2015届福建省宁德市高三普通高中毕业班5月质检数学文试卷

2015届福建省宁德市高三普通高中毕业班5月质检数学文试卷数学（文科）试卷本试卷分第I 卷（选择题）和第Ⅱ卷（非选择题）两部分．本卷满分150分，考试时间120分钟． 注意事项:1．答题前，考生先将自己的姓名、准考证号填写在答题卡上．2．考生作答时，将答案答在答题卡上．请按照题号在各题的答题区域（黑色线框）内作答，超出答题区域书写的答案无效．在草稿纸、试题卷上答题无效．3．选择题答案使用2B 铅笔填涂，如需改动，用橡皮擦干净后，再选涂其他答案标号；非选择题答案使用0．5毫米的黑色中性（签字）笔或碳素笔书写，字体工整、笔迹清楚．4．保持答题卡卡面清楚,不折叠、不破损．考试结束后，将本试卷和答题卡一并交回． 参考公式：第I 卷（选择题 共60分）一、选择题：本大题共12小题，每小题5分，共60分．在每小题给出的四个选项中，只有一项是符合题目要求的．1．已知集合2{|320}M x x x =-+=，{2,1,1,2}N =--，则M N =IA ．{2,1}--B ．{1,2}C ．{2,1}-D ．{2,1,1,2}-- 2．若x ∈R ，则“21x <”是“10x -<<”的A ．充分不必要条件B ．必要不充分条件C ．充要条件D ．既不充分也不必要条件3．某全日制大学共有学生5400人，其中专科生有1500人，本科生有3000人，研究生有900人. 现采用分层抽样的方法调查学生利用因特网查找学习资料的情况，抽取的样本为180人，则应在专科生、本科生与研究生这三类学生中分别抽取 A ．55人，80人，45人 B ．40人，100人，40人 C ．60人，60人，60人D ．50人，100人，30人4．经过圆22(2)1x y -+=的圆心且与直线210x y -+=平行的直线方程是，，(n x x ++-A ．240x y --=B ．240x y -+=C ．220x y +-=D ．220x y ++= 5．设m ，n 是两条不同的直线，α，β是两个不同的平面，下列正确的是A ．若m ∥α，n ⊂α，则m ∥nB ．若m ∥α，m ∥β，则α∥βC ．若m ∥α，α⊥β，则m ⊥βD ．若m ∥n ，m ⊥α，则n ⊥α 6．已知sin α=(0,)2απ∈，则tan 2α= A ．43- B ．43C ．12-D ．27．下列函数中，既为奇函数又在(0,)+∞内单调递减的是A ．()sin f x x x =B ．12()f x x -=C ．1()1xxe f x e-=+ D ．3()f x x x=-8．运行如图所示的程序，若输出y 的值为1， 则可输入x 的个数．．为 A ．0 B ．1C ．2D ．39．已知实数,x y 满足122x x y x y ≥⎧⎪+≤⎨⎪-≤⎩，若不等式3ax y -≤恒成立，则实数a 的取值范围为A ．(,4]-∞B ．3(,]2-∞ C3[ D ．[2,4]10．已知四棱锥P ABCD -的三视图如图所示,则此四棱锥的侧面积为A ．6+B ．9+C ．12+D ．20+11．已知点P 是ABC ∆所在平面上一点，AB 边的中点为D ，若23PD PA CB =+，则ABC ∆与ABP ∆的面积比为A ．3B ．2C ．1D ．1212. O 为坐标原点，,A B 为曲线y =上的两个不同点，若6OA OB ⋅=，则直线AB 与圆2249x y +=的位第8题图侧视图俯视图第10题图置关系是A. 相交B. 相离C. 相交或相切D. 相切或相离第Ⅱ卷（非选择题 共90分）二、填空题：本大题共4小题，每小题4分，共16分．把答案填写在答题卡的相应位置． 13．复数i(12i)z =+（i 为虚数单位），则z = .14．在区间(0,4)内任取一个实数x ，则使不等式2230x x --<成立的概率为 . 15．关于x 的方程2log 0x a -=的两个根为1212,()x x x x <，则122x x +的最小值为 .16．已知函数()sin cos (22x x f x a a =+∈R)，且()()3f x f 2π≤恒成立. 给出下列结论：①函数()y f x =在[0,]32π上单调递增；②将函数()y f x =的图象向左平移3π个单位，所得图象对应的函数为偶函数；③若2k ≥，则函数()(2)3g x kx f x π=--有且只有一个零点.其中正确的结论是 .（写出所有正确结论的序号）三、解答题：本大题共6小题，共74分，解答应写出文字说明、证明过程或演算步骤． 17．（本小题满分12分）已知等比数列{}n a 的前n 项和2n n S r =+. （Ⅰ）求实数r 的值和{}n a 的通项公式；（Ⅱ）若数列{}n b 满足11b =，121log n n n b b a ++-=，求n b .18．（本小题满分12分）某中学刚搬迁到新校区，学校考虑，若非住校生．．．．上学路上单程所需时间人均超过20分钟，则学校推迟5分钟上课. 为此，校方随机抽取100个非住校生，调查其上学路上单程所需时间(单位：分钟)，根据所得数据绘制成如下频率分布直方图，其中时间分组为[0,10)，[10,20)，[20,30)，[30,40)，[40,50].（Ⅰ）求频率分布直方图中a 的值；（Ⅱ）从统计学的角度说明学校是否需要推迟5分钟上课； （Ⅲ）若从样本单程时间不小于30分钟的学生中， 随机抽取2人，求恰有一个学生的单程时间落在[40,50]上的概率.19．（本小题满分12分）已知函数()2sin()f x x ωϕ=+(0,)2ωϕπ><在一个周期内的图象如图所示，其中M (,2)12π，N (,0)3π．（Ⅰ）求函数()f x 的解析式；（Ⅱ）在ABC ∆中，角,,A B C 的对边分别是a,b,c且3,()2Aa c f ==ABC ∆的面积．