 1 分
所以
2
1
2
2 2 co
5
s4
x
x
=
+
−
， ································································ 2 分
所以
2
2
2
2 2
1
2
x
x
+
−
= ，
所以
2
2
1
0
x
x
−
+ =
，
解得
1
x = ． ························································································ 3 分
所以
2
2
2
2
SA
AB
SB
+
=
=
，所以SA
AB
⊥
． ················································ 4 分
因为BC ⊥平面SAB ，SA 平面SAB ，
所以BC
SA
⊥
， ··················································································· 5 分
又BC
AB
B
=
，
所以SA ⊥平面ABCD ； ········································································· 6 分
（2）略，同解法一．
【解法三】（1）设AB
x
=
，在△SAB 中，
因为
1
SA = ，
o
45
SBA
=
，
2
SB =
，
由余弦定理，得
2
2
2
2
cos
SA
SB
AB
SB
S
AB
BA
=
+
−
， ······················································ 1 分
所以
2
1
2
2 2 co
5
s4
x
x
=
+
−
， ································································ 2 分
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
数学试题 第7页（共14 页）
所以
2
2
2
2 2
1
2
x
x
+
−
= ，
所以
2
2
1
0
x
x
−
+ =
，
解得
1
x = ． ························································································ 3 分
所以
2
2
2
2
SA
AB
SB
+
=
=
，所以SA
AB
⊥
． ················································ 4 分
因为BC ⊥平面SAB ，BC 平面ABCD ，
所以平面ABCD ⊥平面SAB ； ································································· 5 分
又平面ABCD
平面SAB
AB
=
，SA
AB
⊥
，SA 平面SAB ，
所以SA ⊥平面ABCD ； ········································································· 6 分
（2）由（1）知SA ⊥平面ABCD ，过B 作BM
SA ，则BM ⊥平面ABCD ，
又
,
AB BC 平面ABCD ，所以BM
AB
⊥
，BM
BC
⊥
，
因为BC ⊥平面SAB ， ··········································································· 7 分
又
AB
平面SAB ，所以BC
AB
⊥
，
所以
,
,
BM BA BC 两两垂直． ·································································· 8 分
以点B 为原点，分别以BA ，BC ，BM 所在直线为x 轴，y 轴，z 轴建立如图所示的
空间直角坐标系， ················································································ 9 分
则
1
(0,1,0),
1,
(
,0 ,
2
1,0,1),C
S
D
所以
(
)
1,1, 1
SC = −
−
，
1
1,
,0
2
CD
=
−
，
设平面SCD 的法向量为
1
( , , )
x y z
=
n
，
则
1
1
,
,
SC
CD
⊥
⊥
n
n
即
1
1
0,
1
0,
2
SC
x
y
z
CD
x
y
= −+
−
=
=
−
=
n
n
取
2
y =
，则
(
)
1
1,2,1
=
n
， ···································································· 11 分
显然平面SAB 的一个法向量
(
)
2
0,1,0
=
n
， ··············································· 12 分
所以cos
=
1
2
1
2
1
2
n
n
n ,n
n
n
····································································· 13 分
2
2
2
2
1
2
1
=
+
+
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
数学试题 第8页（共14 页）
6
3
=
， ········································································· 14 分
所以平面SCD 与平面SAB 的夹角的余弦值为
6
3
． ··································· 15 分
【解法四】（1）略，同解法一或解法二或解法三；
（2）延长CD 、BA 交于点M ，连接SM ，
则平面SCD
平面SAB
SM
=
， ······························································· 7 分
在
SBM
△
中，
2
