(参考答案)福州市年高中毕业班综合质量检测数学

1、2021 年 3 月福州市高中毕业班质量检测评分说明：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那么。2对计算题，当考生的解答在某一步出现错误时，如果后继局部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局部的给分，但不得超过该局部正确解容许给分数的一半；如果后继局部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。一、单项选择题：此题共 8 小题，每题5 分，共 40 分1C2B3B4A5D6C7D8C二、多项选择题：此题共 4 小题，每题5 分，共 20

2、 分9AC10ABD11BCD12BCD三、填空题：本大题共 4 小题，每题5 分，共 20 分13-2，414515251612四、解答题：本大题共 6 小题，共 70 分17. 本小题总分值10分【命题意图】本小题主要考查等比数列、 an 与 Sn 的关系、数列求和等根底知识；考查推理论证能力、运算求解能力；考查化归与转化思想、函数与方程思想；考查逻辑推理、数学运算等核心素养，表达根底性、综合性总分值10 分【解答】1选，即 Sn=2an+1那么当n =1时，S1 =2a1 +1，故a1 =-1；········

3、;···············································1 分当n2时，Sn-1=2an-

4、1+1，两式相减得an=2an-1，···············································

5、·············3 分所以an为等比数列，其中公比为2，首项为-1·································

6、;····4 分所以an =-2n-1···········································

7、83;······································5 分选，即a1=-1,log2(anan+1)=2n-1所以当n2时，log2(anan+1)-log2(an-1an)=2，·

8、;······································1 分即 an+1 = 4，·········

9、83;·················································

10、83;···························2 分an-1所以a2k-1k ÎN* 为等比数列，其中首项为 a =-1 ，公比为 4，12k -1所以a=-1´ 4k -1 =-2(2k-1)-1 ········

11、··················································

12、····3 分由a1=-1,log2(a1a2)=1，得a2=-2，n2k同理可得，a=-2´4k-1=-22k-1kÎN*··································

13、3;·········4 分综上，a =-2n-1······································&#

14、183;········································5 分选，即a2 =aa，S =-3，a =-4n+1n n+223所以an为等比数列，设其公比为q，·

15、;··················································

16、;·1 分ìïa(1+q)=-3，ìa =-1，ìa1 =-9 ，那么í1ïa q2 =-4 ，解得1q = 2，ïíq =-2 .·······························

17、83;···········3 分î1îïî3í或又因为a 为单调数列，所以 q 0 ，故ìa1 =-1，·····4 分îníq = 2 ，··················&

18、#183;·············n所以a =-2n-1··································&#

19、183;···············································5 分n2由1知， -na

20、=n × 2n-1 ，所以Tn=1+2´2+3´22+(n-1)×2n-2+n×2n-1， ···································· 6分2Tn =2+2&#

21、180;22+(n-2)×2n-2+(n-1)×2n-1+n×2n， ······················7 分两式相减得-Tn=1+2+22+2n-2+2n-1-n×2n·············&#

22、183;··························8 分=(2n-1)-n×2n····················

23、········································9 分所以Tn=(n-1)×2n+1 ······

24、··················································

25、···············10 分18. 本小题总分值12分【命题意图】本小题主要考查解三角形等根底知识；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；考查直观想象、逻辑推理、数学运算等核心素养，表达根底性、综合性总分值12 分【解答】解法一：1因为 a+b=ccosB-bcosC，由正弦定理得sinA+sinB=sinCcosB-sinBcosC，······

26、····························· 2分因为sin(B+C)=sin(-A)=sinA，所以sin(B+C)+sinB= sinCcosB-sinBcosC，··········

27、3;······························3 分所以2sinBcosC+sinB=0，················

28、··················································

29、 4分因为BÎ(0,)，所以sinB ¹0，所以cosC=-1，······································5 分2又CÎ(0,)，所以C=2.·&

30、#183;·················································&

31、#183;················ 6分32因为CD 是ABC 的角平分线，且C =2 ，3所以ÐACD=ÐBCD =························

32、83;···················7 分3在ABC 中， S ABC =S ACD +SBCD ，那么由面积公式得1CA×CBsin2=1CA×CDsin+1CD×CBsin， ················

33、···················· 10分232323即CA×CB=CA×CD+CD×CB.······················

34、3;········································11分两边同时除以CA×CB×CD得1 +1 =1 .···

35、83;········································12 分CACBCD解法二：1因为 a+b=ccosB-bcosC，a2 +c2-b2a2 +b2 -c2由余弦定

36、理得a+b=c×-b×，··································· 2分2ac2ab整理得2a(a+b)=2c2 -2b2，即a2 +b2 -c2 +ab=0，···

