福建省泉州市2024届高三质量监测三数学答案

2024-03-06 · 16页 · 632.5 K

保密★使用前泉州市2024届高中毕业班质量监测(三)2024.03高三数学试卷共19题,满分150分,共8页。考试用时120分钟。一、选择题答案:1.B2.C3.C4.B5.C.6.D.7.A8.D二、选择题答案:9.AD10.BCD11.ACD三、填空题:本题共3小题,每小题5分,共15分。1612.513.x24y14.e33四、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)如图,在三棱锥PABC中,PAPCABAC2,BC22,E为PC的中点,点F在PA上,且EF平面PAB,PMPB(R).(1)若MF‖平面ABC,求;1(2)若,求平面PAB与平面MAC夹角的正弦值.2【命题意图】本题考查空间点、直线与平面间的位置关系等知识;考查推理论证、运算求解等能力;考查数形结合思想、化归与转化思想等;体现应用性、创新性、综合高三数学试题第1页(共8页){#{QQABDY4QgggAAAAAAAhCAwVqCAKQkAEACAoGwFAMsAABCBFABAA=}#}性,导向对直观想象、数学运算等核心素养的关注.解法一:(1)依题意得,△PAC为正三角形,取PA中点N,连结CN,则CNPA,因为EF平面PAB,所以EFPA,EF∥CN,·················································································(2分)1又因为E为PC的中点,所以F为PN中点,则PFPA,·················(3分)4因为MF∥平面ABC,MF平面PAB,平面PAB平面ABCAB,即MF∥AB,······································(5分)11也即PMPB,.·······························································(6分)44(2)因为EF∥CN,则CN平面PAB,且CN3·························(7分)又因为EF平面PAB,所以EFAB,···········································(8分)由ABAC2,BC22可知BC2AB2AC2,则△ABC为等腰直角三角形,ABAC,又因为EF与AC相交于平面PAC,所以AB平面PAC,·················(10分)△PAB为等腰直角三角形,M为斜边PB中点,则△PAM也为等腰直角三角形,2取AM中点H,则NHAM,且NH,2连结CH,AM平面CNH,则AMCH,CHN为二面角PAMC的平面角,·····················(11分)114在Rt△CHN中,CHCN2NH2322高三数学试题第2页(共8页){#{QQABDY4QgggAAAAAAAhCAwVqCAKQkAEACAoGwFAMsAABCBFABAA=}#}CN342sinCHN.则CH147····················································(13分)2解法二:(1)同解法一;(2)因为EF平面PAB,所以EFAB,······································(7分)由ABAC2,BC22可知BC2AB2AC2,则△ABC为等腰直角三角形,ABAC,取AC中点O,连结PO,则POAC,取BC中点Q,连结OQ,则OQ∥AB,又因为EF与AC相交于平面PAC,所以AB平面PAC,···················(9分)也即OQ平面PAC,所以OQ,OC,OP两两相互垂直,以O为原点,OQ,OC,OP所在直线分别为x轴,y轴,z轴,建立空间直角坐标系.13133A(0,1,0),B(2,1,0),C(0,1,0),P(0,0,3),E(0,,),F(0,,),2244131333M(0,,),AM(1,,),AC(0,2,0),EF(0,,),······(10分)222244设平面MAC的一个法向量n(x,y,z),则nAM0,nAC0,即13xyz0,22取z2,则x3,2y0.所以n(3,0,2)为平面MAC的一个法向量.···································(11分)3EFn7cosEF,n2,37EFn74142记平面PAB与平面MAC夹角为,sin1.····················(13分)77解法三:(1)因为EF平面PAB,所以EFAB,高三数学试题第3页(共8页){#{QQABDY4QgggAAAAAAAhCAwVqCAKQkAEACAoGwFAMsAABCBFABAA=}#}由ABAC2,BC22可知BC2AB2AC2,则△ABC为等腰直角三角形,ABAC,△PAC为正三角形,取AC中点O,连结PO,则POAC,取BC中点Q,连结OQ,则OQ∥AB,又因为EF与AC相交于平面PAC,所以AB平面PAC,也即OQ平面PAC,··································(3分)所以OQ,OC,OP两两相互垂直,以O为原点,OQ,OC,OP所在直线分别为x轴,y轴,z轴,建立空间直角坐标系.13A(0,1,0),B(2,1,0),C(0,1,0),P(0,0,3),E(0,,),PA(0,1,3),22··································································································(4分)因为MF∥平面ABC,MF平面PAB,平面PAB平面ABCAB,则MF∥AB,·····································(6分)因为PMPB,所以PFPA(0,,3),F(0,,33),13EF(0,,3),EFPA410,221所以.