发布时间 : 星期六 文章2018年北京市海淀区高三一模理科数学试题及参考答案更新完毕开始阅读6eb60f9b541810a6f524ccbff121dd36a22dc478
海淀区高三年级第二学期期中练习
数学(理)参考答案与评分标准 2018.4
一、选择题共8小题，每小题5分，共40分。在每小题列出的四个选项中，选出符合题目
要求的一项。
题号 答案 1 C 2 A 3 D 4 B 5 A 6 D 7 D 8 B 二、填空题共6小题，每小题5分，共30分。
题号 答案 9 1?i10 5211 2 3312 1313 48 (?3,14 3] x??1 注：第12、14题第一空均为3分，第二空均为2分。
三、解答题共6小题，共80分。解答题应写出解答步骤。 15. （本题满分13分） （Ⅰ）f(?6)?23sin?6cos?6?2cos2?62?1
?23?12?32?2?(32················································ 2分 )?1
··················································································· 3分 ?2 ·
（Ⅱ）f(x)?3sin2x?cos2x ······························································· 7分
?2sin(2x??6) ····································································· 9分
因为函数y?sinx的单调递增区间为[2kπ?令2k???2?2x?π2,2kπ?π2·········· 10分 ]（k?Z）， ·
?6?2k???2········································· 11分 （k?Z）， ·
解得 k???3?x?k???6················································ 12分 （k?Z）， ·
π3,kπ?π6故f(x)的单调递增区间为[kπ?····························· 13分 ]（k?Z） ·
注：第（Ⅰ）题先化简f(x)，再求f(?6)，求值部分3分；第（Ⅱ）题y=sinx的单调
区间及“解得”没写但最后答案正确，则不扣分；最后结果没写成区间扣1分，写开区间不扣分；全部解答过程没有写k?Z，扣1分。
16. （本题满分13分）
（Ⅰ）设事件A：从上表12个月中，随机取出1个月，该月甲地空气月平均相对湿度有利····························· 1分 于病毒繁殖和传播. 用Ai表示事件抽取的月份为第i月，则 ·
,A12}个基本事件， ??{A1,A2,A3,A4,A5,A6,A7,A8,A9,A10,A1，共···· 2分 1 A={A2,A6,A8,A9,A10,A11}，共6个基本事件， ·········································· 3分 所以，P(A)?612?12. ···································································· 4分
（Ⅱ）在第一季度和第二季度的6个月中，甲、乙两地空气月平均相对湿度都有利于病毒繁················ 5分 殖和传播的月份只有2月和6月，故X所有可能的取值为0，1，2.·
P(X?0)?C4C622?615?25，P(X?1)?C2C4C6211?815，P(X?2)?C2C622?115
··································································································· 8分 随机变量X的分布列为
X 0 251 8152 115P ······································································································ 9分 ················································· 13分 （Ⅲ）M的最大值为58%，最小值为54%.
注：第（Ⅰ）题没列Ω和A包含的基本事件，若后面出现
612612则不扣分，若后面没出现
则扣2分，结果不化简不扣分；第（Ⅱ）题X所有可能的取值没列扣1分，概率值错一
个扣1分，分布列不写扣1分；第（Ⅲ）题漏写“%”不扣分。
17.（本题满分14分） （Ⅰ）方法1：
PA
COB设AC的中点为O，连接BO，PO. 由题意
PA?PB?PC?2，PO?1，AO?BO?CO?1
因为 在?PAC中，PA?PC，O为AC的中点
····································································· 1分 所以 PO?AC， ·
因为 在?POB中，PO?1，OB?1，PB?2 ········································································ 2分 所以 PO?OB ·因为 ACOB?O，AC,OB?平面ABC ···································· 3分
所以 PO?平面ABC
······························································· 4分 因为 PO?平面PAC ·所以 平面PAC?平面ABC 方法2：
PA
COB设AC的中点为O，连接BO，PO.
因为 在?PAC中，PA?PC，O为AC的中点
····································································· 1分 所以 PO?AC， ·
因为 PA?PB?PC，PO?PO?PO，AO?BO?CO
所以 ?POA≌?POB≌?POC 所以 ?POA??POB??POC?90?
········································································ 2分 所以 PO?OB ·因为 ACOB?O，AC,OB?平面ABC ···································· 3分
所以 PO?平面ABC
································································· 4分 因为 PO?平面PAC 所以 平面PAC?平面ABC
方法3：
设AC的中点为O，连接PO， 因为 在?PAC中，PA?PC，
······································· 1分 所以 PO?AC ·
设AB的中点Q， 连接PQ，OQ及OB. 因为 在?OAB中，OA?OB，Q为AB的中点 所以
OQ?ABPA
QCOB.
?PB因为 在?PAB中，PA所以 因为 所以 因为 所以 因为
PQ?ABPQ，Q为AB的中点
.
，PQ,OQ?OQ?Q平面OPQ
AB?PO?平面OPQ 平面OPQ
································································ 2分
，AB,AC?PO?ABABAC?A····························· 3分 平面ABC ·
所以 PO?平面ABC
······················································ 4分 因为 PO?平面PAC ·
所以 平面PAC 法4：
设AC的中点为O，连接BO，PO.
因为 在?PAC中，PA?PC，O为AC的中点 ·································· 1分 所以 PO?AC， ·
因为 在?ABC中，BA?BC，O为AC的中点 ·································· 2分 所以 BO?AC， ·因为
POBO?O?平面ABC
PA
COB，PO?平面PAC，BO?平面ABC，
····························· 3分 所以∠POB为二面角P-AC-B的平面角。 ·因为 在?POB中，PO?1，OB?1，PB?2
······························································· 4分 所以 PO?OB ·
故二面角P-AC-B为直二面角，即平面PAC
?平面ABC。