发布时间 : 星期五 文章2015届福州高三上学期教学质量检查数学理试题更新完毕开始阅读b56ab032be1e650e52ea99de
福州市2014－2015学年度第一学期高三质量检查
理科数学试卷参考答案及评分细则
一、选择题：本大题共12小题，每小题5分，共60分． 1．C 2．B 3．A 4．A 5．B 7．B 8．C 9．C 10．D 11．B 二、填空题：本大题共4小题，每小题4分，共16分， 13．?2
14．32
15．3?6 6．D 12．C 16．(??,2]
三、解答题：本大题共6小题，共74分．
17． 本题主要考查一元二次方程的根、等比数列的通项公式、错位相减法求数列的和等基础知识，考查应用能力、运算求解能力，考查函数与方程思想． 解：（Ⅰ）方程x2?3x?2?0的两根分别为1,2， ························································ 1分 依题意得a1?1，a2?2． ································································································ 2分 所以q?2， ······················································································································ 3分 所以数列{an}的通项公式为an?2n?1． ········································································· 4分 （Ⅱ）由（Ⅰ）知2n?an?n?2n， ··················································································· 5分 所以Sn?1?2?2?22?????n?2n， ············································ ① ························ ② 2?Sn?1?22?2?23?????(n?1)?2n?n?2n?1， ·
由①－②得
·············································································· 8分 ?Sn?2?22?23?????2n?n?2n?1， ·
n2?2?2即 ?Sn?······················································································· 11分 ?n?2n?1，
1?2所以Sn?2?(n?1)?2n?1． ···························································································· 12分
18．本题主要考查离散型随机变量的概率、分布列、数学期望等基础知识，考查运算求解能力以及应用意识，考查必然与或然思想等．
解法一：（Ⅰ）这3个人接受挑战分别记为A、B、C，则A,B,C分别表示这3个人不接受挑战． 这3个人参与该项活动的可能结果为：?A,B,C?，A,B,C，A,B,C，A,B,C，A,B,C，A,B,C，················································································· 2分 ?A,B,C?，?A,B,C?．共有8种； ·
其中，至少有2个人接受挑战的可能结果有：?A,B,C?，?A,B,C?，?A,B,C?，?A,B,C?，共有4种． ································································································································· 3分
41根据古典概型的概率公式，所求的概率为P??． ·················································· 4分
82（说明：若学生先设“用?x,y,z?中的x,y,z依次表示甲、乙、丙三人接受或不接受挑战的情况”，再
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???????????A,B,C?,
?A,B,C?,
?A,B,C?，
?A,B,C?,?A,B,C?,?A,B,C?,?A,B,C?,?A,B,C?，不扣分．）
（Ⅱ）因为每个人接受挑战与不接受挑战是等可能的，
11···································· 5分 所以每个人接受挑战的概率为，不接受挑战的概率也为．
220651630?1??1?1?1??1??所以P?X?0??C6?????，P?X?1??C6??????，
?2??2?64?2??2?6432?1?P?X?2??C???2?262?1?153?1?????，P?X?3??C6?2?64?2???43?1?205?????， ?2?64163
4251?1?P?X?4??C???2?46?1?155?1?????，P?X?5??C6?2?64?2???063?1??????， 26432??16?1??1?P?X?6??C6?. ······················································································· 9分 ?2??2?64????故X的分布列为：
6X P 10分
0 1 641 3 322 15 643 5 164 15 645 3 326 1 64131551531?1??2??3??4??5??6??3． 64326416643264故所求的期望为3． ······································································································· 12分 解法二：因为每个人接受挑战与不接受挑战是等可能的，
11···································· 1分 所以每个人接受挑战的概率为，不接受挑战的概率也为．
22（Ⅰ）设事件M为“这3个人中至少有2个人接受挑战”， 所以E?X??0??1?则P(M)?C???2?231?1?3?1?????C3?·········································································· 4分 ?2?2． 2????（Ⅱ）因为X为接下来被邀请的6个人中接受挑战的人数，
?1?········································································································· 5分 所以X~B?6,?． ·
?2?163?1??1?1?1??1????所以P?X?0??C?????，P?X?1??C6?2??2?6432，
?2??2?64????0606523?1?P?X?2??C???2?262?1?153?1?????，P?X?3??C6?2?64?2???243?1?205?????， 26416??63?1??????， ?2?643213?1?P?X?4??C???2?46664?1?155?1?????，P?X?5??C6?2?64?2???051?1??1?P?X?6??C?????. ······················································································· 9分
64?2??2?6故X的分布列为：
X P 10分
0 1 641 3 322 15 643 5 164 15 645 3 326 1 641?3． 2故所求的期望为3． ······································································································ 12分
19．本题主要考查反比例函数、三角函数的图象与性质、三角函数的定义、同角三角函数的基本关系式、二倍角公式、两角和的正弦公式等基础知识，考查运算求解能力，考查数形结合思想、化归与转化思想、函数与方程思想．
解法一：（Ⅰ）?OPQ为等边三角形． ·········································································· 1分 所以E?X??6?理由如下：
???因为函数f(x)?23sin?x?，
?4?