20．（本小题满分12分）如图四棱锥P ABCD -中，平面PAD ⊥平面ABCD ，//AB CD ， 90ABC ︒∠=，且2,1CD AB BC PA ====，PD =（Ⅰ）求三棱锥A PCD -的体积；（Ⅱ）问：棱PB 上是否存在点E ，使得//PD 平面ACE ？ 若存在，求出BEBP的值，并加以证明；若不存在，请说明理由．21．（本小题满分12分）已知点(A B ，动点E 满足直线EA 与直线EB 的斜率之积为12-.（Ⅰ）求动点E 的轨迹C 的方程；（Ⅱ）设过点()1,0F 的直线1l 与曲线C 交于点,P Q ，记点P 到直线2:2l x =的距离为d .（ⅰ）求PF d的值；（ⅱ）过点F 作直线1l 的垂线交直线2l 于点M ，求证：直线OM 平分线段PQ .22．（本小题满分14分）C已知函数1()ln (1)2f x x a x =--（a ∈R ）．（Ⅰ）若2a =-，求曲线()y f x =在点(1,(1))f 处的切线方程； （Ⅱ）若不等式()0f x <对任意(1,)x ∈+∞恒成立．（ⅰ）求实数a 的取值范围；（ⅱ）试比较2a e -与2e a -的大小，并给出证明（e 为自然对数的底数， 2.71828e ≈）．2015年宁德市普通高中毕业班第二次质量检查数学（文科）试题参考答案及评分标准说明：一、本解答指出了每题要考察的主要知识和能力，并给出了一种或几种解法供参考，如果考生的解法与本解法不同，可根据试题的主要考察内容比照评分标准指定相应的评分细则。

二、对计算题，当考生的解答在某一部分解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分。

三、解答右端所注分数，表示考生正确做到这一步应得的累加分数。

四、只给整数分数，选择题和填空题不给中间分。

一、选择题：本题考查基础知识和基本运算．本大题共12小题，每小题5分，共60分．1．B 2．B 3．D 4．A 5．D 6．A 7．C 8．D 9．B 10．C 11．C 12．A二、填空题：本题考查基础知识和基本运算．本大题共4小题，每小题4分，共16分．13．2i --； 14．34； 15．； 16．①③．三、解答题：本大题共6小题，共74分．17.本题主要考查等差数列、等比数列、数列求和等基础知识；考查推理论证与运算求解能力，满分12分．解：（Ⅰ）∵2n n S r =+,∴112a S r ==+，2212a S S =-=，3324a S S =-=. ············································ 3分 ∵数列{}n a 是等比数列,∴2213a a a =⋅，即224(2)r =+， ·········································································· 4分∴1r =- . ············································································································· 5分 ∴数列{}n a 是以1为首项，2为公比的等比数列， ·············································· 6分 ∴ 12()n n a n -*=∈N . ······························································································ 7分 （Ⅱ）∵12n n a +=，∴12log 2n n n b b n +-==， ······················································ 8分 当2n ≥时，121321()()()n n n b b b b b b b b -=+-+-++-112(1)n =++++- ···································································· 9分(1)12n n-=+211122n n =-+ ··············································································· 11分 又11b =符合上式，∴2111()22n b n n n *=-+∈N . ················································································· 12分18．本题主要考查概率、统计等基础知识，考查数据处理能力、抽象概括能力、运算求解能力以及应用意识，考查或然与必然思想、化归与转化思想．满分12分． 