1、高 三 診 斷 性 測(cè) 試數(shù)學(xué)參考答案及評(píng)分細(xì)則評(píng)分說明：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)則。 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)，如果后繼部分的解答未改變?cè)擃}的內(nèi)容和難度，可視影響的程度決定后繼部分的給分，但不得超過該部分正確解答應(yīng)給分?jǐn)?shù)的一半；如果后繼部分的解答有較嚴(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)。選擇題和填空題不給中間分。一、選擇題：本大題考查基礎(chǔ)知識(shí)和基本運(yùn)算。每小題 5 分，滿分 40 分。 1B 2B 3D 4B 5D 6C

2、7A 8D二、選擇題：本大題考查基礎(chǔ)知識(shí)和基本運(yùn)算。每小題 5 分，滿分 20 分。全部選對(duì)的得 5分，部分選對(duì)的得 2 分，有選錯(cuò)的得 0 分。9ABD 10AC 11BD 12ACD三、填空題：本大題考查基礎(chǔ)知識(shí)和基本運(yùn)算。每小題 5 分，滿分 20 分。ì - 1 >æ ö 3 1ï1 , x 1,13 ；14 - ,1 1 ,ç ÷ f x f x x；15答案不唯一，如： ( ) = - ( ) = íç ÷2 3 2xè ø ï -îx 1, x1

3、等；165 ； 54 四、解答題：本大題共 6 小題，共 70 分。解答應(yīng)寫出文字說明、證明過程或演算步驟。17本小題主要考查等差數(shù)列、等比數(shù)列、遞推數(shù)列及數(shù)列求和等基礎(chǔ)知識(shí)，考查運(yùn)算求解能力、邏輯推理能力和創(chuàng)新能力等，考查化歸與轉(zhuǎn)化思想、分類與整合思想、函數(shù)與方程思想、特殊與一般思想等，考查邏輯推理、數(shù)學(xué)運(yùn)算等核心素養(yǎng)，體現(xiàn)基礎(chǔ)性和創(chuàng)新性滿分 10 分解法一：（1）因?yàn)?S +2 ,S ,S +1 成等差數(shù)列，所以 Sn - Sn+2 = Sn+1 - Sn ，···········

4、········1 分n n n所以 a + a + a +- - = ，·····································&

5、#183;·····································3 分n 2 n 1 n 1a + = - a + ，設(shè) a 的公比為 q ，則 q = -2 ，····

6、······································4 分 即 2 2 1n nn所以 ( ) ( )a = -2´ -2 = -2 ····&#

7、183;·················································&#

8、183;···········6 分n-1 nné2 -1+1ù é2 +1ùk k（2）依題意， ( ) ( )b - = ê ú = k k Î b = ê ú = k k ÎN , N ， ················&

9、#183;7 分* * 2k 1 2kë 2 û ë 2 û所以T = (a b + a b + ································8 分10 1 1 3 3= (a1 + 2a3 += (a + a + &#

10、183;······································9 分1 2 3= -(a1 + 2a3 +=1´ 2 + 2´ 2 +3所以 4T =1´ 23 + 2

11、´ 25 + 1 ，10數(shù)學(xué)參考答案及評(píng)分細(xì)則 第 1頁（共 16頁）兩式相減得 ( ) 2´ 1- 4 14 25-3T = 2 + 2 + 2 + 5´ 2 = - 5´ 2 = - ´ 2 -3 5 11 11 11 ，101- 4 3 3 14 2所以 2 + 3186T = ´ = ····················

12、83;············································ 10 分11 109 9解法二：（1）因?yàn)?S +2 ,S ,S

13、 +1 成等差數(shù)列，所以 Sn+1 + Sn+2 = 2Sn ，n n n············································

14、;··················································

15、;············1 分 設(shè)a 的公比為 q ，n若 q =1，則 a = -2,S = -2n ，S + + S + = - n - S = - n ，所以 S + + S + ¹ S ，n n n n n n n n1 2 4 6, 2 4 1 2 2S +1 + S +2 = 2S 矛盾，不合題意； ·············

16、;·············································2 分 與n n n若 q ¹1，則 ( )(

17、) ( ) a1 1- q a 1- q a 1- qn n+1 n+2S = S = ,S =， ， ···············3 分1 1n - - -n+1 n+21 q 1 q 1 q( ) ( ) ( ) a q a q a q1 1- n+1 1 1- n+2 2 1 1- n 所以 + = 1- q 1- q 1- q，整理得，qn+1 + qn+2 = 2qn ，即 q2 + q - 2 = 0 ，解得 q =1（舍去）或 q

18、 = -2 ， ················································

19、3;·················4 分所以 a = -2´(-2) = (-2) ···························

20、3;······································6 分n-1 nné2k -1+1ù é2k +1ù（2）依題意， ( ) ( )b - k k N

21、 ,b k k N ， ·················7 分= ê ú = Î * = ê ú = Î *2k 1 2kë 2 û ë 2 û所以T = (a b + a b )+ (a b + a b )+ (a b + a b )+ (a b + a b )+ (a b + a b ) ···

22、;· 8 分10 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10= ( 1 + 2 )+ 2( 3 + 4 )+ 3( 5 + 6 )+4( 7 + 8 )+ 5( 9 + 10 ) ···························9 分a a a a a a a a a a=

