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 an 满足 a1 = 2， an+1 = 3an + 2 ．
（1）证明：数列 an +1 是等比数列；
（2）求 an 的前 n项和 S ． n
【解法一】（1）证明：因为 a ，且n+1 = 3an + 2 a1 = 2，
所以 a +1 0 ， ··················································································· 1 分 n
a +1 3a + 2 +1
所以 n+1 = n ········································································ 3 分
an +1 an +1
3(a +1)
= n = 3， ···································································· 5 分
an +1
又 a1 +1= 3，
所以数列 an +1 是以3为首项，3为公比的等比数列． ······························· 6 分
（2）由（1）得 an +1= 3
n，所以 a = 3nn 1， ············································· 8 分
所以 S 2 nn = (3 1) + (3 1) + + (3 1)
数学试题 第1页（共 14 页）
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
2 3 n
= (3+ 3 + 3 + + 3 ) n ···························································· 10 分
3 3n+1
= n ············································································ 12 分
1 3
3n+1 3
= n. ············································································ 13 分
2
【解法二】（1）证明：因为 an+1 = 3an + 2 ，
所以 an+1 +1= 3an + 3 = 3(an +1)， ····························································· 2 分
因为 a = 2，所以 a +1= 3 0 ，所以 a +1 0 ， ········································ 4 分 1 1 n
a
所以 n+1
+1
= 3 ,
an +1
所以数列 an +1 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 2acosC = 3bcosC + 3c cos B ．
（1）求角C ；
（2）若 a = 4， b = 3 ， D 为 AB 中点，求CD 的长．
【解法一】（1）因为 2acosC = 3bcosC + 3c cos B ，
由正弦定理，
得 2sin AcosC = 3 sin BcosC + 3 cos Bsin C ·············································· 2 分
= 3sin (B +C) ·································································· 4 分
= 3sin (π A)
= 3 sin A ， ······································································ 6 分
3
因为 0 A π，则 sin A 0，所以 cosC = ， ·········································· 7 分
2
π
由于 0 C π，则C = ； ···································································· 8 分
6
1
（2）因为 D 为 AB 中点，故CD = (CA +CB)， ······································ 10 分
2
数学试题 第2页（共 14 页）
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
2 1 2
所以 CD = (CA +CB) ······································································ 11 分
4
1 2 1 2 1 π
= CA + CB + CA CB cos ············································ 13 分
4 4 2 6
1 1 1 3
= 3+ 16 + 3 4
4 4 2 2
31
= ，················································································· 14 分
4
31
所以CD 的长为 ． ······································································ 15 分
2
【解法二】（1）因为 2acosC = 3bcosC + 3c cos B ，
a2 + b2 c2 a2 + c2 b2
由余弦定理，得 2a cosC = 3b + 3c ··························· 2 分
2ab 2ac
= 3a， ···························································· 4 分
3
所以 cosC = ， ················································································ 6 分
2
π
由于 0 C π，则C = ； ···································································· 8 分
6
π
（2）由（1）知， ACB = ，
6
在△ABC 中，由余弦定理，得
c2 = a2 + b2 2abcos ACB ··································································· 10 分
3
= 42 + ( 3)2 2 4 3
2
= 7， ··························································································· 11 分
