【2015厦门5月质检】福建省厦门市2015届高三毕业班适应性考试数学理试题 Word版含答案

2015年厦门市高三适应性考试数学（理科）试卷注意事项：1．本科考试分试题卷和答题卷，考生须在答题卷指定位置上作答，答题前，请在答题卷的密封线内填写学校、班级、考号、姓名．2．本试卷分为第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，全卷满分150分，考试时间120分钟．第Ⅰ卷（选择题共50分）一、选择题：本大题共10小题，每小题5分，共50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．复数i(1i)+（i为虚数单位）的共轭复数是A. 1i+B. 1i-C. 1i--D2．随机变量~(0,1)Nξ，则()12Pξ≤≤=Ａ.0.0215 B. 0.1359 C. 0.1574 D （参考数据：()0.6Pμσξμσ-≤≤+=，(22)Pμσξμσ-≤≤+=(33)0.9974Pμσξμσ-≤≤+=）3．直线22+-=xy恰好经过椭圆12222=+byaxA.55B.21C.552D4．已知函数()f x的图像如图所示，则()f x的解析式可能是()31.21A f x xx=--()31.21B f x xx=+-()31.21C f x xx=-+()31.21D f x xx=---5．已知实数,x y满足2,20y xx y⎧≥⎨-+≥⎩，则z x y=+的取值范围是.[0,6]A1.[,6]4B-1.[,0]4C-6．命题:p函数2y xx=+在[]1,4上的值域为93,2⎡⎤⎢⎥⎣⎦；命题:q()()1122log1log0a a a+>>.下列命题中，真命题的是A．p q∧B．p q∨C．()p q∧⌝D．()p q⌝∨7．已知数列{}n a满足: 当()*11,,p q p q N p q+=∈<时，2pp qa a+=，则{}na的前10项和10S=.31A.62B.170C.1023D8．如图，圆O与x轴的正半轴的交点为A，点B，C在圆O上，点B的坐标为(1,2)-，点C位于第一象限，AOCα∠=．若BC=则2sincos22ααα=.A.B CD9． 如图1，已知正方体ABCD －A 1B 1C l D 1的棱长为a ，动点M 、N 、Q 分别在线段1111,,AD B C C D 上.当三棱锥Q-BMN 的俯视图如图2所示时， 三棱锥Q-BMN 的正视图面积等于A. 212a B. 214aC. 2D. 210．如图所示，由直线()2,10,x a x a a y x ==+>=及x小矩形与大矩形的面积之间，即⎰++<<1222)1(a aa dx x a .类比之，n ∀∈*N ,111111122121A n n n nn n +++<<++++++-恒成立, 则实数A 等于A.21B.53C.2lnD.25ln第Ⅱ卷 （非选择题 共100分）二、填空题：本大题共5小题，每小题4分，共20分．11．阅读如图所示的程序，该程序输出的结果是 ▲ ．12．设52501251(1)(1)(1)x a a x a x a x +=+-+-++-,则125a a a +++= ▲ ．13．一个口袋内有5个不同的红球，4个不同的白球.若取一个红球记2分，取一个白球记1分，从中任取4个球，使总分不少于7分的取法有 ▲ 种.14．如图，在ABC △中，0AD BC ⋅=,3BC BD =，过点D 的直线分别交直线AB ，AC 于点M ，N .若(),0,0AM AB AN AC λμλμ==>>， 则μλ2+的最小值是 ▲ ．15．十八世纪，法国数学家布丰和勒可莱尔提出投针问题：在平面上画有一组间距为a 的平行线，将一根长度为l 的针任意掷在这个平面上，求得此针与平行线中任一条相交的概率2lp aπ=（π为圆周率）. 已知 3.14,6l a ==， 3.14π≈，现随机掷14根相同的针（长度为l ）在这个平面上，记这些针与平行线（间距为a ）相交的根数为m ，其相应的 概率为()P m .当()P m 取得最大值时，m = ▲ ．三、解答题：本大题共6小题，共80分.解答应写出文字说明、证明过程或演算步骤. 16．（本小题满分13分）如图，平面直角坐标系xOy 中，,3ABC π∠= 6ADCp?,AC =，BCD ∆的面积（Ⅰ）求AB 的长；（Ⅱ）若函数()sin()(0,0,)2f x M x M πωϕωϕ=+>><的图象经过,,A B C 三点，其中,A B 为()f x 的图象与x 轴相邻的两个交点，求函数()f x 的解析式.17．（本小题满分13分）如图，梯形ABCD 中，AB ⊥AD ，AD ∥BC ，AD ＝6，BC ＝2AB ＝4，E ，F 分别在线段BC ，AD 上，EF ∥AB ．将四边形ABEF 沿EF 折起，连接AD ，AC .（Ⅰ）若BE ＝3，在线段AD 上一点取一点P ，使12AP PD =，求证：CP ∥平面ABEF ； （Ⅱ）若平面ABEF ⊥平面EFDC ，且线段F A ,FC ,FD 的长成等比数列，求二面角E －AC －F 的大小．18．（本小题满分13分）某茶厂现有三块茶园，每块茶园的茶叶估值为6万元.根据以往经验：今年5月12日至14日是采茶的最佳时间，在此期间，若遇到下雨，当天茶园的茶叶估值减少为前一天的一半.现有两种采摘方案：方案①：茶厂不额外聘请工人，一天采摘一块茶园的茶叶；方案②：茶厂额外聘请工人，在12日采摘完全部茶叶，额外聘请工人的成本为3.2万元.A D E根据天气预报，该地区5月12日不降雨，13日和14日这两天降雨的概率均为40%.每天是否下雨不相互影响.（Ⅰ）若采用方案①，求茶厂14日当天采茶的预期收益； （Ⅱ）从统计学的角度分析，茶厂采用哪种方案更合理. 19．（本小题满分13分）如图，抛物线E :()220y px p =>的焦点为F ，其准线l 与x 轴交于点A过抛物线E 上的动点．．P 作PD l ⊥于点D . 当23DPF π∠=时， 4PF =. （Ⅰ）求抛物线E 的方程；（Ⅱ）过点P 作直线m ⊥DF ，求直线m 与抛物线E 的交点个数；（Ⅲ）点C 是DPF ∆的外心，是否存在点P ，使得CDP ∆若存在，请求出面积的最小值及P 的坐标；若不存在， 20．