解：（Ⅰ）时间分组为[0,10)的频率为110(0.060.020.0030.002)0.15-+++=， ····························································· 2分∴0.150.01510a ==， 所以所求的频率直方图中a 的值为0.015. ···························································· 3分 （Ⅱ）100个非住校生上学路上单程所需时间的平均数： 0.1550.6150.2250.03350.0245x =⨯+⨯+⨯+⨯+⨯ ············································· 4分 0.7595 1.050.9=++++ 16.7=. ··················································································································· 5分 因为16.720<，所以该校不需要推迟5分钟上课. ······································································· 6分（Ⅲ）依题意满足条件的单程所需时间在[30,40)中的有3人，不妨设为123,,a a a ， 单程所需时间在[40,50]中的有2人，不妨设为12,b b ， ····································· 7分 从单程所需时间不小于30分钟的5名学生中，随机抽取2人共有以下10种情况：12(,)a a ，13(,)a a ，11(,)a b ，12(,)a b ，23(,)a a ，21(,)a b ，22(,)a b ，31(,)a b ，32(,)a b ，12(,)b b ；··············································································· 10分 其中恰有一个学生的单程所需时间落在[40,50]中的有以下6种：11(,)a b ，12(,)a b ，21(,)a b ，22(,)a b ，31(,)a b ，32(,)a b ； ···························· 11分 故恰有一个学生的单程所需时间落在[40,50]中的概率63105P ==. ··············· 12分 19．本题主要考查解三角形，三角函数的图象与性质等基础知识；考查运算求解能力，考查化归与转化思想、数形结合思想．满分12分． 解：（Ⅰ）由图像可知：函数()f x 的周期4()312T πππ=⨯-=， ································ 1分∴22ωπ==π.·········································································································· 2分 又()f x 过点(,2)12π,∴()2sin()2126f ππϕ=+=，sin()16πϕ+=， ························································ 3分∵2πϕ<，2(,)633πππϕ+∈-，∴62ππϕ+=，即3πϕ=. ························································································ 4分∴()2sin(2)3f x x π=+. ··························································································· 5分（Ⅱ）∵()2sin()23A f A π=+即sin()3A π+=，又4(0,),(,)333A A ππππ∈+∈∴233A ππ+=，即3A π=. ······················································································ 7分在ABC ∆中，,33A a c π===，由余弦定理得 2222cos a b c bc A =+-， ······························································· 8分 ∴21393b b =+-，即2340b b --=， 解得4b =或1b =-（舍去）. ··············································································· 10分∴11sin 43sin 223ABC S bc A π∆==⨯⨯⨯=························································· 12分20．本题主要考查空间线与线、线与面的位置关系、体积的计算等基础知识；考查空间想象能力、运算求解能力及推理论证能力，满分12分．解：（Ⅰ）取CD 中点G ，连接AG ，2,//,CD AB AB CD = //,,AB GC AB GC ∴=∴四边形AGCB 为平行四边形，090AGD DCB ABC ∴∠=∠=∠=在Rt AGD ∆中，11,1,2AG BC DG CD ====AD ∴········································································ 1分 2223,123,PD PA AD ∴=+=+=222,PD PA AD =+090,PAD ∴∠= 即,PA AD ⊥ ················································································· 2分平PAD ⊥面平ABCD 面，平PAD 面平ABCD AD =面PA ∴⊥平ABCD 面································································································· 3分 