故 c = 7 ， ······················································································· 12 分
因为 D 为 AB 中点，
所以 cos ADC + cos BDC = 0 ，
AD2 +CD2 AC 2 BD2 +CD2 BC 2
故 + = 0 ， ·········································· 13 分
2 AD CD 2 BD CD
2 2
7 2
+CD
2
( 3 )
7
+CD2 42
2
2
所以 + = 0，
7 7
2 CD 2 CD
2 2
数学试题 第3页（共 14 页）
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
31
解得CD
2 = ， ················································································ 14 分
4
31
故CD 的长为 ． ··········································································· 15 分
2
【解法三】（1）略，同解法一或解法二；
π
（2）由（1）知， ACB = ，
6
在△ABC 中，由余弦定理，得
c2 = a2 + b2 2abcos ACB ··································································· 10 分
2
= 42 + ( ) 33 2 4 3
2
= 7， ··························································································· 11 分
故 c = 7 ， ······················································································· 12 分
b2 + c2 a2
所以 cos A =
2bc
2 2
( 3 ) + ( 7 ) 42
=
2 3 7
3
= ， ············································································· 13 分
7
在△ACD中，由余弦定理，
得CD2 = AC2 + AD2 2AC ADcos A
2
2 7 7 3
= ( 3 ) + 2 3
2 2
7
31
= ， ······················································································· 14 分
4
31
故CD 的长为 ． ··········································································· 15 分
2
17. （15 分）
如图，在四棱锥 S ABCD 中，BC ⊥平面 SAB ，AD∥BC ，
SA = BC =1， SB = 2 ， SBA = 45o ．
（1）求证： SA ⊥平面 ABCD ；
数学试题 第4页（共 14 页）
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
1
（2）若 AD = ，求平面 SCD 与平面 SAB 的夹角的余弦值．
2
【解法一】（1）在△SAB 中，
因为 SA =1， SBA = 45o ， SB = 2 ，
由正弦定理，得
SA SB
= ， ········································································· 1 分
sin SBA sin SAB
1 2
所以 = ， ······································································ 2 分
sin 45 sin SAB
所以 sin SAB =1，
因为 0 SAB 180 ，所以 SAB = 90 ，
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
1
则 S(0,0,1),C(1,1,0), D ,0,0 ,
2
1
所以 SC = (1,1, 1)， SD = ,0, 1 ，
2
设平面 SCD 的法向量为n1 = (x, y, z)，
n1 SC = x + y z = 0, n1 ⊥ SC, 则 即 1
n1 ⊥ SD, n1 SD = x z = 0,
2
取 x = 2，则 n1 = (2, 1,1)， ·································································· 11 分
数学试题 第5页（共 14 页）
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
显然平面 SAB 的一个法向量n2 = (1,0,0)， ················································ 12 分
n n
所以 cos n ,n = 1 2 ····································································· 13 分 1 2
n n
1 2
2
=
22
2
+ ( 1) +12
6
= ， ········································································· 14 分
3
6
所以平面 SCD 与平面 SAB 的夹角的余弦值为 ． ··································· 15 分
3
【解法二】（1）证明：设 AB = x ，在△SAB 中，
因为 SA =1， SBA = 45o ， SB = 2 ，
由余弦定理，得
SA2 = SB2 + AB2 2SB ABcos SBA， ······················································ 1 分
所以1= 2 + x2 2 2xcos45 ， ································································ 2 分
所以 x2
2
+ 2 2 2x =1，
2
所以 x2 2x +1= 0，
解得 x =1． ························································································ 3 分
所以 SA2 + AB2 = SB2 = 2 ，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 SA =1， SBA = 45o ， SB = 2 ，
由余弦定理，得
