2022屆青島一模參考答案

1、數(shù)學(xué)高三練習(xí)參考答案及評(píng)分標(biāo)準(zhǔn)一、單項(xiàng)選擇題：本題共 8 小題，每小題 5 分，共 40 分。B B C A C D A D二、多項(xiàng)選擇題：本題共 4 小題，每小題 5 分，共 20 分。9BC 10AD 11ABC 12ACD三、填空題：本題共 4 個(gè)小題，每小題 5 分，共 20 分。 10 23 213 -80； 14 ； 15 ； 16(1) (4, 4) ；(2) 44 10 12四、解答題：本題共 6 小題，共 70 分。解答應(yīng)寫(xiě)出文字說(shuō)明，證明過(guò)程或演算步驟。17. （本小題滿分10分）解：（1）由題意知a1 = 2,a2 = 4,a3 = 8 ···

2、;··················································

3、;········ 2分所以等比數(shù)列a 的公比 q = 2，na = a1q - = 2 ····································

4、;·········· 3分n 1 nn設(shè)等差數(shù)列b 公差為 d ，則n7(b +b )2 = b - 2b = 2d -b ， S = 1 7 = 7b = 7a3 1 1 7 4 32所以b4 = 8 = b1 + 3d ，所以b1 = 2,d = 2 ，b = 2n ···················

5、;·························· 6分n（2） cn = lg(2n)T = c + c +L+ c100 1 2 100= lg 2+lg 4+Llg 8+lg10+L+lg 98+lg100+L+lg 200 = 4´0 + 45´1+ 51´2 =147 ··&

6、#183;·················································&

7、#183;·············10分18（本小題滿分12分）解：（1）因?yàn)?sin B -sinC) = sin A-sin BsinC2 2所以sin B +sin C -sin A = sin BsinC2 2 2所以由正弦定理得b + c - a = bc ················

8、3;·················································2分2

9、2 2所以由余弦定理得cos因?yàn)?AÎ(0, )Ab2 + c2 - a2 bc 1= = = ·········································&

10、#183;···· 4分 2bc 2bc 2所以A = ··········································

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

12、······ 6分3（2）由三角形面積公式得1 1 10 7 5 7SD = ah = ´ a = a ····························· 7分ABC2 2 7 7數(shù)學(xué)答案 第 1頁(yè)（共 6頁(yè)）1 1 5 3S bc

13、 sin A 5c sin cD = = ´ ´ = ············································

14、;··········· 8分ABC2 2 3 4所以5 7 5 3a = c ，即7 421a = c ································

15、·································9分4由余弦定理得 a2 = 25+ c2 -5c ············

16、3;·················································

17、3;······· 11分21將 a = c 代入上式得c2 +16c -80 = 0 ，解得 c = 4或 -20（舍）4所以邊 c = 4································&#

18、183;·················································&#

19、183;············ 12分19（本小題滿分12分）解：（1）證明：在圖中DC1因?yàn)?DC / AB,CD = AB ， E 為 AB 中點(diǎn)2所以 DC / AE,DC = AEA E B所以 ADCE 為平行四邊形，所以 AD = CE = CD = AE = 2 ················

20、3;················ 1分同理可證 DE = 2在圖中，取 DE 中點(diǎn)O ，連接OA,OC ，OA = OC = 3 ·························

21、83;·········2分因?yàn)?AD = AE = CE = CD所以 DE OA,DE OC ，··································&

22、#183;········································ 4分因?yàn)镺AIOC = O ，所以 DE 平面 AOC因?yàn)?AC Ì 平面 AOC ，所以 DE

23、AC ·················································

24、83;··········5分 （2）若選擇：z因?yàn)?DE 平面 AOC ， DE Ì 平面 BCDEA所以平面 AOC 平面 BCDE 且交線為OC所以過(guò)點(diǎn) A 作 AH OC ，則 AH 平面 BCDE ·······················

25、83;··················6分因?yàn)?2 3S =BCDE所以四棱錐 A- BCDE 的體積DC1V - = 2 = ´2 3× AH ·····················