················································································(9分)4补充说明:第一小题利用坐标法解答得9分,其中证明AB平面PAC得3分,请把这3分的分值写到第二小题的得分栏。131333(2)M(0,,),AM(1,,),AC(0,2,0),EF(0,,),(10分)222244设平面MAC的一个法向量n(x,y,z),则nAM0,nAC0,即高三数学试题第4页(共8页){#{QQABDY4QgggAAAAAAAhCAwVqCAKQkAEACAoGwFAMsAABCBFABAA=}#}13xyz0,22取z2,则x3,2y0.所以n(3,0,2)为平面MAC的一个法向量.···································(11分)3EFn7cosEF,n2,37EFn74142记平面PAB与平面MAC夹角为,sin1.····················(13分)77解法四:(1)同解法一;(2)设点C在平面PAB的射影为N,则CN平面PAB,过点N做NHAM,垂足为H,连结CH,AM平面CNH,则AMCH,CHN为PAMC的平面角,因为EF平面PAB,所以EFAB,··············································(7分)由ABAC2,BC22可知BC2AB2AC2,则△ABC为等腰直角三角形,ABAC,△PAC为正三角形,取AC中点O,连结PO,则POAC,又因为EF与AC相交于平面PAC,所以AB平面PAC,···················(9分)ABPO,ABACA,PO平面ABC11因为VCPABVPACB,则S△PABCNS△ABCPO,所以CNPO3,(10分)33在△PBC中,PC2,BCBP22,高三数学试题第5页(共8页){#{QQABDY4QgggAAAAAAAhCAwVqCAKQkAEACAoGwFAMsAABCBFABAA=}#}(22)2(22)2223cosPBC,222224CM2BM2BC22BMBCcosPBC4,CM2,······················(11分)在△MCA中,AM2,ACMC2,△MCA∽△PBC111714S△AMCHACMCsinACM22,CH,(12分)AMC22242CN342sinCHN.CH147·······················································(13分)216.(15分)淄博烧烤、哈尔滨冬日冰雪、山河四省梦幻联动、鄂了赣饭真湘……,2023年全国各地的文旅部门在网络上掀起了一波花式创意宣传,带火了各地的文旅市场,很好地推动国内旅游业的发展.已知某旅游景区在手机APP上推出游客竞答的问卷,题型为单项选择题,每题均有4个选项,其中有且只有一项是正确选项.对于游客甲,在知道答题涉及的内容的条件下,可选出唯一的正确选项;在不知道答题涉及的内容的条件下,则随机选择一个选项.已知甲知道答题涉及内容的题数占问卷总题数的1.3(1)求甲任选一题并答对的概率;(2)若问卷答题以题组形式呈现,每个题组由2道单项选择题构成,每道选择题答对得22分,答错扣1分,放弃作答得0分.假设对于任意一道题,甲选择作答的概率均为,且两3题是否选择作答及答题情况互不影响,记每组答题总得分为X.(i)求P(X4)和P(X2);(ii)求E(X).【命题意图】本小题主要考查全概率、独立事件、分布列、数学期望、等基础知识,考查运算求解、推理论证、数据处理能力,体现综合性、创新性、应用性,导向对发高三数学试题第6页(共8页){#{QQABDY4QgggAAAAAAAhCAwVqCAKQkAEACAoGwFAMsAABCBFABAA=}#}展数学运算、逻辑推理、数据分析、数学建模等核心素养的关注.【试题简析】(1)记“甲任选一道题并答对”为事件M,“甲知道答题涉及内容”为事件A.········································································································1分121依题意,P(A),P(A),P(MA)1,P(MA).·······················2分334因为事件MA与MA互斥,所以P(M)P(MAMA)P(MA)P(MA)(没说明互斥不扣分)分·········································································31P(MA)P(A)P(MA)P(A).·······················(公式1分,结果1分)5分221211(2)(ⅰ)P(X4).···················································7分3232921211P(X2);··(本小题答对任意一个2分,答对两个3分)8分32329(ⅱ)依题
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