SA2 = SB2 + AB2 2SB ABcos SBA， ······················································ 1 分
所以1= 2 + x2 2 2xcos45 ， ································································ 2 分
数学试题 第6页（共 14 页）
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
2
所以 x2 + 2 2 2x =1，
2
所以 x2 2x +1= 0，
解得 x =1． ························································································ 3 分
所以 SA2 + AB2 = SB2 = 2 ，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
1
则 S(1,0,1),C(0,1,0), D 1, ,0 , 所以
2
1
SC = ( 1,1, 1)，CD = 1, ,0 ，
2
设平面 SCD 的法向量为n1 = (x, y, z)，
n1 ⊥ SC,则
n1 ⊥ CD,
n1 SC = x + y z = 0,
即 1
n1 CD = x y = 0,
2
取 y = 2 ，则 n1 = (1,2,1)， ···································································· 11 分
显然平面 SAB 的一个法向量n2 = (0,1,0)， ··············································· 12 分
n n
所以 cos n1 ,n2 =
1 2 ····································································· 13 分
n n
1 2
2
=
12 + 22 +12
数学试题 第7页（共 14 页）
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
6
= ， ········································································· 14 分
3
6
所以平面 SCD 与平面 SAB 的夹角的余弦值为 ． ··································· 15 分
3
【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
S
在△SBM 中，
SB = 2 ， SBA = 45 ， BM = 2 ，
C
B
由余弦定理，得 A
SM 2 = SB2 + MB2 2SB MBcos SBM ， D
M
2
所以 SM 2 = ( 2 ) 2 2+ 2 2 2 2 = 2， ·············································· 9 分
2
所以 SM 2 + SB2 = BM 2 ，
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 SB = 2, BC =1，得 SC = 3 ， ······················································· 13 分
SB 2 6
所以 cos BSC = = = ，
SC 3 3
6
所以平面 SCD 与平面 SAB 的夹角的余弦值为 ． ··································· 15 分
3
18. （17 分）
2 2 1
已知椭圆W ： x y+ =1(a b 0)的离心率为 ，且过点 (2,0)．
a2 b2 2
（1）求W 的方程；
（2）直线 x my +1= 0(m 0)交W 于 A ， B 两点．
数学试题 第8页（共 14 页）
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
k
（i）点 A关于原点的对称点为C ，直线 BC 的斜率为 k ，证明： 为定值；
m
（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．
c 1
=a 2
【解法一】(1)依题意，得 a = 2 ， ··················································· 3 分
b2 = a2 c2
a = 2
解得 ， ···················································································· 4 分
b = 3
x2 y2
所以W 的方程为 + =1； ································································ 5 分
4 3
(2)依题意可设点 A(x1, y1) ， B(x2 , y2 )，且 x1 x2 ，
(ⅰ) 证明：因为点 A 关于原点的对称点为C ，所以C( x1, y1) ，
x2 y21 1
+ =1
4 3
因为点 A ， B 在W 上，所以 ， ················································ 6 分
2
x2 y
2
+ 2 =1
4 3
2 2 2x2 x
2
1 y
2 2 y y 3
所以 = 2
y1 ，即 2 1 = ，··············································· 8
x2 2
分
4 3 2 x1 4
1
因为直线 AB ： x my +1= 0(m 0)的斜率为 ，直线 BC 的斜率为 k ， ········· 9 分
m
2 2
k y
所以 = 2
y1 y + y 2 1
y
= 2
y1 3= ，
m x 22 x1 x2 + x1 x2 x
2
1 4
k 3
即 为定值 ； ············································································· 11 分
m 4
(ⅱ)设弦 AB 的中点 D 的坐标为 (xD , yD ) ，点 P 的坐标为 (xP , yP )，△PAB 的重心G 的
x2 y2
+ =1 2 2
坐标为 (x , y ) ，由 4 3 ，得 (3m + 4) y 6my 9 = 0G G ， ················ 11 分
x my +1= 0
2
所以 = 36m +36(3m2 +4) =144(m2 +1) 6m 0 ，且 y1 + y2 = ··············· 12
3m2
， 分
+ 4
数学试题 第9页（共 14 页）
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
x + x + x
因为△PAB 的重心G 在 y 轴上，所以
1 2 P = 0， ······························ 13 分
