校对稿福州市年高三月份质检数学答案文科

2015年福州市高中毕业班质量检测 文科数学能力测试参考答案及评分细则一、选择题：本题共有12个小题，每小题5分，满分60分． 1．A 2．A 3．C 4．D 5．D 6．C 7．C 8．B 9．D 10．B 11．B 12．D 二、填空题：本大题共 4 小题，每小题 4 分，满分 16 分．13．{}1- 14．1315．2 16．55三、解答题：本大题共6小题，共74分．17． 本小题主要考查三角函数的图象与性质（对称性、周期性、单调性）、二倍角的余弦公式等基础知识，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．满分12分．【解析】（Ⅰ）因为()cos (0,)f x x x x ωωω=->∈R ，所以()2sin()6f x x πω=-． ··········································································· 2分所以max ()2f x =．因为函数()f x 与直线2y =的相邻两个交点之间距离为π， 所以T π=， ····························································································· 3分所以2ππω=，解得2,ω=············································································ 4分 所以()2sin(2)6f x x π=-．令222,262k x k k πππππ--+∈Z 剟， ···························································· 5分 解得,63k xk k ππππ-+∈Z 剟． ··································································· 6分 所以函数()f x 的单调递增区间是[,],63k k k ππππ-+∈Z ． ·································· 7分 （Ⅱ）由（Ⅰ）知，2sin()26f απα⎛⎫=- ⎪⎝⎭，因为223f α⎛⎫= ⎪⎝⎭，所以1sin()63πα-=． ·················································································· 8分所以cos 2cos236ππαα⎛⎫⎛⎫-=- ⎪ ⎪⎝⎭⎝⎭································································· 10分212sin 6πα⎛⎫=-- ⎪⎝⎭ ····························································· 11分79=． ············································································· 12分 18．本小题主要考查概率、统计等基础知识，考查数据处理能力、运算求解能力、应用意识，考查必然与或然思想．满分12分．·········由上表可知：成就感为三级（即01w 剟）的只有9A 一位，其频率为110． ··········· 3分用样本的频率估计总体的频率，可估计该群体中成就感等级为三级的人数有12002010⨯=．·············································································································· 5分 （Ⅱ）设事件A 为 “从成就感等级是一级的被采访者中随机抽取两人，他们的综合指标w 均为4”．由（Ⅰ）可知成就感是一级的（4w …）有：123568,,,,,A A A A A A ，共6位，从中随机抽取两人，所有可能的结果为：{}{}{}{}{}{}{}{}{}{}{}1213151618232526283536,,,,,,,,,,,,,,,,,,,,,,A A A A A A A A A A A A A A A A A A A A A A {}{}{}{}38565868,,,,,,,A A A A A A A A ，共15种． ··················································· 9分 其中综合指标4w =有：125,,A A A ，共3名，事件A 发生的所有可能结果为：{}{}{}121525,,,,,A A A A A A ，共3种， ··························· 10分 所以3()15P A ==15．················································································· 12分 19．本小题主要考查直线与直线、直线与平面的位置关系，以及几何体的体积等基础知识，考查空间想象能力、推理论证能力、运算求解能力，考查数形结合思想、化归与转化思想．满分12分． 证明：（Ⅰ）连结BD ．因为1111ABCD A B C D -是长方体，且2AB BC ==， 所以四边形ABCD 是正方形， ······································································· 1分 所以AC BD ⊥． ························································································ 2分 因为在长方体1111ABCD A B C D -中，1BB ⊥平面ABCD ，AC ⊂平面ABCD ， 所以1AC BB ⊥． ··························································· 4分 因为BD ⊂平面11BB D D ，1BB ⊂平面11BB D D ， 且1BD BB B =，所以AC ⊥平面11BB D D ． ················································ 5分 因为BP ⊂平面11BB D D ， 所以AC BP ⊥ ······························································· 6分 （Ⅱ）点P 到平面ABC 的距离14AA =，ABC ∆的面积122ABC S AB BC ∆=⋅⋅=， ····························································· 7分 所以111824=333P ABC ABC V S AA -∆=⋅=⨯⨯． ························································· 8分在Rt △1BB P中，114,BB B P ==所以BP =， ·········································· 9分同理CP =又=2BC ，所以PBC ∆的面积122PBC S ∆=⨯ ··· 10分设三棱锥A PBC -的高为h ，则因为A PBC P ABC V V --=，所以183PBC S h ⋅=， ····················································· 11分83=，解得h =． 