（本小题满分14分）已知函数()e x f x ax =-（e 为自然对数的底数）.（Ⅰ）讨论()f x 的单调性；（Ⅱ）定义：函数()F x 的定义域为D ，若0x D ∃∈，使00()F x x =成立，则称x 0为()F x 的不动点.当1a =时， （ⅰ）证明：函数1(0)()y x f x =>存在唯一的不动点0x ，且0(ln 2,1)x ∈； （ⅱ）已知数列{}n a 满足*111ln 2,()()n n a a n N f a +==∈， 求证：*200020()(),()1n n f a f x n N f x x a x -∀∈>+--，（其中0x 为1(0)()y x f x =>的不动点）.21．本题有（1）、（2）、（3）三个选答题，每小题7分，请考生任选2题作答，满分14分，如果多做，则按所做的前两题计分．作答时，先用2B 铅笔在答题卡上把所选题目对应的题号涂黑，并将所选题号填入括号中． （1）（本小题满分7分）选修4-2：矩阵与变换已知矩阵214aA ⎛⎫=⎪-⎝⎭，A 的一个特征值2λ=. （Ⅰ）求矩阵A ；（Ⅱ）在平面直角坐标系中，点(1,1)P 依次．．在矩阵A 所对应的变换σ和关于x 轴的反射变换τ的作用下得到点P '，写出复合变换τσ⋅的变换公式，并求出点P '的坐标.（2）（本小题满分7分）选修4-4：坐标系与参数方程已知曲线C 的极坐标方程为2cos 4sin ρθθ=-.以极点为原点，极轴为x 轴的正半轴，建立平面直角坐标系，直线l 的参数方程为1cos ,1sin x t y t αα=+⎧⎨=-+⎩(t 为参数).（Ⅰ）判断直线l 与曲线C 的位置关系，并说明理由；（Ⅱ）若直线l 和曲线C 相交于,A B 两点，且AB =l 的斜率.（3）（本小题满分7分）选修4-5：不等式选讲已知0,0,0a b c >>>，3331113abc a b c +++的最小值为m . （Ⅰ）求m 的值；（Ⅱ）解关于x 的不等式|1|2x x m +-<.2015年厦门市高中毕业班适应性考试 数学（理科）试题参考答案及评分标准二、填空题：三、解答题：16．本题考查解三角形和三角函数图象及性质等知识，考查学生运算求解能力、数据处理能力及推理论证能力，考查学生数形结合思想、函数与方程思想及转化与化归思想，属于中档偏易题.本题满分13分.解：（Ⅰ）∵3ABCp?，6ADC p ?，∴6BCD p ?，23CBD p ?， BC BD = ···················· 1分又∵BCD ∆12sin 23BCD S BD BC p D =鬃2BC == ······························· 2分 ∴2BC =. ············································································································································· 3分在ABC D 中，AC =，3ABC p?，由余弦定理得：2222cos 3AC AB BC AB BC p =+- ， ································································· 4分即2174222AB AB =+-创，整理得2230AB AB --=， ∴3AB =，或1AB =-(舍去)，∴AB 的长为3. ············································································· 6分（Ⅱ）由（Ⅰ）知，(2,0),(1,0),A B C -， ···················································································· 7分∵函数()sin()(0,0,)2f x M x M πωϕωϕ=+>><的图象经过,,A B C 三点，其中,A B为()f x 的图象与x 轴相邻的两个交点， ∴函数()f x 的半个周期32T =，对称轴为12x =， ·········································································· 9分 ∴26T pw==， ∵0ω>，∴3p w =， ························································································································ 10分 ∴1,232k k Z p p j p ?=+ ，∴,3k k Z pj p =+ ， 又∵2πϕ<，∴3pj =， ·················································································································· 11分∴()sin()33f x Mx p p =+，又∵(0)sin 32f M M p ===，∴2M =， ······································································· 12分 ∴函数()f x 的解析式是()2sin()33f x x p p=+. ············································································· 13分 17．本题以翻折的图形为载体，考查空间点、线、面位置关系、线面平行证明及求二面角大小等有关基础知识，同时结合考查数列知识．本题考查空间想象能力、运算求解能力和推理论证能力，考查数形结合和化归与转化等数学思想方法. 本题满分13分.