2π?8，所以函数f(x)的半周期为4， ?4所以OQ?4． ·················································································································· 2分
所以T?又因为P为函数f(x)图象的最高点，
所以点P坐标为(2，··································································· 4分 23)，所以OP?4， ·又因为Q坐标为(4,0)，所以PQ?(2?4)2?(23?0)2?4，
所以?OPQ为等边三角形． ··························································································· 6分 （Ⅱ）由（Ⅰ）知，OP?OQ?4，
????????4sin?)， 所以点P?,Q?的坐标分别为?4cos????，4sin?????,(4cos?,················ 7分
33???????????2k?代入y?，得k?16cos????sin?????8sin(2??π)，
3??3?3x?且k?16sin?cos??8sin2?， ························································································· 9分
2?所以sin2??sin(2??π)，结合sin2(2?)?cos2(2?)?1，0???，
321解得sin2??， ············································································································ 11分
2所以k?4，所以所求的实数k的值为4． ···································································· 12分 解法二：（Ⅰ）?OPQ为等边三角形． ········································································ 1分 理由如下：
???因为函数f(x)?23sin?x?，
?4?2π?8，所以函数f(x)的半周期为4，所以OQ?4， ·所以T?·································· 2分 ?4因为P为函数f(x)的图象的最高点，
所以点P坐标为(2，····································· 4分 23)，所以OP?4，所以OP?OQ． ·23?3，所以?POQ?60?， 2所以?OPQ为等边三角形． ··························································································· 6分
又因为直线OP的斜率k?（Ⅱ）由（Ⅰ）知，OP?OQ?4，
????????4sin?)， ·4sin?????,(4cos?,所以点P?,Q?的坐标分别为?4cos????,················ 7分
33??????k因为点P?,Q?在函数y?(x?0)的图象上，
x????????k?16cos????sin????,所以?··············································································· 8分 3??3?, ·??k?16sin?cos??2??k?8sin(2??π),所以?································································································· 9分 3, ·
?k?8sin2??2消去k得， sin2??sin(2??π)，
3
22所以sin2??sin2?cosπ?cos2?sinπ，
33333cos2?，所以tan2??所以sin2??， ···························································· 10分
223??1又因为 0???，所以2??，所以sin2??， ···················································· 11分
262所以k?4．所以所求的实数k的值为4． ···································································· 12分 解法三：（Ⅰ）同解法一或同解法二；
（Ⅱ）由（Ⅰ）知，?OPQ为等边三角形．
k因为函数y?(x?0)的图象关于直线y?x对称， ······················································· 8分
x?k由图象可知，当??时，点P?,Q?恰在函数y?(x?0)的图象上． ······················ 10分
x12??此时点Q?的坐标为(4cos,········································································ 11分 4sin)， ·
1212???所以k?16sincos?8sin?4，所以所求的实数k的值为4． ···························· 12分
1212620． 本题主要考查分段函数模型的应用问题、一元二次函数的最值、解不等式等基础知识，考查应用意识、运算求解能力，考查化归与转化思想、分类讨论思想等．
?30,0≤x?6,??4?x解：（I）因为m?3，所以y??． ···················································· 1分
3x?12?,6≤x≤8?2?30当0≤x?6时，由······································ 3分 ≥2，解得x≤11，此时0≤x?6； ·
4?x3x2020当6≤x≤8时，由12?≥2，解得x≤，此时6≤x≤． ····························· 5分
23320综上所述，0≤x≤．
320故若一次服用3个单位的药剂，则有效治疗的时间可达小时． ······························ 6分
311010m]?8?x?（Ⅱ）当6≤x≤8时，y?2?(4?x)?m[， ······················· 8分
24?(x?6)x?210m因为8?x?≥2对6≤x≤8恒成立，
x?2x2?8x?12即m≥对6≤x≤8恒成立，
10x2?8x?12等价于m≥(···································································· 9分 )max，6≤x≤8． ·
102x?8x?12(x?4)2?4令g(x)?，则函数g(x)?在[6,8]是单调递增函数， ·············· 10分
101026x?8x?12当x=8时，函数g(x)?取得最大值为， ················································ 11分
51066所以m≥，所以所求的m的最小值为． ································································ 12分
55解法二：（Ⅰ）同解法一；
11010m]?8?x?（Ⅱ）当6≤x≤8时，y?2?(4?x)?m[， ······················· 8分
24?(x?6)x?2