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。