112ACD S CD AG ∆=⋅=, ·························································································· 4分 A PCD P ACD V V --∴= ····································································································· 5分13ACD S PA ∆=⋅⋅ 111133=⨯⨯=. ····························································································· 6分 （II ）棱PB 上存在点E ，当13BE BP =时，//PD 平面ACE .···································· 7分 证明：连结BD 交AC 于点O ，连结OE .∵//,2AB CD CD AB =∴1,2BO AB OD CD == ···················································· 8分 ∴13BO BD =,又13BE BP = ∴BO BEBD BP=， ∴//,OE DP ····························································· 10分 又,OE ACE PD ACE ⊂⊄面，面 //PD ACE ∴面.········································································································ 12分21．本题主要考查直线、椭圆、轨迹等基础知识及直线与圆锥曲线的位置关系；考查运算求解能力、推理论证能力；考查特殊与一般的思想、化归与转化思想．满分12分． 解法一：（Ⅰ）设(,)E x y ，依题意得 1,2EA EB k k ⋅==-(x ≠, ·········································· 1分 整理得2212xy +=,∴动点E 的轨迹C 的方程为 221(2x y x +=≠. ············································ 3分（Ⅱ）（ⅰ）(1,0)F ，设11(,),P x y 则 221112x y =-, ·············································· 4分∴1||PF d =······················································································· 5分1=1=. ········································································································ 7分 (说明：直接给出结论正确，没有过程得1分)（ⅱ）依题意,设直线22:1,(,)PQ x my Q x y =+,联立221,12x my x y =+⎧⎪⎨+=⎪⎩可得22(2)210m y my ++-=, ················································ 8分 显然12220,,2my y m ∆>+=-+ ··············································································· 9分 所以线段PQ 的中点T 坐标为222(,),22mm m -++ ·················································· 10分 又因为1,FM l ⊥故直线FM 的方程为(1)y m x =--, 所以点M 的坐标为(2,)m -,所以直线OM 的方程为:,2my x =-···································································· 11分 因为222(,)22m T m m -++满足方程,2my x =-故OM 平分线段.PQ ···························································································· 12分 解法二：（Ⅰ）（Ⅱ）（ⅰ）同解法一（ⅱ）当直线1l 的方程为1x =时，显然OM 平分线段PQ ； ···························· 8分当直线1l 的方程为(1)(0)y k x k =-≠时，设22(,)Q x y 联立22(1),12y k x x y =-⎧⎪⎨+=⎪⎩可得2222(21)4220k x k x k +-+-=, 显然212240,,21k x x k ∆>+=+·················································································· 9分 所以线段PQ 的中点坐标为2222(,)2121k kT k k -++ ················································· 10分 又因为1,FM l ⊥故直线FM 的方程为1(1)y x k=--,所以点M 的坐标为1(2,)k-,。