3
6m 8
所以 xP = (x1 + x2 ) = (my1 1+ my2 1) = m ( y1 + y2 ) + 2 = m + 2 =2 2 ， 3m + 4 3m + 4
x1 + x2 4 y1 + y2 3m所以 xD = = ， yD = = ， ······························ 14 分
2 3m2 + 4 2 3m2 + 4
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 y yD = m(x xD )，
3m 8 4 9m
所以 yP = yD m(xP xD ) = m + = 2 2 2 2 ， ······· 15 分 3m + 4 3m + 4 3m + 4 3m + 4
8 9m
所以点 P ,
3m2 2
，
+ 4 3m + 4
2 2
8 9m
又点 P 在W 上，所以 3m
2 + 4 3m2 + 4 ， ································ 16 分
+ =1
4 3
即m
2 (3m2 1) = 0，
3
又因为m 0，所以m = ，所以直线 AB 的方程为3x 3y + 3 = 0． ······· 17 分
3
【解法二】（1）略，同解法一；
(2)依题意可设点 A(x1, y1) ， B(x2 , y2 )，且 x1 x2 ，
(ⅰ) 证明：因为点 A 关于原点的对称点为C ，所以C( x1, y1) ，
x2 y2
+ =1 2
由 4 3 ，得 (3m + 4) y2 6my 9 = 0， ······································· 6 分
x my +1= 0
2 2 2 6m
所以 = 36m +36(3m +4) =144(m +1) 0 ，且 y1 + y2 = ， ················· 72 分 3m + 4
6m
所以 y1 + y2 = ，
3m2 + 4
6m 8
所以 x1 + x2 = my1 1+ my2 1= m ( y1 + y2 ) 2 = m 2 = ， ······· 8
3m2
分
+ 4 3m2 + 4
数学试题 第10页（共 14 页）
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
因为直线 BC 的斜率为 k ，
6m
y2 ( y1 ) y + y 3m2k = = 2 1 = + 4
3m
所以 = ， ······································ 10 分
x2 ( x1 ) x2 + x 81 4
3m2 + 4
k 3
所以 为定值 ； ·········································································· 11 分
m 4
(ⅱ)设弦 AB 的中点 D 的坐标为 (xD , yD ) ，点 P 的坐标为 (xP , yP )，△PAB 的重心G 的
坐标为 (xG , yG ) ，
6m
由(ⅰ)知， y1 + y2 = ， ······························································· 12
3m2
分
+ 4
x
因为 的重心G 在 y 轴上，所以 1
+ x2 + xP
△PAB = 0， ······························ 13 分
3
6m 8
所以 xP = (x1 + x2 ) = (my1 1+ my2 1) = m ( y1 + y2 ) + 2 = m + 2 = ，
3m2 + 4 3m2 + 4
x + x 4 y + y 3m
所以 x = 1 2 = ， y = 1 2D D = ， ······························ 14 分
2 3m2 + 4 2 3m2 + 4
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 y yD = m(x xD )，
3m 8 4 9m
所以 yP = yD m(xP xD ) = m + = ， ······· 15 分
3m2 + 4 3m
2 + 4 3m2 + 4 3m
2 + 4
8 9m
所以点 P , ，
3m
2 + 4 3m2 + 4
2 2
8 9m
又点 P 在W 上，所以 2
3m + 4 3m2
+ 4 ， ································ 16 分
+ =1
4 3
m2即 (3m2 1) = 0，
3
又因为m 0，所以m = ，所以直线 AB 的方程为3x 3y + 3 = 0． ······· 17 分
3
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
数学试题 第11页（共 14 页）
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
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 f (x) = ax4 + x3 3(2a +1) x2 x + 3．
（i）当 a 0 时，讨论 f ( x)的凹凸性；
（ii）当 a = 0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 g (x) = xex ，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 f (x) = ax4 + x3 3(2a +1) x2 x + 3，
所以 f (x) = 4ax3 + 3x2 6(2a +1) x 1， ·················································· 1 分
令 h (x) = 4ax3 + 3x2 6(2a +1) x 1，
所以 h (x) =12ax2 + 6x 6(2a +1) = 6(2ax + 2a +1)(x 1)． ····························· 2 分
（i）当 a = 0时， h (x) = 6(x 1)，令 h ( x) 0 ，解得 x 1；
令 h ( x) 0 ，解得 x 1；
故 f ( x)为区间 1,+ )上的凹函数，为区间 ( ,1 上的凸函数； ····················· 3 分
1 2a +1
当 a 0 时，令 h ( x) 0 ，解得1 x ，
4 2a
2a +1
令 h ( x) 0 ，解得 x 1或 x ，
2a
2a +1 2a +1
故 f ( x)为区间 1, 上的凹函数，为区间 ( ,1 和 ,+ 上的凸函数；
2a 2a
·············································································································· 4 分
1 2
当 a = 时， h (x) = 3(x 1) 0，故 f ( x)为区间 ( ,+ )上的凸函数； ······ 5 分
