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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。