福州市2020届高中毕业班期末质量检测参考答案(文科数学)

2019—2020学年度第一学期福州市高三期末质量检测数学（文科）参考答案及评分细则评分说明：1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制定相应的评分细则。

2．对计算题，当考生的解答在某一步出现错误时，如果后继部分的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。

3．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。

4．只给整数分数。

选择题和填空题不给中间分。

一、选择题：本大题考查基础知识和基本运算．每小题5分，满分60分．1．B 2．D 3．C 4．D 5．C 6．A 7．D8．C9．A10．C11．B12．B二、填空题：本大题考查基础知识和基本运算．每小题5分，满分20分．13．3214．5- 1516．①③④三、解答题：本大题共6小题，共70分．解答应写出文字说明，证明过程或演算步骤． 17．本题主要考查用样本估计总体、平均数、标准差、古典概型等基础知识，考查考生运用概率与统计知识解决实际问题的能力、数据处理能力，考查的核心素养主要有数学建模、数学抽象、数学运算、数据分析．满分12分． 【解答】（1）参与该活动的网友单次游戏得分的平均值 ()1351045405560654075308520200x =⨯⨯+⨯+⨯+⨯+⨯+⨯ ························ 2分60=． ···························································································· 3分标准差s =················· 5分13.60=≈．······································································· 6分 （2）用分层抽样抽取7人，其中得分在[30,40)的有1人，得分在[60,70)的有4人，得分在[80,90]的有2人， ······································································· 7分 分别记为a ,1234,,,b b b b ,12,c c ，7人中任选2人，有21种结果， ······················ 8分 其中2人得分在同一组的有7种，分别是{}12,b b 、{}13,b b 、{}14,b b 、{}23,b b 、{}24,b b 、{}34,b b 、{}12,c c ， ············································································· 10分 故2人得分不在同一组内的概率721213P =-=.·········································· 12分 18．本小题考查数列的通项与前n 项和的关系、等比数列、裂项相消法求数列的和等基础知识，考查运算求解能力，考查函数与方程思想、化归和转化思想等，考查的核心素养主要有逻辑推理、数学运算等．满分12分．【解答】（1）当1n =时,111a S p ==+， ···················································· 1分 当2n ≥时，221[(1)(1)]21n n n a S S n pn n p n n p -=-=+--+-=-+， ·············· 3分 经检验，1n =时也满足上式，所以21n a n p =-+． ···································· 4分 因为248,,a a a 成等比数列，所以2284a a a =， ··················································································· 5分即()()()23157p p p ++=+，解得1p =． ················································· 6分 所以2.n a n = ························································································ 7分 （2）由（1）及题设得，14111(1)1n n n b a a n n n n +===-⋅++， ·························· 9分所以12n n T b b b =++⋅⋅⋅+11111(1)()()2231n n =-+-+⋅⋅⋅+-+ ················································· 11分 111n =-+ =1nn +． ················································································ 12分 19．本小题主要考查空间直线与直线、直线与平面、平面与平面的位置关系、空间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，考查化归与转化思想等，考查的数学素养主要有逻辑推理、直观想象等．满分12分． 【解答】解法一：（1）平面AEF 与平面PBC 互相垂直， ····························· 1分 理由如下：因为PA ⊥底面ABCD ，BC ⊂平面ABCD ，所以PA BC ⊥． ············································· 2分因为ABCD 为正方形，所以AB BC ⊥又PA AB A =，且,PA AB ⊂平面PAB ，所以BC ⊥平面PAB ．因为AE ⊂平面PAB ，所以AE BC ⊥． ····················································· 3分 因为PA AB =，E 为线段PB 的中点，所以AE PB ⊥， 又PBBC B =，且,PB BC ⊂平面PBC ，所以AE ⊥平面PBC ， ·········································································· 4分 因为AE ⊂平面AEF ，所以平面AEF ⊥平面PBC ． ··································· 6分 （2）因为PA ⊥底面ABCD ，E 为线段PB 的中点，所以点E 到底面ABCD 的距离为12PA =32， ············································· 7分则133393P ABCD V -=⨯⨯⨯=， ··································································· 8分PABCDEF又F 为线段BC 的三等分点，当113BF BC ==时，1133313224E ABF V -⎛⎫=⨯⨯⨯⨯= ⎪⎝⎭，所以多面体PAEFCD 的体积为333944P ABCD E ABF V V ---=-=； ······················· 10分 当223BF BC ==时，1133323222E ABF V -⎛⎫=⨯⨯⨯⨯= ⎪⎝⎭，所以多面体PAEFCD 的体积为315922P ABCD E ABF V V ---=-=．