解：（Ⅰ）证法一：在梯形ABCD 中，AD ∥BC ， EF ∥AB ，BE ＝3，∴AF =3，又AD ＝6，BC ＝4，∴EC =1，FD =3， ····················································· 1分在线段AF 上取点Q ，使12AQ QF =，连接,PQ QE , ··························· 2分∵12AP PD =，∴1//3PQ DF ，∵1//3CE DF ，∴//CE PQ ， ············································································································· 3分 ∴四边形ECPQ 为平行四边形，∴//CP EQ ， ··································· 4分∵CP ⊄平面ABEF ，EQ ⊂平面ABEF ，∴CP ∥平面ABEF . ············ 5分证法二：同证法一，EC =1，FD =3， ····················································· 1分延长DC 交FE 的延长线于点M ，连接AM ，则12MC CD =，······· 2分 ∵12AP PD =，∴//CP MA ， ··························································· 4分 ∵CP ⊄平面ABEF ，MA ⊂平面ABEF ，∴CP ∥平面ABEF . ············ 5分证法三：同证法一，EC =1，FD =3， ····················································· 1分 在线段DF 上取点R ，使12FR RD =，连接PR ，CR ， ∵12AP PD =，∴//PR AF ， ∵PR ⊄平面ABEF ，AF ⊂平面ABEF ，∴PR ∥平面ABEF ; ··············· 2分 ∵12FR RD =， ∴1,EC FR ==∵//EC FR ， ∴四边形ECRF 为平行四边形，∴//CR EF ， ∵CR ⊄平面ABEF ，EF ⊂平面ABEF ，∴CR ∥平面ABEF ; ···························································· 3分 ∵PR CR R =，∴平面//PRC 平面ABEF ， ·················································································· 4分 ∵CP ⊂平面PRC ，∴CP ∥平面ABEF . ································································································· 5分（Ⅱ）解法一：在梯形ABCD 中，AB ⊥AD ，AD ∥BC ，∴EF ⊥AF ， EF ⊥FD ， ∵平面ABEF ⊥平面EFDC ，平面ABEF 平面EFDC=EF ，AF ⊂平面EFDC ， ∴AF ⊥平面EFDC ， ······························································································································· 6分 设(04)AF x x =<<，∵EF ＝BA ＝2，∴6,4FD xEC x =-=-， ∴FC =······················································································································· 7分∵线段AF ，FC ，FD 的长成等比数列，∴2FC AF FD =⋅，24(4)(6)x x x +-=-，化简得27100x x -+=， ∴2x =或5x =（舍）， ························································································································ 9分 以F 为原点，FE ，FD ，F A 分别为,,x y z 轴建立空间直角坐标系，如图， 则(0,0,0)F ，(2,0,0)E ，(2,2,0)C ，(0,4,0)D ，(0,0,2)A ···· 10分∴(0,2,0)EC =，(2,0,2)EA =-，设1111(,,)n x y z =是平面ACE 的一个法向量，则110,0n EC n EA ⎧⋅=⎪⎨⋅=⎪⎩，即11120,220y x z =⎧⎨-+=⎩，取11z =，则111,0x y ==，∴1(1,0,1)n =； ······························· 11分又(2,2,0)FC =，(0,0,2)FA =，设2222(,,)n x y z =是平面ACF 的一个法向量，则220,0n FC n FA ⎧⋅=⎪⎨⋅=⎪⎩，即222220,20x y z +=⎧⎨=⎩，取21x =，则221,0,y z =-=∴2(1,1,0)n =-； ··································· 12分∴1212121cos 2||||2n n n n n n ⋅<⋅>===⋅，∵ 二面角E-AC-F 为锐角， ∴二面角E-AC-F 为060. ··························· 13分解法二：同解法一得2x =或5x =（舍）， ············································· 9分 2,2AF EC ==，∵2EF =，∴EF EC =，设点G 为FC 的中点，连接EG ，则EG FC ⊥， ······························· 10分∵AF ⊥平面EFDC ，AF ⊂平面AFC ，∴平面AFC ⊥平面EFDC ， ∵平面AFC 平面EFDC =FC ，∴EG ⊥平面AFC ， ∵AC ⊂平面AFC ， ∴EG ⊥AC ， ···································································································· 11分 过G 作GH ⊥AC 交AC 于H ，连接EH ， ∵EG GH =G ， ∴AC ⊥平面EGH ， ∵EH ⊂平面EGH ，∴AC ⊥EH ，∴EHG ∠是二面角E-AC-F 的平面角， ····························································································· 12分在Rt CEF ∆中，EG =， 在Rt ACE ∆中，,2A E EC ==，∴AC =，EH ，∴sin 3EG EHG EH ∠=∵EHG ∠为锐角， ∴060EHG ∠=，即二面角E －AC －F 为060. ················································ 13分 18．本题考查概率概念及其意义，独立事件，对立事件和互斥事件的概率，随机变量的分布列及数学期望等相关知识；考查运算求解能力，数据处理能力和应用意识，考查转化与化归和必然与或然等数学思想方法. 本题满分13分. 解：(Ⅰ)设茶厂14日当天采茶的预期收益为ξ万元，则ξ的可能取值为6，3，1.5 ································· 1分339322312224(6),(3),( 1.5)55255555255525P P P ξξξ==⨯===⨯+⨯===⨯=······························· 4分所以ξ的分布列为·················································································································· 5分 所以ξ的数学期望为9124()63 1.5 3.84252525E ξ=⨯+⨯+⨯=， 即茶厂14日当天采茶的预期收益为3.84万元. ················································································ 6分(Ⅱ)茶厂若采用方案①，设茶厂第二天采茶的预期收益为η万元，则η的可能取值为6和3，因为32(6),(3)55P P ηη====， ····································································································· 7分 所以η的分布列为·································································································································· 8分所以η的数学期望为3263 4.855E η=⨯+⨯=，··············································································· 9分 所以茶厂若采用方案①则其采茶总收益为16 4.8 3.8414.64y =++=， ····································· 11分茶厂若采用方案②则其采茶总收益为263 3.214.8y =⨯-=， ····················································· 12分因为14.64<14.8，所以茶厂应该采用方案②收益高，风险小，和谐社会，提供就业岗位. ······································································································································································ 13分 19．本题考查抛物线定义，直线方程，直线方程与抛物线、圆的位置关系等知识，考查学生运算求解能力、推理论能力、抽象概括能力，考查数形结合、函数与方程、化归与转化等数学思想. 本题满分13分. 解：(Ⅰ)过点P 作PQ x ⊥轴于点Q ，当23DPF π∠=时，PF =∴6FPQ π∠=， ················································································ 1分 Rt PQF ∆中，sin26QF PF π=⋅=， ········································ 2分DP PF =.6AF DP QF ∴=+=，即 6p =， ·················· 3分 ∴抛物线E 的方程：212y x =， ······················································ 4分（也可由余弦定理求得DF =Rt DAF ∆中，6AF =，即6p =） (Ⅱ) 解法一：当点P 为原点O 时，直线m 的方程：0x =与抛物线E 设()00,P x y ，则()03,D y -，()3,0F ，06DF y k =-，06m k y ∴=， 直线()0006:m y y x x y -=-，化简得：200066x y y y x =-+， 代入212y x =得()2200026y y y y x =-+， (22)0020y y y y ∴-+=，0y y ∴=（0∆=），·······································∴直线m 与抛物线E 有且只有一个交点P .············································解法二：由(Ⅰ)得()()3,0,3,0A F -，设()23,6P t t ，则()3,6D t -，。