ID: 22946935

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)

日期:2025-05-10 科目:数学 类型:高中试卷 查看:49次 大小:2168863B 来源:二一课件通
预览图 1/4
2025年,安徽省,合肥市,示范,中学,5月
  • cover
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× ... ...

~~ 您好,已阅读到文档的结尾了 ~~