(參考答案)福州市年高中畢業(yè)班綜合質(zhì)量檢測(cè)數(shù)學(xué)

1、2021 年 3 月福州市高中畢業(yè)班質(zhì)量檢測(cè)評(píng)分說(shuō)明：1本解答給出了一種或幾種解法供參考，如果考生的解法與本解答不同，可根據(jù)試題的主要考查內(nèi)容比照評(píng)分標(biāo)準(zhǔn)制定相應(yīng)的評(píng)分細(xì)那么。2對(duì)計(jì)算題，當(dāng)考生的解答在某一步出現(xiàn)錯(cuò)誤時(shí)，如果后繼局部的解答未改變?cè)擃}的內(nèi)容和難度，可視影響的程度決定后繼局部的給分，但不得超過(guò)該局部正確解容許給分?jǐn)?shù)的一半；如果后繼局部的解答有較嚴(yán)重的錯(cuò)誤，就不再給分。3解答右端所注分?jǐn)?shù)，表示考生正確做到這一步應(yīng)得的累加分?jǐn)?shù)。4只給整數(shù)分?jǐn)?shù)。一、單項(xiàng)選擇題：此題共 8 小題，每題5 分，共 40 分1C2B3B4A5D6C7D8C二、多項(xiàng)選擇題：此題共 4 小題，每題5 分，共 20

2、 分9AC10ABD11BCD12BCD三、填空題：本大題共 4 小題，每題5 分，共 20 分13-2，414515251612四、解答題：本大題共 6 小題，共 70 分17. 本小題總分值10分【命題意圖】本小題主要考查等比數(shù)列、 an 與 Sn 的關(guān)系、數(shù)列求和等根底知識(shí)；考查推理論證能力、運(yùn)算求解能力；考查化歸與轉(zhuǎn)化思想、函數(shù)與方程思想；考查邏輯推理、數(shù)學(xué)運(yùn)算等核心素養(yǎng)，表達(dá)根底性、綜合性總分值10 分【解答】1選，即 Sn=2an+1那么當(dāng)n =1時(shí)，S1 =2a1 +1，故a1 =-1；········

3、;···············································1 分當(dāng)n2時(shí)，Sn-1=2an-

4、1+1，兩式相減得an=2an-1，···············································

5、·············3 分所以an為等比數(shù)列，其中公比為2，首項(xiàng)為-1·································

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

7、83;······································5 分選，即a1=-1,log2(anan+1)=2n-1所以當(dāng)n2時(shí)，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* 為等比數(shù)列，其中首項(xiàng)為 a =-1 ，公比為 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 分綜上，a =-2n-1······································&#

14、183;········································5 分選，即a2 =aa，S =-3，a =-4n+1n n+223所以an為等比數(shù)列，設(shè)其公比為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?yàn)閍 為單調(diào)數(shù)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 分兩式相減得-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. 本小題總分值12分【命題意圖】本小題主要考查解三角形等根底知識(shí)；考查推理論證能力、運(yùn)算求解能力； 考查函數(shù)與方程思想、數(shù)形結(jié)合思想；考查直觀想象、邏輯推理、數(shù)學(xué)運(yùn)算等核心素養(yǎng)，表達(dá)根底性、綜合性總分值12 分【解答】解法一：1因?yàn)?a+b=ccosB-bcosC，由正弦定理得sinA+sinB=sinCcosB-sinBcosC，······

26、····························· 2分因?yàn)閟in(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?yàn)锽Î(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?yàn)镃D 是ABC 的角平分線，且C =2 ，3所以ÐACD=ÐBCD =························

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

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

34、3;········································11分兩邊同時(shí)除以CA×CB×CD得1 +1 =1 .···

35、83;········································12 分CACBCD解法二：1因?yàn)?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?yàn)镃D 是ABC 的角平分線，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. 本小題總分值12分【命題意圖】本小題主要考查空間直線與直線、直線與平面、平面與平面的位置關(guān)系等根底知識(shí)；考查推理論證能力、運(yùn)算求解能力與空間想象能力；考查數(shù)形結(jié)合思想；考查直觀想象、邏輯推理、數(shù)學(xué)運(yùn)算等核心素養(yǎng)，表達(dá)根底性、綜合性總分值12 分【解答】1依題意，四邊形 ACC1A1為等腰梯形，過(guò) A1，C1分別

50、引 AC的垂線，垂足分別為 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?yàn)锳1CAB，ABAA1=A，且AB，AA1Ì平面ABB1A1，所以A1C平面ABB1A1，······················

52、83;···············································4 分因?yàn)锳1CÌ平

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

54、183;·············5 分A1C = C2因?yàn)锳BAC，A1CAB，AC，且AC，A1CÌ平面ACC1A1，所以AB平面ACC1A1，結(jié)合1可知AB，AC，A1D三條直線兩兩垂直·····6 分以A為原點(diǎn)，分別以AB，AC，DA1的方向?yàn)閤，y，z軸的正方向，建立空間直角坐標(biāo)系A(chǔ) -xyz ，如下圖，那么各點(diǎn)坐標(biāo)為 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)為平面ABBA的法向量11ç，÷1133è22ø····················