SB =
，
45
SBA
=
，
2
BM =
，
由余弦定理，得
2
2
2
2
cos
SM
SB
MB
SB
S
MB
BM
=
+
−
，
所以
(
)
2
2
2
2
2
2
2
2
2
2
2
SM =
+
−
=
， ·············································· 9 分
所以
2
2
2
SM
SB
BM
+
=
，
所以SM
SB
⊥
， ················································································ 10 分
因为BC ⊥平面SAB ，SM 平面SAB ，
所以SM
BC
⊥
，又SM
SB
⊥
，SB
BC
B
=
，
所以SM ⊥平面SBC ， ········································································ 11 分
又SC 平面SBC ，所以SM
SC
⊥
，
所以
BSC
为平面SCD 与平面SAB 的夹角， ············································ 12 分
因为BC ⊥平面SAB ，SB 平面SAB ，
所以BC
SB
⊥
，
因为
2,
1
SB
BC
=
= ，得
3
SC =
， ······················································· 13 分
所以
2
6
cos
3
3
SB
BSC
SC
=
=
=
，
所以平面SCD 与平面SAB 的夹角的余弦值为
6
3 ． ··································· 15 分
18. （17 分）
已知椭圆W ：
(
)
2
2
2
2
1
0
x
y
a
b
a
b
+
=
的离心率为1
2 ，且过点(
)
2,0 ．
（1）求W 的方程；
（2）直线
(
)
1
0
0
−
+ =
x
my
m
交W 于A ，B 两点．
A
B
C
D
M
S
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
数学试题 第9页（共14 页）
（i）点A 关于原点的对称点为C ，直线BC 的斜率为k ，证明：k
m 为定值；
（ii）若W 上存在点P 使得AP ，PB 在AB 上的投影向量相等，且△PAB 的重心在y
轴上，求直线AB 的方程．
【解法一】(1)依题意，得
2
2
2
1
2
2
c
a
a
b
a
c
=
=
=
−
， ··················································· 3 分
解得
2
3
a
b
=
=
， ···················································································· 4 分
所以W 的方程为
2
2
1
4
3
x
y
+
= ； ································································ 5 分
(2)依题意可设点
1
1
( ,
)
A x y
，
2
2
(
,
)
B x
y
，且
1
2
x
x
，
(ⅰ) 证明：因为点A 关于原点的对称点为C ，所以
1
1
(
,
)
C
x
y
−
−
，
因为点A ，B 在W 上，所以
2
2
1
1
2
2
2
2
1
4
3
1
4
3
x
y
x
y
+
=
+
=
， ················································ 6 分
所以
2
2
2
2
2
1
2
1
4
3
x
x
y
y
−
−
= −
，即
2
2
2
1
2
2
2
1
3
4
y
y
x
x
−
= −
−
，··············································· 8 分
因为直线AB ：
(
)
1
0
0
−
+ =
x
my
m
的斜率为1
m ，直线BC 的斜率为k ， ········· 9 分
所以k
m
2
1
2
1
2
1
2
1
2
2
2
1
2
2
2
1
3
4
−
+
=
=
−
= −
−
−
+
y
y
y
y
x
x
x
x
y
y
x
x
，
即k
m 为定值
3
4
−
； ············································································· 11 分
(ⅱ)设弦AB 的中点D 的坐标为(
,
)
D
D
x
y
，点P 的坐标为(
,
)
P
P
x
y
，
PAB
△
的重心G 的
坐标为(
,
)
G
G
x
y
，由
2
2
1
1
0
4
3
−
+
=
+
=
x
my
x
y
，得(
)
2
2
3
4
6
9
0
m
y
my
+
−
−
=
， ················ 11 分
所以
(
)
(
)
2
2
2
36
+36 3
+4
144
1
0
m
m
m
=
=
+
，且
1
2
2
6
3
4
m
y
y
m
+
=
+
， ··············· 12 分
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
数学试题 第10页（共14 页）
因为
PAB
△
的重心G 在y 轴上，所以
1
2
0
3
P
x
x
x
+
+
=
， ······························ 13 分
所以
(
)
(
)
(
)
1
2
1
2
1
2
2
2
8
1
1
2
2
3
4
6
3
4
P
x
x
x
my
my
m y
y
m
m
m
m
= −
+
= −
−+
−
= −
+
+
= −
+
=
+
+
，
所以
1
2
2
4
2
3
4
D
x
x
x
m
+
=
= −
+
，
1
2
2
3
2
3
4
D
y
y
m
y
m
+
=
=
+
， ······························ 14 分
因为AP ，PB 在AB 上的投影向量相等，所以PA
PB
=
，且PD
AB
⊥
，
所以直线PD 的方程为
(
)
D
D
y
y
m x
x
−
= −
−
，
所以
(
)