························ 11分 综上，多面体PAEFCD 的体积为334或152. ·············································· 12分 解法二：（1）平面AEF 与平面PBC 互相垂直， ········································· 1分 理由如下：因为PA ⊥底面ABCD ，PA ⊂平面PAB ，所以平面PAB ⊥底面ABCD ， ········ 2分 又平面PAB 底面ABCD AB =，BC AB ⊥，BC ⊂平面ABCD ，所以BC ⊥平面PAB ． ·········································································· 3分 因为AE ⊂平面PAB ，所以AE BC ⊥， ····················································· 3分 因为PA AB =，E 为线段PB 的中点，所以AE PB ⊥， 又PBBC B =，且,PB BC ⊂平面PBC ，所以AE ⊥平面PBC ， ·········································································· 4分 因为AE ⊂平面AEF ，所以平面AEF ⊥平面PBC . ···································································· 6分 （2）同解法一． ················································································ 12分 20．本小题考查圆与椭圆的标准方程及其几何性质、直线与圆的位置关系、直线椭圆的位置关系等基础知识，考查推理论证能力、运算求解能力，考查数形结合思想、化归与转化的思想、分类与整合思想等，考查的数学素养主要有逻辑推理、直观想象、数学运算等．满分12分．【解答】解法一：（1）依题意，圆O··································1分因为椭圆的短轴长等于圆O所以2b=b ·················································2分因为C，所以ca=① ·············································3分又因为222a c b-=，所以222a c-=，②联立①②，解得24a=， ······································································4分所以C的方程为22142x y+=. ···································································5分（2）证明：①当直线l斜率不存在时, 直线l的方程为x=x=当x=A B，则4433OA OB⋅=-=，故OA OB⊥．同理可证，当x=OA OB⊥． ···················································6分②当直线l斜率存在时，设其方程为1122,(,),(,),y kx m A x y B x y=+因为直线l=，即223440m k--=, ·····················7分由22,142y kx mx y=+⎧⎪⎨+=⎪⎩得，222(12)4240k x kmx m+++-=,······································8分所以()()()2222221681228420k m k m k m∆=-+-=-+＞，且12221224,1224,12kmx xkmx xk⎧+=-⎪⎪+⎨-⎪=⎪+⎩ ········································································································9分所以1212()()OA OB x x kx m kx m ⋅=+++221212(1)()k x x km x x m =++++ ·············································· 10分2222222(1)(24)4(12)12k m k m m k k +--++=+22234412k m k --=+0=， ·············································································· 11分 所以.OA OB ⊥综上，.OA OB ⊥ ················································································ 12分 解法二：（1）同解法一． ······································································· 5分 （2）①当直线方程为y =, (A B ，则 44033OA OB ⋅=-+=，故.OA OB ⊥同理可证，当直线方程为y =.OA OB ⊥ ······································ 6分 ②当直线l 不与x 轴平行时,设其方程为1122,(,),(,),x ty m A x y B x y =+ 因为直线l=，即223440.m t --= ······················ 7分由22,142x ty m x y =+⎧⎪⎨+=⎪⎩得，222(2)240.t y tmy m +++-= ········································· 8分所以()()()22222244248240t m t m t m ∆=-+-=-+＞，且12221222,24.2tm y y t m y y t ⎧+=-⎪⎪+⎨-⎪=⎪+⎩···· 9分 1212()()OA OB y y ty m ty m ⋅=+++221212(1)()t y y tm y y m =++++ ····················································· 10分2222222(1)(4)2(2)2t m t m m t t +--++=+2223442m t t --=+ 0=， ···················································································· 11分 所以，OA OB ⊥．