37、83;····························· 3分所以ab(1+2cosC)=0，·················&#

38、183;·················································&#

39、183;···4 分所以cosC=-1，············································&

40、#183;···································5 分2又CÎ(0,)，所以C=2.··········&#

41、183;·················································&#

42、183;······· 6分32因为CD 是ABC 的角平分线，且C =2 ，3所以ÐACD=ÐBCD =·································

43、3;········7 分3在ABC 中，由正弦定理得CAsinB=CB =sin AABsin 23， ··································

44、83;···········8 分即 CAsinB=CB =sin AADsin + DBsin ································

45、3;·····························9 分33同理在CAD 和CBD 中，得CD =sin AADsin 3， CDsinB=DB ，sin 3所以 CAsin B=CD +sin ACDsinB，即 CA-CD=sin BCDsinA， ···

46、3;···································10 分故CA-CD=CD，即1=CD+CD，···········

47、;··········································11分CACBCBCA故 1 +1 =1 ····&#

48、183;·················································&#

49、183;·····················12 分CACBCD19. 本小题总分值12分【命题意图】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等根底知识；考查推理论证能力、运算求解能力与空间想象能力；考查数形结合思想；考查直观想象、逻辑推理、数学运算等核心素养，表达根底性、综合性总分值12 分【解答】1依题意，四边形 ACC1A1为等腰梯形，过 A1，C1分别

50、引 AC的垂线，垂足分别为 D，E，那么AD =1 (AC -A C )=1 ´(2 - 1)=1 =1 AA ，故ÐA AC = 60°21 122211在ACA 中， A C2 =A A2 +AC2 - 2A A ×AC cos ÐA AC = 12 + 22 - 2 ´1´ 2 ´1 = 3 ，111112所以AC2+AA2 =AC2，故ÐAAC=90°，即ACAA··········

51、83;·················2 分11111因为A1CAB，ABAA1=A，且AB，AA1Ì平面ABB1A1，所以A1C平面ABB1A1，······················

52、83;···············································4 分因为A1CÌ平

53、面ACC1A1 ，所以平面ACC1A1 平面ABB1A1·············································&#

54、183;·············5 分A1C = C2因为ABAC，A1CAB，AC，且AC，A1CÌ平面ACC1A1，所以AB平面ACC1A1，结合1可知AB，AC，A1D三条直线两两垂直·····6 分以A为原点，分别以AB，AC，DA1的方向为x，y，z轴的正方向，建立空间直角坐标系A -xyz ，如下图，那么各点坐标为 3A(0，0，0)，B(1，0，0)，C(0，2，0)，æ1

55、6;，A1 ç0，÷è22 ø3æ3ö ·········································

56、3;··········7 分C1ç0，÷è22 ø 3ö由1知，n=2AC=2æ03-=(0，3，-1)为平面ABBA的法向量11ç，÷1133è22ø····················

57、83;·················································

58、83;·······································8 分æ13öBC=(-1，2，0)，C1C=çç0，-÷，è22

59、ø设 n2=(x，y，z)为平面 BCC1B1的法向量，那么ìïnBC，ìn2 ×BC =-x + 2 y = 0 ，故2ïíí1取 n2=(23，3，1)， ····················10 分ïîn2C1C，ïn2 ×C1C=y- 3î22z = 0 ，

60、所以cosn1，n2=n1 ×n2=3 -1 =1， ············································

61、3;········11分2´44n1 n2设二面角 A -BB1 -C 的大小为q，那么sinq=20. 本小题总分值12 分= ··················12 分1 - ç - 4 ÷æ 1 ö2èø154【命题意图】本小题主要考查直线与椭圆的位置关系等根

62、底知识；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；考查直观想象、逻辑推理、数学运算等核心素养，表达根底性、综合性与创新性总分值12 分2【解答】解法一：1依题意，a=·································

63、;················1 分a2 + b23由椭圆的对称性可知，四边形A1B2A2B1为菱形，其周长为4=4 ···· 3 分所以b =1，····················

64、3;·················································

65、3;·················4 分x22所以E的方程为+y2=1 ·····························

66、83;··································· 5分2设P(x，y )，那么2y2 =2-x2，··········&#

67、183;············································6 分00直线 A1P 的方程为 y =00x0 + 2y0(x

68、+2)，故Cç0，2y0ö÷， ························7 分æèx0 + 2øy0æ2 y0 öx0 - 2由A1DPA2知A1D的方程为y=(x+2)，故Dç0，÷，·····

69、83;····8 分èx0 - 2ø假设存在Q(t，0)，使得QC×QD=3，那么QC ×QDæt ， 2 y0öæt ， 2 y0 ö=ç-èx0 +÷×ç-øèx0 -÷22ø2 y2=t2+ 00x2 -22 -x2=t2+ 00x2 -2············