即三棱锥A PBC -·································································· 12分20．本小题主要考查数列、等差数列、等比数列、不等式等基础知识，考查运算求解能力、抽象概括能力、应用意识，考查函数与方程思想、化归与转化思想、分类与整合思想．满分12分． 解：（Ⅰ）设第n 年初M 的价值为n a ，依题意，当13n 剟时，数列{}n a 是首项为20，公比为12的等比数列，所以13120522n n n a --⎛⎫=⨯=⨯ ⎪⎝⎭．故210a =，35a =，所以321a a a <<．·············································································································· 2分 当4n …时，数列{}n a 是以4a 为首项，公差为4的等差数列，又4349a a =+=，所以()94447n a n n =+-⨯=-． ········································································· 3分1A A令20n a >，得274n >，又因为*n ∈N ，所以7n …． ·········································· 4分 因此，第7年初M 开始的价值n a 超过原购买的价值． ······································· 5分（Ⅱ）设n S 表示前n 年初M 的价值的和，则n n ST n=由（Ⅰ）知，当13n 剟时，3120124052112n nn S -⎛⎫⎛⎫⋅- ⎪⎪ ⎪⎝⎭⎝⎭==-⨯-，34052n n T n --⨯=①； ·············································································································· 7分当4n ...时，由于335S =，故()()()23453947 (3525322)n n n n S S a a a n n -+-=++++=+=-+，225323225n n n T n n n-+==+-．② ································································ 9分 当13n 剟时，由①得，120T =，215T =，3353T =，所以123T T T >> ················ 10分当4n …时，由②知，3225511n T n n =+-=…，当且仅当322n n=，即4n =时等号成立．即()4min 11n T T ==． ····················································································· 11分 由于34T T >，故在第4年初n T 的值最小，其最小值为11． ································ 12分21．本小题主要考查函数、导数、不等式等基础知识，考查推理论证能力、抽象概括能力、运算求解能力，考查函数与方程思想、数形结合思想、化归与转化思想．满分12分．解：（Ⅰ）因为()e x xf x m=+，所以22e e e (1)()(e )(e )x x x x x m x x m f x m m +-⋅-+'==++， ············· 1分由1x =是()f x 的极值点，得2(1)0(e )mf m '==+， ············································· 2分 解得0m =， ····························································································· 3分此时()ex xf x =，经检验，1x =是()f x 的极值点．所以所求的实数m 的值为0． ······································································· 4分（Ⅱ）证明：取1m =-时，()e 1x xf x =-，此时2e (1)1()(e 1)xx x f x --'=-． ···················· 6分 构造函数()e (1)1x h x x =--， ········································································ 7分 所以()e (1)e (1)e x x x h x x x '=-+-=-在(0,)+∞上恒负，所以()h x 在(0,)+∞上单调递减， ··································································· 8分 所以()(0)0h x h <=, ···················································································· 9分故()0f x '<在(0,)+∞恒成立，说明()e 1x xf x =-在(0,)+∞上单调递减． ··············· 10分所以当01a b <<<时，e 1e 1b ab a<--，又因为e e 1b a >>，所以e 10,e 10b a ->->， 所以(e 1)(e 1)a bb a -<-， ··········································································· 11分所以e e a b b a a b +<+成立． ········································································ 12分22．本小题主要考查直线、椭圆等基础知识，考查推理论证能力、运算求解能力，考查函数与方程思想、数形结合思想、化归与转化思想、分类与整合思想．满分14分． 【解析】（Ⅰ）设椭圆Γ的半焦距为c ，则1,23,c e a AF a c ⎧==⎪⎨⎪=+=⎩·································································· 2分解得2,1a c ==， ······················································································· 3分所以2223b a c =-=， ················································································· 4分所以椭圆Γ的方程为22143x y+=． ································································· 5分 （Ⅱ）方法一：依题意得， PQ 与坐标轴不垂直．设()()1122,,,P x y Q x y ．因为点Q 与点Q '关于x 轴对称,所以()22,Q x y '-．由（Ⅰ）讨论可知，()()2,0,1,0A F -． 