2
2
2
2
3
8
4
9
3
4
3
4
3
4
3
4
P
D
P
D
m
m
y
y
m x
x
m
m
m
m
m
=
−
−
=
−
+
= −
+
+
+
+
， ······· 15 分
所以点
2
2
8
9
,
3
4
3
4
m
P
m
m
−
+
+
，
又点P 在W 上，所以
2
2
2
2
8
9
4
3
3
4
1
4
3
+
+
−
+
=
m
m
m
， ································ 16 分
即
(
)
2
2
3
1
0
m
m −
=
，
又因为
0
m
，所以
3
3
m =
，所以直线AB 的方程为3
3
3
0
x
y
+
=
． ······· 17 分
【解法二】（1）略，同解法一；
(2)依题意可设点
1
1
( ,
)
A x y
，
2
2
(
,
)
B x
y
，且
1
2
x
x
，
(ⅰ) 证明：因为点A 关于原点的对称点为C ，所以
1
1
(
,
)
C
x
y
−
−
，
由
2
2
1
1
0
4
3
−
+
=
+
=
x
my
x
y
，得(
)
2
2
3
4
6
9
0
m
y
my
+
−
−
=
， ······································· 6 分
所以
(
)
(
)
2
2
2
36
+36 3
+4
144
1
0
m
m
m
=
=
+
，且
1
2
2
6
3
4
m
y
y
m
+
=
+
， ················· 7 分
所以
1
2
2
6
3
4
m
y
y
m
+
=
+
，
所以
(
)
1
2
1
2
1
2
2
2
8
1
1
2
2
3
6
3
4
4
x
x
my
my
m y
y
m
m
m
m
+
=
−+
−=
+
−
=
−
= −
+
+
， ······· 8 分
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
数学试题 第11页（共14 页）
因为直线BC 的斜率为k ，
所以
(
)
(
)
2
1
2
1
2
1
2
2
2
1
6
3
3
4
8
3
4
4
y
y
y
y
k
x
x
x
x
m
m
m
m
−−
+
=
=
=
−−
+
= −
−
+
+
， ······································ 10 分
所以k
m 为定值
3
4
−
； ·········································································· 11 分
(ⅱ)设弦AB 的中点D 的坐标为(
,
)
D
D
x
y
，点P 的坐标为(
,
)
P
P
x
y
，
PAB
△
的重心G 的
坐标为(
,
)
G
G
x
y
，
由(ⅰ)知，
1
2
2
6
3
4
m
y
y
m
+
=
+
， ······························································· 12 分
因为
PAB
△
的重心G 在y 轴上，所以
1
2
0
3
P
x
x
x
+
+
=
， ······························ 13 分
所以
(
)
(
)
(
)
1
2
1
2
1
2
2
2
8
1
1
2
2
3
4
6
3
4
P
x
x
x
my
my
m y
y
m
m
m
m
= −
+
= −
−+
−
= −
+
+
= −
+
=
+
+
，
所以
1
2
2
4
2
3
4
D
x
x
x
m
+
=
= −
+
，
1
2
2
3
2
3
4
D
y
y
m
y
m
+
=
=
+
， ······························ 14 分
因为AP ，PB 在AB 上的投影向量相等，所以PA
PB
=
，且PD
AB
⊥
，
所以直线PD 的方程为
(
)
D
D
y
y
m x
x
−
= −
−
，
所以
(
)
2
2
2
2
3
8
4
9
3
4
3
4
3
4
3
4
P
D
P
D
m
m
y
y
m x
x
m
m
m
m
m
=
−
−
=
−
+
= −
+
+
+
+
， ······· 15 分
所以点
2
2
8
9
,
3
4
3
4
m
P
m
m
−
+
+
，
又点P 在W 上，所以
2
2
2
2
8
9
4
3
3
4
1
4
3
+
+
−
+
=
m
m
m
， ································ 16 分
即
(
)
2
2
3
1
0
m
m −
=
，
又因为
0
m
，所以
3
3
m =
，所以直线AB 的方程为3
3
3
0
x
y
+
=
． ······· 17 分
19. （17 分）
阅读以下材料：
①设
( )
f
x
为函数
( )
f x 的导函数．若
( )
f
x
在区间D 上单调递增，则称
( )
f x 为区间D
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
数学试题 第12页（共14 页）
上的凹函数；若
( )
f
x
在区间D 上单调递减，则称
( )
f x 为区间D 上的凸函数．
②平面直角坐标系中的点P 称为函数
( )
f x 的“ k 切点”，当且仅当过点P 恰好能作曲
线
( )
y
f x
=
的k 条切线，其中k N ．
（1）已知函数
( )
(
)
4
3
2
3 2
1
3
f x
ax
x
a
x
x
=
+
−
+
−
+ ．
（i）当
0
a
时，讨论
( )
f x 的凹凸性；
（ii）当
0
a =
时，点P 在y 轴右侧且为
( )
f x 的“3切点”，求点P 的集合；
（2）已知函数( )
ex
g x
x
=
，点Q 在y 轴左侧且为( )
g x 的“3切点”，写出点Q 的集合
（不需要写出求解过程）．
【解析】(1)因为
( )
(
)
4
3
2
3 2
1
3
f x
ax
x
a
x
x
=
+
−
+
−
+ ，
所以
( )
(
)
3
2
4
3
6 2
1
1
f
x
ax
x
a
x
=
+
−
+
−， ·················································· 1 分
令( )
(
)
3
2
4
3
6 2
1
1
h x
ax
x
a
x
=
+
−
+