70、··················································

71、··············9 分= t2 -1= 3 ·································

72、3;·················································

73、3;10 分解得t = ±2 ··············································

74、3;······································11 分所以当Q的坐标为(±2，0)时，QC×QD=3·····

75、83;······································12 分解法二1同解法一·········

76、83;·················································

77、83;············5 分2当点P与点B1重合时，C点即B1(0，1)，而点D即B2(0，-1)，假设存在Q(t，0)，使得QC×QD=3，那么(-t，1)×(-t，-1)=3，即t2-1=3，解得t=±2···················6 分

78、以下证明当Q为(±2，0)时， QC×QD=3设P(x，y )，那么2y2 =2-x2，·········································

79、······················7 分0000x0 + 2直线 AP的方程为 y=1y0(x+2 )，故C 0 ，2y0ö，···················

80、83;·····8 分æ2èøçx0 +÷由 AD PA知 AD的方程为 y=y0(x+2 )，故 D 0 ，2y0öæ2ø， ··········9 分12所以QC×QD1æt ， 2 y0x0 - 222èøöæt ， 2 y0 öçx0 -÷=ç-èx0

81、 +2 y2÷×ç-øèx0 -÷=t2+ 00x2 -22 -x2·······································&#

82、183;································10 分=4+ 0 ···············

83、3;·················································

84、3;·······11分0x2 - 2= 4 -1=3·······································

85、3;·········································12 分说明： Q只求出(2，0)或(-2，0)，不扣分21. 本小题总分值12分【命题意图】本小题主要

86、考查古典概型、概率分布列、等差数列、导数等根底知识；考查数据处理能力、推理论证能力、运算求解能力与创新意识；考查函数与方程思想、化归与转化思想、分类与整合思想、必然与或然思想；考查数学建模、逻辑推理、数学运算等核心素养，表达综合性、应用性与创新性总分值12 分【解答】1设恰好有 3 个股东同时选择同一款理财产品的事件为 A，由题意知，5个股东共有45 种选择，而恰好有 3 个股东同时选择同一款理财产品的可能情况为C3×(A2+A3)种，544C3×(A2+A3)45所以P(A)=54454 =128 ······&

87、#183;·················································&

88、#183;···4 分22021 年全年该公司从协定存款中所得的利息为：éë(550+500+450+50)+ 50ù´0.0 168+ 100 +û12=é550+50´11+50ù´0.0014=4.69万元······················&#

89、183;·····················6 分êë2úû由条件，高新工程投资可得收益频率分布表投资收益 t3-x+ 0.02x2 + 0.135x 30 0000-0.27xP0.60.20.2············&

90、#183;·················································&

91、#183;·········································7 分所以，高新工程投资所得收益的期望为：æx32ö32+ 0.027x

92、32;30000ø所以，存款利息和投资高新工程所得的总收益的期望为：L (x)=-0.000 02x3 +0.012x2 + 0.027x + 0.036 ´(500 -x)+ 0.018 ´6 x + 4.6912=-0.00002x3+0.012x2+22.690x500 ························&#

93、183;··············9 分L¢(x)=-0.00006(x2-400x)令 L¢(x)= 0 ，得 x = 400 ，或 x = 0 由L¢(x)0，得0x400；由L¢(x)0，得400x500·················11

94、分由条件可知，当x=400时，L(x)取得最大值为：L(400)=662.69万元所以当x=400时，该公司2021年存款利息和投资高新工程所得的总收益的期望取得最大值662.69万元··································

95、3;··············································12 分22. 本小题总分值12分【解答】解

96、法一：1依题意，f¢(x)=x(x+2)ex，那么································1 分当xÎ(-¥,-2)(0，+¥)时，f¢(x)0；当xÎ(-2，0)时，f¢(x)0；&#

97、183;···········2 分所以f(x)在区间(-¥,-2)，(0，+¥)上单调递增，在区间(-2，0)上单调递减····· 3 分因为 f(-2)=4 -10 ， f (1)= e -10 ，e2所以f(x)有且只有1个零点···············

98、83;···············································5 分2令 F (x)=x2e

99、x -a (2 ln x +x)-1 ，那么¢( ) = ( +x -=)a(x+2)(x+2)(x2ex-a)Fxxx2ex0 ··························6 分xx假设a 0 ，那么F ¢(x) 0 ， F (x)为增函数，eeF æ1ö=-1-aæ2ln1+1

100、ö=-1-aæ1-ln4ö0，不合题意；··············7 分ç2÷4ç22÷4ç2÷èøèøèø假设a0，令 h(x)=x2ex(x0)，易知 h(x)单调递增，且值域为(0，+¥)，那么存在x0，使得x2ex0=a，即2lnx+x=lna····

101、;··············································8 分0000当 xÎ(0, x0)时， F¢(x) 0， F(x)单调递减； 当 xÎ(x0，+¥)时， F¢(x)0， F(x)单调递增F (x)=F(x)=x2ex0-a(2lnx+x)-1=a-alna-1， ·························