4
1 2a +1
当 a 时，令 h ( x) 0 ，解得 x 1，
4 2a
数学试题 第12页（共 14 页）
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
2a +1
令 h ( x) 0 ，解得 x 1或 x ，
2a
2a +1 2a +1
故 f ( x)为区间 ,1 上的凹函数，为区间 , 和 1,+ )上的凸函数；
2a 2a
1 2a +1 2a +1
综上所述，当 a 时， f ( x)为区间 ,1 上的凹函数，为区间 , 4 2a 2a
和 1,+ )上的凸函数；
1
当 a = 时， f ( x)为区间 ( ,+ )上的凸函数；
4
1 2a +1 2a +1
当 a 0 时，f ( x)为区间 1, 上的凹函数，为区间 ( ,1 和 ,+ 4 2a 2a
上的凸函数；
当 a = 0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 a = 0时， f (x) = x3 3x2 x + 3， f (x) = 3x2 6x 1，
故在点 (t, f (t))处的切线方程为 y = (3t2 6t 1)(x t ) + t3 3t2 t + 3． ············ 7 分
设 P (u,v)(u 0)为 f ( x)的“3切点”，
则关于 t 的方程 v = (3t2 6t 1)(u t ) + t3 3t2 t + 3 有三个不同的解，
即关于 t 的方程 v = 2t3 + (3+ 3u )t 2 6ut + 3 u 有三个不同的解，
令 F (t ) = 2t3 + (3+ 3u )t 2 6ut + 3 u ，
所以直线 y = v与曲线 y = F (t )恰有三个不同的交点． ·································· 8 分
F (t ) = 6t2 + 6(1+ u )t 6u = 6(t 1)(t u )． ············································ 9 分
当u 1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u3 3u2 u + 3 减
故 4 4u v u3 3u2 u + 3； ······························································ 11 分
数学试题 第13页（共 14 页）
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}
2
当u =1时， F (t ) = 6(t 1) 0， F (t )单调递减，不符合题意； ················· 12 分
当 0 u 1时， F (t ) , F (t )随 t 变化情况如下：
t ( ,u) u (u,1) 1 (1,+ )
F (t) 0 + 0
F(t) 减 极小值u3 3u2 u + 3 增 极大值 4 4u 减
故u3 3u2 u + 3 v 4 4u；
综上所述，点 P 的集合为
x 1 0 x 1
(x, y) 或 ；
4 4x y x3 3x2
x + 3 x3 3x2
x + 3 y 4 4x
······································································································ 14 分
4 x 2 2 x 0
x 4
（2）点Q的集合为 (x, y) 或 x x + 4或 x + 4 ．
xe
x y 0 xe y y xe
x
e2 2
e
······································································································ 17 分
数学试题 第14页（共 14 页）
{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}(在此卷上答题无效)
6.已知圆2+y2+4mx-2my+m=0(meR)与x轴相切，则m=
2024-2025学年高三年级第一次质量检测
发0或
1
A.1
C.0或1
0.4
数学试题
7.已知圆锥S0的底面半径为1，过高线的中点且垂直于高线的平面将圆锥S0截成
本试卷共4页，考试时间120分钟；总分150分。
上、下两部分，若截得小圆锥的体积为没，则圆锥50的侧面积为
注意事项：
1.答卷前，考生务必将自己的姓名、准考证号填在答题卡上。
A.4π
B.2
C.2m
D.T
2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题月的答策标号
8.大气压强p(单位：kPa)与海拔h(单位：m)之间的关系可以由p=Pe协近似描
涂黑。如需改动，用橡皮擦千净后，再选涂其它答案标号。回答非选择题时，将答案
述，其中o为标淮大气压强，k为常数.已知海拔为5000m,8000m两地的大气压
写在答题卡上。写在本试卷上无效。
强分别为54kPa,36kPa.若测得某地的大气压强为8OkPa,则该地的海拔约为
3.考试结束后，将答题卡交回。
(参考数据：1g2≈0.301,1g3≈0.477)
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
A.295m
B.995m
C.2085m
D.3025m
中，只有一项是符合题目要求的。
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
=4:5≤0明，B=x0≤x≤6，侧A
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
A.{x-5≤x≤6
B.{-1≤x≤6}
得0分。
C.{x0≤x≤1}
D.{x0≤x≤5}
5
9.已知(1-2x)9=a0+a1x+2x2+…+ox9,则
2已知氨数：2则z与
A.6=1
.B.a,=18
9
1+39
B.
5
C.5
D.5
C.a1+a2+…+ag=-1
D.a+asas a+ag =
2
3.以坐标原点为顶点，x轴非负半轴为始边的角α，其终边落在直线y=2x上，则
10.如图是函数∫(x)=sin(wx+p)的部分图象，则
A.sino =2i5 B.cosc =5 C.lancr =2
5
D.sin2a=-4
A.π是∫(x)的一个周期
5
4.以y=±3x为渐近线的双曲线可以是
R.）=r）
背-y1R-苦=1C号-e1
c》
5.如图，梯形ABGD的腰CD的巾点为R,且BC=3AD,
D.f(x)在[0.3π]上恰有6个零点
记AB=元，AD=九，则B2=
11.已知函数∫(x),g(x)均为定义在R上的非常值函数，且g(x)为f(x)的导函数.
A-+2
1
B.2元+2列
对Vx,y∈R,f(x+y)+f(x-y)=2f(x)f(y)且f(1)=0,则
A.f(0)=0
B.f(x)为偶函数
c-2m+
3￥
C.g(x)+g(2024-x)=0
D.[f(x)]2+[f(1-x)]2=1
高三数学一1一（共4页）
高三数学一2一（共4页)