因为PF AQ '∥，所以直线FQ 与直线AQ '的斜率相等，故222212y y x x -=+-， ············· 7分 解得212x =． ···························································································· 8分 又因为点()22,Q x y 在椭圆Γ上，所以2y =，或2y = ························· 9分由椭圆对称性，不妨取2y =PQ的斜率221y k x ==+ 所以直线PQ方程为)1y x =+． ····························································· 10分由)221,3412,y x x y ⎧+⎪⎨⎪+=⎩得点P坐标为7,4⎛- ⎝⎭． ·············································· 11分 所以()()()()()2222222211111811111164PF x y x k x k x =++=+++=++=， ················ 12分 ()()()()()2222222222222812221216AQ x y x k x k x '=-+=-+-=+-=． ·················· 13分 所以12PF AQ '=． ·················································································· 14分方法二：依题意得， PQ 与坐标轴不垂直．设l 方程为()1y k x =+（0k ≠），()()1122,,,P x y Q x y ． 因为点Q 与点Q '关于x 轴对称,所以()22,Q x y '-．又因为椭圆关于x 轴对称，所以点Q '也在椭圆Γ上． ········································ 6分 由()221,3412,y k x x y ⎧=+⎨+=⎩消去y 得()22223484120k x k x k +++-=． 所以2212122284120,,3434k k x x x x k k -∆>+=-⋅=++． ················································· 7分 因为PF AQ '∥，所以直线AQ '的方程为()2y k x =-．由()222,3412,y k x x y ⎧=-⎨+=⎩消去y 得()2222341616120k x k x k +-+-=． 因为直线AQ '交椭圆于()()222,0,,A Q x y '-两点，所以2221612234k x k -⋅=+，故2228634k x k -=+． ························································· 9分所以22221211212222868864120,,34343434k k k k x x x x x x k k k k ---∆>+=+=-⋅=⋅=++++， 解得2157,44k x ==-．所以222861342k x k -==+． ·············································································· 11分 所以()()()()()2222222211111811111164PF x y x k x k x =++=+++=++=， ················ 12分()()()()()2222222222222812221216AQ x y x k x k x'=-+=-+-=+-=． ·················· 13分 所以12PF AQ '=．·················································································· 14分 方法三：依题意，得PQ 与坐标轴不垂直． 设l 方程为()1y k x =+（0k ≠），()()1122,,,P x y Q x y ． 因为点Q 与点Q '关于x 轴对称,所以()22,Q x y '-．又因为椭圆关于x 轴对称，所以点Q '也在椭圆Γ上． ········································ 6分 由()221,3412,y k x x y ⎧=+⎨+=⎩消去x 得()22234690k y ky k +--=． 所以12260,34ky y k∆>+=+． ········································································ 7分 因为PF AQ '∥，所以直线AQ '的方程为()2y k x =-．由()222,3412,y k x x y ⎧=-⎨+=⎩消去x 得，()2234120k y ky ++=． 因为直线AQ '交椭圆于()()222,0,,A Q x y '-两点，所以221234k y k --=+，即221234ky k =+． ····························································· 9分 设FP AQ λ'=（0λ>），则()()11221,2,x y x y λ+=--，所以1221234ky y k λλ-=-=+． · 11分所以()122212163434k k y y k k λ-+==++，解得12λ=， ············································ 13分 所以12FP AQ '=，即12PF AQ '=． ···························································· 14分方法四：依题意，得PQ 与坐标轴不垂直． 设l 方程为()1y k x =+（0k ≠），()()1122,,,P x y Q x y ．因为点Q 与点Q '关于x 轴对称,所以()22,Q x y '-． ············································· 6分 因为,,P F Q 三点共线，所以()111,FP x y =+与()221,FQ x y =+共线，所以()()1221110x y x y +-+=． ····································································· 7分 因为PF AQ '∥，所以可设FP AQ λ'=（0λ>），即()()11221,2,x y x y λ+=--，所以()121212,x x y y λλ+=-=-． ·································································· 8分 所以()()2222210x y x y λλ-++=，即()22210y x λ-=． ····································· 9分依题意，120y y ⋅≠，所以212x =． ······························································· 10分 因为点21,2Q y ⎛⎫⎪⎝⎭在椭圆22143x y +=上，所以2211163y +=，解得2y =2y = ···································································· 11分由椭圆对称性，不妨取2y =()()221,02,OP OF FP x y =+=-+--λ31,2⎛⎫=-- ⎪⎝⎭λ，因为点31,2P λ⎛⎫-- ⎪⎝⎭在椭圆22143x y +=上，所以22312143λ⎛⎫⎛⎫-- ⎪ ⎪⎝⎭⎝⎭+=，解得 12λ=或1λ=-（舍去）． ························ 13分。