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

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

59、ø設(shè) n2=(x，y，z)為平面 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設(shè)二面角 A -BB1 -C 的大小為q，那么sinq=20. 本小題總分值12 分= ··················12 分1 - ç - 4 ÷æ 1 ö2èø154【命題意圖】本小題主要考查直線與橢圓的位置關(guān)系等根

62、底知識(shí)；考查推理論證能力、運(yùn)算求解能力；考查函數(shù)與方程思想、數(shù)形結(jié)合思想、化歸與轉(zhuǎn)化思想；考查直觀想象、邏輯推理、數(shù)學(xué)運(yùn)算等核心素養(yǎng)，表達(dá)根底性、綜合性與創(chuàng)新性總分值12 分2【解答】解法一：1依題意，a=·································

63、;················1 分a2 + b23由橢圓的對(duì)稱性可知，四邊形A1B2A2B1為菱形，其周長(zhǎng)為4=4 ···· 3 分所以b =1，····················

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

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

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

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

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

69、83;····8 分èx0 - 2ø假設(shè)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 分所以當(dāng)Q的坐標(biāo)為(±2，0)時(shí)，QC×QD=3·····

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

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

77、83;············5 分2當(dāng)點(diǎn)P與點(diǎn)B1重合時(shí)，C點(diǎn)即B1(0，1)，而點(diǎn)D即B2(0，-1)，假設(shè)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、以下證明當(dāng)Q為(±2，0)時(shí)， QC×QD=3設(shè)P(x，y )，那么2y2 =2-x2，·········································

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

80、83;·····8 分æ2èøçx0 +÷由 AD PA知 AD的方程為 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 分說(shuō)明： Q只求出(2，0)或(-2，0)，不扣分21. 本小題總分值12分【命題意圖】本小題主要

86、考查古典概型、概率分布列、等差數(shù)列、導(dǎo)數(shù)等根底知識(shí)；考查數(shù)據(jù)處理能力、推理論證能力、運(yùn)算求解能力與創(chuàng)新意識(shí)；考查函數(shù)與方程思想、化歸與轉(zhuǎn)化思想、分類與整合思想、必然與或然思想；考查數(shù)學(xué)建模、邏輯推理、數(shù)學(xué)運(yùn)算等核心素養(yǎng)，表達(dá)綜合性、應(yīng)用性與創(chuàng)新性總分值12 分【解答】1設(shè)恰好有 3 個(gè)股東同時(shí)選擇同一款理財(cái)產(chǎn)品的事件為 A，由題意知，5個(gè)股東共有45 種選擇，而恰好有 3 個(gè)股東同時(shí)選擇同一款理財(cái)產(chǎn)品的可能情況為C3×(A2+A3)種，544C3×(A2+A3)45所以P(A)=54454 =128 ······&

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

88、#183;···4 分22021 年全年該公司從協(xié)定存款中所得的利息為：éë(550+500+450+50)+ 50ù´0.0 168+ 100 +û12=é550+50´11+50ù´0.0014=4.69萬(wàn)元······················&#

89、183;·····················6 分êë2úû由條件，高新工程投資可得收益頻率分布表投資收益 t3-x+ 0.02x2 + 0.135x 30 0000-0.27xP0.60.20.2············&

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

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

92、32;30000ø所以，存款利息和投資高新工程所得的總收益的期望為：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、分由條件可知，當(dāng)x=400時(shí)，L(x)取得最大值為：L(400)=662.69萬(wàn)元所以當(dāng)x=400時(shí)，該公司2021年存款利息和投資高新工程所得的總收益的期望取得最大值662.69萬(wàn)元··································

95、3;··············································12 分22. 本小題總分值12分【解答】解

96、法一：1依題意，f¢(x)=x(x+2)ex，那么································1 分當(dāng)xÎ(-¥,-2)(0，+¥)時(shí)，f¢(x)0；當(dāng)xÎ(-2，0)時(shí)，f¢(x)0；&#

97、183;···········2 分所以f(x)在區(qū)間(-¥,-2)，(0，+¥)上單調(diào)遞增，在區(qū)間(-2，0)上單調(diào)遞減····· 3 分因?yàn)?f(-2)=4 -10 ， f (1)= e -10 ，e2所以f(x)有且只有1個(gè)零點(diǎn)···············

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假設(shè)a 0 ，那么F ¢(x) 0 ， F (x)為增函數(shù)，eeF æ1ö=-1-aæ2ln1+1

100、ö=-1-aæ1-ln4ö0，不合題意；··············7 分ç2÷4ç22÷4ç2÷èøèøèø假設(shè)a0，令 h(x)=x2ex(x0)，易知 h(x)單調(diào)遞增，且值域?yàn)?0，+¥)，那么存在x0，使得x2ex0=a，即2lnx+x=lna····

101、;··············································8 分0000當(dāng) xÎ(0, x0)時(shí)， F¢(x) 0， F(x)單調(diào)遞減； 當(dāng) xÎ(x0，+¥)時(shí)， F¢(x)0， F(x)單調(diào)遞增F (x)=F(x)=x2ex0-a(2lnx+x)-1=a-alna-1， ·························