2025年安徽省合肥市示范中学5月质检数学试题（图片版，含答案）

2025 届高三年级 5 月教学质量检测
数学试题参考答案及评分标准
一、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。
1.A 2.D 3.C 4.B 5.B 6.C 7.D 8.C
二、选择题：本题共 3 小题，每小题 6 分，共 18 分。在每小题给出的选项中，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。
9.ACD 10.AD 11.BCD
三、填空题：本题共 3 小题，每小题 5 分，共 15 分。
12. 2 13.5+ 2 14. ( ∞，1)
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）
【解析】
（1）因为 3asinC + acosC = b + c ，即 3sinAsinC + sinAcosC = sinB + sinC
即 3sinAsinC = sinCcosA+ sinC
π
因为 sinC > 0 所以 3sinA = cosA+1即 2sin A =1
6
因为 A∈(0,π )所以 A π= . ·················································································································· 6 分
3
（2）由 S 1 ABC = bcsinA =10 3 得bc = 40① 2
由 BC 2 = b2 + c2 2bccosA得49 = b2 + c2 bc②
由①②得b + c =13
2
2 2 2 b + c bc 129
由 AD 1= (AB + AC ) AD 1= (c2 + b2 bc) ( ) 129+ = = 得 AD = . 4 4 4 4 2
·····················································································································1 3 分
16.（15 分）
【解析】
（1）连接 DE ，因为 DB = DC ， E 是 BC 中点，所以 DE ⊥ BC
因为面 ACB ⊥面 DCB，面 ACB∩面 DCB = BC ，
数学试题答案 第 1 页（共 4 页）
且 DE 面DCB ， BC ⊥ DE
所以 DE ⊥面 ACB ，又因为 AE 面ACB 所以 DE ⊥ AE
由 AD = 7, DE = 3 AE = 2又 AB = 5 , BE =1得 AE ⊥ BC . ··········································· 7 分
（2）由（1）知， ED, EB, EA两两垂直，建立如图坐标系，则
E (0,0,0) , D ( 3,0,0) , B (1,0,0) , A(0,0,2)
( ) 所以 DB = 3,1,0 , AB = (0,1, 2) , AF = AE + EF = AE + DA = ( 3,0,0)

设m = (x, y, z ) ,n = (a,b,c)分别是面 DAB 和面 FAB的法向量，二
面角 D AB F 记为θ

m DB = 3x + y = 0
由 得m = (2,2 3, 3 )是面 DAB 的一个法

m AB = y 2z = 0
向量
同理n

= (0,2,1)是面 FAB的一个法向量
5 3 15 2 2 19
所以 cosθ = cos m,n = = ，所以 sinθ = =
4+12+ 3 4+1 19 19 19
2 19
故二面角 D AB F 的正弦值为 . ··························································································· 15 分
19
17.（15 分）
【解析】
（1）记 A：按照方式 A分类；M：最终被填埋
则 P (M ) = P (M |A) + P (M |A) = 0.6×0.15×0.6+ 0.4×0.25 = 0.154 ········································ 4 分
P (AM )
（2） P (A|M ) 0.4×0.25 50= = = ···················································································· 8 分 P (M ) 0.154 77
(3)由题意，X 的可能取值为0,200,500,700
且 P (X = 200) = 0.6×0.15×0.4 = 0.036 , P (X = 500) = 0.4×0.25 = 0.1
P (X = 700) = 0.6×0.15×0.6 = 0.054 , P (X = 0) =1 0.1 0.036 0.054 = 0.81
故X 分布列如下：
X 0 200 500 700
P 0.81 0.036 0.1 0.054
E (X) = 0×0.81+ 200×0.036+ 500×0.1+ 700×0.054 = 95 . ···················································· 15 分
数学试题答案 第 2 页（共 4 页）
18.（17 分）
【解析】
x 3 y p（1）由题意， AB 方程为 = + ，设 A(x1, y1 ) , B (x2 , y2 ) . 4 2
y2 = 2 px

联立方程 3 p ，得2y2 3py 2 p2 = 0 .
x = y + 4 2
所以 y 1 3 3M = ( y1 + y2 ) = p = 所以 p = 2 ， 2 4 2
抛物线C 方程为 y2 = 4x . ······················································································································ 5 分
1
（2）因为点 A在第一象限，所以由（1）可得 A(4,4) , B , 1

.
4
设 P (a2 , 2a)，因为点 E 在点 D 的上方 ，所以，a 1 , 1∈ .
2 2
AP : y y y1 2a 4+ 2ay1 2+ 4a直线 1 = x a2
(x x1 )令x = 1 yD = =
1 2a y a 2
.
+ 1 +
y 4+ 2ay2 4 2a同理 E = =2a y 2a 1 . ········································································································· 8 分 + 2
8
（i）延长 AB 交准线于Q，易知 yQ = ，则 S = S 等价于 S = S . 3 PAB PDE ADQ BEQ
S 1因为 ADQ = ( y 1 52 D yQ )×5, S BEQ = ( yE y2 Q )× 44 ( yD yQ ) = ( yE yQ )
解得a = 0故存在点 P (0,0)，使得 S PAB = S PDE . ··········································································· 12 分
（ii） DE y y 5 5 3a 4= E D = + . 2a2 + 3a 2
10(3a +1)(a 3)
令 h (a) = 5+ 5 3a 4 1 12 ,a∈

, ，则h '(a) = 2 . 2a + 3a 2 2 2 (2a2 + 3a 2)
所以h (a) 1 , 1 1 1 1在

上单减，在 ,2 3 3 2
上单增， h (a) ≥ h = 4 ，此时 DE ≥ 4；
3
故 DE = 4min . ········································································································································ 17 分
19.（17 分）
【解析】
（1）令 g (x) = f (x) f ′(m)(x m) f (m)
则 g′(x) = f ′(x) f ′(m) = ex em故 g (x)在 ( ∞,m)上递减，在 (m,+∞)上递增
所以 g (x) ≥ g (m) = 0 即 f (x) ≥ f ′(m)(x m) + f (m) . ······························································ 4 分
数学试题答案 第 3 页（共 4 页）
（2）由（1）知ai = f (i) ≥ f ′(m)(i m) + f (m)
n n n
所以∑ai ≥∑( f ′(m)(i m) + f (m)) = f ′(m) ∑i nm + nf (m)
i=1 i=1 i=1
n
∑ i n n+1令 i=1 1 ( )得∑ai ≥ nf
1
(n +1)

= n e 2 ································································ 10 分 m = = n +1
n 2 i=1
2
k + k + + k
（3）要证 1 2 n ≥ n k1 k1 kn n
lnk1 + lnk2 + + lnkn ln k1 + k2 + + kn 只要证 ≤ n n
令 g (x) = lnx 1 (x m) lnm,则 g '(x) 1 1=
m x m
故 g (x)在 (0, m)上单增， (m,+∞)上单减，所以 g (x) ≤ g (m) = 0
1
故 lnx ≤ (x m) + lnm
m
n n
所以 lnk
1
i ≤ (ki m) + lnm (i =1,2,
1
,n) ∑lnki ≤ ∑ki nm + nlnm m i=1 m i=1
n
∑ k n令 i
m = i=1
则∑lnki ≤ nlnm
n i=1
lnk1 + lnk2 + + lnkn
即 ≤ ln
k1 + k2 + + kn
. ············································································ 17 分 n n
数学试题答案 第 4 页（共 4 页）