26、;············ 7分A BCDE3OxE B所以 AH = 3 = OAy所以 AO 與 AH 重合，所以 AO 平面 BCDE ····························&

27、#183;······················ 8分建系如圖，則O(0, 0, 0),C(- 3,0,0),E(0,1, 0), A(0, 0, 3)平面 DAE 法向量為CO = ( 3，0, 0) ··············&

28、#183;·················································&

29、#183;· 9分設(shè)平面 AEC 法向量為 n = (x, y, z)因?yàn)镃E = ( 3,1, 0),CA = ( 3, 0，3)數(shù)學(xué)答案 第 2頁(yè)（共 6頁(yè)）所以ì + =3x y 0ïíï 3x + 3z = 0î得 n = (1,- 3,-1) ························

30、;···································11分設(shè)二面角 D - AE -C 的大小為q ，則cosq CO×n 3 5= uuur r = =| CO | ×| n | 3´ 5 5所以二

31、面角 D - AE -C 的余弦值為若選擇：因?yàn)?DC / EB55············································&#

32、183;··············12分所以 ÐACD 即為異面直線 AC 與 EB 所成角·····························

33、3;·························6分在 DADC 中，cosÐACD=AC2+4 -4AC4=64所以 AC = 6 所以O(shè)A2 + OC = AC ，所以O(shè)A OC ············

34、83;·························7分因?yàn)?DE 平面 AOC ， DE Ì 平面 BCDE2 2所以平面 AOC 平面 BCDE 且交線為OC所以 AO 平面 BCDE ············&

35、#183;·················································&

36、#183;···················8分建系如圖，則O(0, 0, 0),C(- 3,0,0),E(0,1, 0), A(0, 0, 3)平面 DAE 法向量為CO = ( 3，0, 0) ·················&#

37、183;················································ 9分設(shè)平面 A

38、EC 法向量為 n = (x, y, z)因?yàn)镃E = ( 3,1, 0),CA = ( 3, 0，3)所以ì + =3x y 0ïíï 3x + 3z = 0î得 n = (1,- 3,-1) ·····························

39、83;·····························11分設(shè)二面角 D - AE -C 的大小為q ，則cosq CO×n 3 5= uuur r = =| CO | ×| n | 3´ 5 5所以二面角 D - AE -C 的余弦值為55··&#

40、183;·················································&#

41、183;······12分20（本小題滿分12分）1解:（1）設(shè)W (x, y) ，則 F(- , y) ····································&

42、#183;····························· 1分4 uuuur uuur1 3所以 ················

43、3;···············································2分OW = (x, y), E

44、F = (- - , y)4 4因?yàn)镺W × EF = 0所以 (x, y)×(-1, y) = -x + y = 0 ······································&#

45、183;······························ 4分2所以曲線C 的方程為y = x ················

46、··················································

47、··········5分2數(shù)學(xué)答案 第 3頁(yè)（共 6頁(yè)）（2）設(shè) A1(x1, y1), B1(x2 , y2 ), A2 (x3, y3 ), B2 (x4 , y4 ) ，1 y 1直線 A1B1, A2 B2 的方程分別為： x my , x 4 m 4= + = - + ··················&#

48、183;················· 6分1 2 1將 x = my + 代入拋物線 y2 = x 得 y - my - = ·························

49、;·················7分04 4 1所以 y + y = m y y = - ····························

50、················································8分,1 2 1 24所

51、以| A B |= 1+ m | y - y | = 1+ m (y + y ) -4y y = m2 +1··················· 9分2 2 21 1 1 2 1 2 1 21因?yàn)?= + ·················

52、;··································10分A B A B ，同理得：| A B | 11 1 2 22 2 2m1 1 1所以 | | | | (1 )(1 )A A B B 的面積 S = A B × A B =

53、+ m +2 1 2 1 21 1 2 2 22 2 m1 1= + 2 + ³ （當(dāng)且僅當(dāng) m = ±1時(shí)等號(hào)成立） 2 m(2 m ) 22所以四邊形 A A B B 面積的最小值為 2 ································&#