−，
所以
( )
(
)
(
)(
)
2
12
6
6 2
1
6 2
2
1
1
h
x
ax
x
a
ax
a
x
=
+
−
+
=
+
+
−
． ····························· 2 分
（i）当
0
a =
时，
( )
(
)
6
1
h x
x
=
−
，令
( )
0
h
x
，解得
1
x
；
令
( )
0
h
x
，解得
1
x
；
故
( )
f x 为区间
)
1,+上的凹函数，为区间(
,1
−
上的凸函数； ····················· 3 分
当
1
0
4
a
−
时，令
( )
0
h
x
，解得
2
1
1
2
+
−a
x
a
，
令
( )
0
h
x
，解得
1
x
或
2
1
2
+
−a
x
a
，
故
( )
f x 为区间
2
1
1,
2
a
a
+
−
上的凹函数，为区间(
,1
−
和
2
1,
2
a
a
+
−
+
上的凸函数；
·············································································································· 4 分
当
1
4
a = −
时，
( )
(
)
2
3
1
0
h x
x
= −
−
，故
( )
f x 为区间(
)
,
−+上的凸函数； ······ 5 分
当
1
4
a −
时，令
( )
0
h
x
，解得
2
1
1
2
+
−a
x
a
，
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
数学试题 第13页（共14 页）
令
( )
0
h
x
，解得
1
x
或
2
1
2
+
−a
x
a
，
故
( )
f x 为区间
2
1,1
2
a
a
+
−
上的凹函数，为区间
2
1
,
2
a
a
+
−−
和
)
1,+上的凸函数；
综上所述，当
1
4
a −
时，( )
f x 为区间
2
1,1
2
a
a
+
−
上的凹函数，为区间
2
1
,
2
a
a
+
−−
和
)
1,+上的凸函数；
当
1
4
a = −
时，
( )
f x 为区间(
)
,
−+上的凸函数；
当
1
0
4
a
−
时，( )
f x 为区间
2
1
1,
2
a
a
+
−
上的凹函数，为区间(
,1
−
和
2
1,
2
a
a
+
−
+
上的凸函数；
当
0
a =
时，
( )
f x 为区间
)
1,+上的凹函数，为区间(
,1
−
上的凸函数； ········ 6 分
（ii）当
0
a =
时，
( )
3
2
3
3
f x
x
x
x
=
−
−
+
，
( )
2
3
6
1
f
x
x
x
=
−
−，
故在点(
)
,
( )
t f t
处的切线方程为
(
)(
)
2
3
2
3
6
1
3
3
y
t
t
x
t
t
t
t
=
−
−
−
+
−
−+
． ············ 7 分
设
(
)(
)
,
0
P u v
u
为
( )
f x 的“3切点”，
则关于t 的方程
(
)(
)
2
3
2
3
6
1
3
3
v
t
t
u
t
t
t
t
=
−
−
−
+
−
−+
有三个不同的解，
即关于t 的方程
(
)
3
2
2
3
3
6
3
v
t
u t
ut
u
= −
+
+
−
+
−
有三个不同的解，
令
( )
(
)
3
2
2
3
3
6
3
F t
t
u t
ut
u
= −
+
+
−
+
−
，
所以直线y
v
=
与曲线
( )
y
F t
=
恰有三个不同的交点． ·································· 8 分
( )
(
)
(
)(
)
2
6
6 1
6
6
1
F
t
t
u t
u
t
t
u
= −
+
+
−
= −
−
−
． ············································ 9 分
当
1
u 时，
( )
( )
,
F t
F
t
随t 变化情况如下：
t
(
)
,1
−
1
(
)
1,u
u
(
)
,u +
( )
F t
−
0
+
0
−
( )
F t
减
极小值4
4u
−
增
极大值
3
2
3
3
u
u
u
−
−
+
减
故
3
2
4
4
3
3
u
v
u
u
u
−
−
−
+ ； ······························································ 11 分
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
数学试题 第14页（共14 页）
当
1
u = 时，
( )
(
)
2
6
1
0
F t
t
= −
−
，
( )
F t 单调递减，不符合题意； ················· 12 分
当0
1
u
时，
( )
( )
,
F t
F
t
随t 变化情况如下：
t
(
)
,u
−
u
(
)
,1
u
1
(
)
1,+
( )
F t
−
0
+
0
−
( )
F t
减
极小值
3
2
3
3
u
u
u
−
−
+
增
极大值4
4u
−
减
故
3
2
3
3
4
4
u
u
u
v
u
−
−
+
−
；
综上所述，点P 的集合为
(
)
3
2
3
2
1
0
1
,
4
4
3
3
3
3
4
4
x
x
x y
x
y
x
x
x
x
x
x
y
x
−
−
−
+
−
−
+
−
或
；
······································································································ 14 分
（2）点Q 的集合为(
)
2
2
4
2
2
0
4
,
4
4
e
e
e
0
e
e
x
x
x
x
x
x
x y
x
x
x
y
y
x
x
y
−
−
−
−
+
+
−
−
或
或
．
······································································································ 17 分
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