综上，.OA OB ⊥ ················································································ 12分 21．本小题主要考查导数及其应用、函数的零点、函数的最值与值域等基础知识，考查推理论证能力、运算求解能力、抽象概括能力等，考查函数与方程思想、化归与转化思想、分类与整合思想、数形结合思想等，考查的数学素养主要有逻辑推理、直观想象、数学运算等．满分12分．【解答】（1）()()x x x a x f sin cos /-=， ······················································ 1分 因为⎥⎦⎤⎢⎣⎡∈6,0πx ，所以cos sin 0x x ＞≥，又10x ＞≥，所以1cos sin x x x ⋅＞，即0sin cos >-x x x ． ··············································· 2分当0>a 时，()0/>x f ，所以()f x 在区间π0,6⎡⎤⎢⎥⎣⎦上递增，所以()16312366max -=-⋅⋅=⎪⎭⎫⎝⎛=πππa f x f ，解得2=a ． ···························· 3分 当0<a 时，()0/<x f ，所以()f x 在区间π0,6⎡⎤⎢⎥⎣⎦上递减，所以()()10max -==f x f ，不合题意． ······················································· 4分 当0a =，()1f x =-，不合题意．综上，2=a . ······················································································· 5分 （2）设()x x x x g sin cos -=，则()⎪⎭⎫ ⎝⎛<<<--=200cos sin 2/πx x x x x g ， ··················································· 6分所以()x g 在⎪⎭⎫ ⎝⎛2,0π上单调递减，又()022,010<-=⎪⎭⎫⎝⎛>=ππg g ， ······················ 7分所以存在唯一的⎪⎭⎫⎝⎛∈2,00πx ，使得()00=x g ， ············································ 8分当00x x <<时，()0>x g ，即()()02/>=x g x f ，所以()()0,0x x f 在上单调递增；当20π<<x x 时，()0<x g ，即()()02/<=x g x f ，所以()()0,0x x f 在上单调递减， ·· 9分又()012,01424,010<-=⎪⎭⎫⎝⎛>-=⎪⎭⎫ ⎝⎛<-=πππf f f ， ···································· 10分 所以()f x 在π0,4⎛⎫ ⎪⎝⎭与ππ,42⎛⎫⎪⎝⎭上各有一个零点， ········································ 11分综上，函数()x f 在区间⎪⎭⎫⎝⎛2,0π上有且仅有两个零点. ··································· 12分（二）选考题：共10分．请考生在第22，23两题中任选一题作答．如果多做，则按所做第一个题目计分，作答时请用2B 铅笔在答题卡上将所选题号后的方框涂黑． 22．本题考查极坐标方程和直角坐标方程的互化，直线和圆的位置关系，以及直线的参数方程的参数的几何意义等基础知识,考查学生的逻辑推理能力,化归与转化能力．考查的核心素养是直观想象、逻辑推理与数学运算.【解答】（1）由2cos ρθ=，得22cos ρρθ=． ········································· 2分 将cos ,sin x y ρθρθ=⎧⎨=⎩代入得，222x y x +=， ······················································ 4分所以C 的直角坐标方程为22(1)1x y -+=． ················································ 5分（2）设,A B 所对应的参数分别为12,t t ，因为直线l的参数方程为5,(12x t y t ⎧=⎪⎪⎨⎪=+⎪⎩为参数 ，所以M 在l 上, ········· 6分 把l 的参数方程代入22(1)1x y -+=可得2180,t ++= ····························· 7分所以241830∆=-⨯=>， ······························································ 8分所以1212180t t t t +=-=＞， ······························································ 9分 故11MA MB +=12121212||||||||||||||||||||t t t t MA MB MA MB t t t t +++==⋅. ···························· 10分 23．本题考查含有绝对值的函数的最值，基本不等式的应用等基础知识,考查学生的逻辑推理能力,化归与转化能力.考查的核心素养是直观想象、逻辑推理与数学运算. 【解答】（1）根据题意,函数113,,122()|21|312,,22x x f x x x x x ⎧-⎪⎪=-++=⎨⎪-+<⎪⎩≥ ·················· 2分所以()f x 为在1,2⎛⎤-∞ ⎥⎝⎦单调递减，在1,2⎡⎫+∞⎪⎢⎣⎭单调递增， ······························ 3分所以min 1()1, 1.2f x f m ⎛⎫=== ⎪⎝⎭即 ······························································· 5分（2）由(1)知，1m =，所以1,a b c ++= ···················································· 6分 又因为,,a b c 为正实数，222a b ab +≥，222b c bc +≥，222a c ac +≥， ·········································· 8分所以()()22222a b c ab bc ac ++++≥，即222a b c ab bc ac ++++≥， ·············· 9分所以22221()222a b c a b c ab bc ca =++=+++++2223()a b c ++≤，即22213a b c ++≥． ··········································································· 10分。