54、183;··························12分1 2 1 221（本小題滿分12分）解：（1）由題知， X 的取值可能為1, 2,3···············

55、3;··········································· 1分所以1 1P(X =1) = ( ) = ；2C 4121 1 1P(X

56、= 2) = 1- ( ) ( ) = ；2 2C C 12 1 12 31 1 2P(X = 3) = 1- ( ) 1- ( ) = ；2 2C C 3 1 12 3所以 X 的分布列為：X 1 2 31 1 2P4 12 3 ······························

57、3;················································4分所以數(shù)學(xué)期望為1

58、 1 2 3 +2 +24 29E(X ) 1 2 3= ´ + ´ + ´ = = ···························· 5分4 12 3 12 12（2）令xi1= ，則 y = bx + a$ ，由題知：ti5å ···&#

59、183;························ 6分 x y = 315, y = 90i ii=15åx y -5x× yi i315-5´0.46´90 108$ ············

60、;···························· 7分 i= = 270所以 1b = = =5 21.46-5´0.212 0.4å 2x -5xi i=1數(shù)學(xué)答案 第 4頁(yè)（共 6頁(yè)）所以 a$ = 90- 270´0.46 = -34.2 ， $y = 270x -34.2

61、 ··········································· 8分故所求的回歸方程為： $y 270 34.2= - ··

62、··················································

63、··········9分t所以，估計(jì)t = 6時(shí)， y »11；估計(jì)t = 7 時(shí)， y » 4 ；估計(jì)t ³ 8 時(shí)， y < 0；預(yù)測(cè)成功的總?cè)藬?shù)為 450 +11+ 4 = 465 ·······················

64、····································10分（3）由題知，在前 n 輪就成功的概率為P1 1 1 1 1 1 1 1 1 1= + (1- ) + (1- )(1- ) + + (1- )(1- ) (1-

65、 )L L ····11分2 2 3 2 3 4 2 3 n (n +1)2 2 2 2 2 2 2 2 2 2又因?yàn)樵谇?n 輪沒(méi)有成功的概率為1 1 11- P = (1- )´ (1- )´L´1- 2 3 (n +1)2 2 2故1 1 1 1 1 1 1 1= (1- )(1+ )´(1- )´(1+ )L´(1- )(1+ )(1- )(1+ )2 2 3 3 n n n +1 n +1 1 3 2 4 n -1 n +1 n n + 2 n + 2 1= ( )( )´

66、( )´ ( )L´ ( )( )( )( ) = >2 2 3 3 n n n +1 n +1 2n + 2 2 1 1 1 1 1 1 1 1 1 1 1 + (1- ) + (1- )(1- ) +L+ (1- )(1- )(1- ) < ···· 12分2 2 3 2 3 4 2 3 n (n +1) 22 2 2 2 2 2 2 2 2 222（本小題滿分12分）解：（1）由題意知 ( ) cos sinf ¢ x = ex + x + x -a ····

67、3;·············································1分因?yàn)楹瘮?shù) f (x) 在0,+¥) 上單調(diào)

68、遞增，所以 ( ) cos sin 0f ¢ x = ex + x + x -a ³ ，即 cos sina £ ex + x + x 對(duì) xÎ0,+¥)恒成立································

69、83;····················· 2分設(shè) h(x) = e + cos x +sin x ，則 h¢ x = e - x + x = e - x -x x x( ) sin cos 2 sin( )4 當(dāng) 0 £ < 時(shí)， h¢ x = e - x - > - = 2 4x ( ) 2 sin( ) 1 1 0x 當(dāng) x 

70、9; 時(shí)， h¢ x > e - > e - > ( ) 2 2 2 02所以函數(shù) h(x) = e + cos x +sin x 在0,+¥) 上單調(diào)遞增··································

71、;····· 4分x所以a £ h(x) = h(0) = 2 ·······································

72、3;·····································5分min（2）由題知 g(x) = f (x) -ln(1- x) = e +sin x -cos x -ax -ln(1- x) （ x <1）x1所以 ¢ = + + - +g (x) e cos x sin x ax ，