23、 ´ ë - + - û + ´ ë - + - û + ë - + - û1 é( 2) ( 2) ù 2 é( 2) ( 2) ù 3é( 2) ( 2) ù2 3 4 5 6+ ë - + - û + ´ ë - + - û4é( 2)7 ( 2)8 ù 5 é( 2)9 ( 2)10 ù=1´ 2 + 2´ 2 + 3´ 2 +

24、4´ 2 + 5´ 2 = 2 +16 + 96 + 512+25603 5 7 9= 3186 ········································&

25、#183;·············································· 10 分 解法三：（1）因?yàn)?S +

26、2 ,S ,S +1 成等差數(shù)列，所以 2Sn = Sn+2 + Sn+1 ， ·························1 分n n n當(dāng) n =1時(shí)， 2S = S + S ，化簡(jiǎn)得 a = - a ， ···········&

27、#183;···································2 分3 2 2 1 3 2設(shè)a 的公比為 q ，所以 q = -2 ， ········

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

29、···4 分n當(dāng) q = -2 時(shí)， ( )-2 - -2n+1S = ，因此 2Sn n3é- - (- ) ùn+12 2 2ë û= ，3( ) ( ) ( ) ( )+ + + é- - - ùn+1n 3 n 2 n 2 2 2 2-2 - -2 -2 - -2 -4 + -2 ë ûS +S = + = = ，n+2 n+13 3 3 3滿足 2S = S + + S + ，故 q = -2 符合題意n n 2 n 1所以 a = -2´(-2)n-1 = (-2

30、)n ·················································

31、83;·················6 分n數(shù)學(xué)參考答案及評(píng)分細(xì)則 第 2頁（共 16頁）（2）依題意， 10 5b1 =1，b2 =1，b3 = 2 ，b = ，b = ，b = ，b = ，4 2 5 3 6 3 7 4 b = ，8 4 b = ，9 5 b = ，·············

32、;··················································

33、;···········································7 分所以T10 = -2 + (-2)2 + 2´(-2)3 + 2&

34、#180;(-2)4 + 3´(-2)5 + 3´(-2)6 + 4´(-2)7 + 4´(-2)8 + 5´(-2)9 + 5´(-2)10 ·································

35、··················································

36、·······················8 分= - + - 2 + ´ - 3 + - 4 + ´ - 5 + - 6 + ´ - 7 + - 8 + ´ - 9 + - 10 2 ( 2) 2 ( 2) ( 2) 3 ( 2) ( 2) 4 ( 2) ( 2) 5 ( 2) ( 2) .9 分= 2 + 2 + 3&#

37、180; 2 + 4´ 2 + 5´ 2 = 2+16+96+512+ 25604 5 7 9= 3186 .10 分18本小題主要考查獨(dú)立事件的概率、互斥事件的概率，二項(xiàng)分布、數(shù)學(xué)期望等基礎(chǔ)知識(shí)；考查數(shù)學(xué)建模能力，運(yùn)算求解能力，邏輯推理能力，創(chuàng)新能力以及閱讀能力等；考查統(tǒng)計(jì)與概率思想、分類與整合思想等；考查數(shù)學(xué)抽象，數(shù)學(xué)建模和數(shù)學(xué)運(yùn)算等核心素養(yǎng)；體現(xiàn)應(yīng)用性和創(chuàng)新性滿分 12 分解法一：（1）甲滑雪用時(shí)比乙多5´36 =180 秒 = 3分鐘，因?yàn)榍叭紊鋼?，甲、乙兩人?被罰時(shí)間相同，所以在第四次射擊中，甲至少要比乙多命中 4 發(fā)子彈設(shè)“甲勝乙”為事件 A，“在

38、第四次射擊中，甲有 4 發(fā)子彈命中目標(biāo)，乙均未命中目標(biāo)” 為事件 B， “在第四次射擊中，甲有 5 發(fā)子彈命中目標(biāo)，乙至多有 1 發(fā)子彈命中目標(biāo)”為事件 C， ····································

39、3;·················································

40、3;···················1 分依題意，事件 B 和事件 C 是互斥事件，A=B+C， ·························

41、83;···············2 分4 5 5 5 4é ù1 æ 4 ö æ 1 ö æ 4 ö æ 1 ö æ 1 ö 3( ) ( )B = ´ç ÷ ´ç ÷ , C = ç ÷ ´ êç ÷ +

42、 ç ÷ ´ ú1 1P C P C5 5è ø è ø è ø ëê è ø è ø ûú5 5 4 5 4 4 4，························4 分 69

43、P A = P B + P C = . 12500所以， ( ) ( ) ( )69即甲勝乙的概率為 . ·········································&

44、#183;······························5 分12500（2）依題意得，甲選手在比賽中未擊中目標(biāo)的子彈數(shù)為 X ，乙選手在比賽中未擊中目標(biāo)的子彈數(shù)為Y ，則X æ ö æ ö1 1， ····

45、;·······································7 分è 5 ø è 4 ø11´ EX =1´ 20´ =

46、 4 （分鐘），································8 分 所以甲被罰時(shí)間的期望為 511´ EY =1´ 20´ = 5（分鐘）， ·······&#

47、183;······························9 分 乙被罰時(shí)間的期望為4 又在賽道上甲選手滑行時(shí)間慢 3 分鐘，所以甲最終用時(shí)的期望比乙多 2 分鐘. ··········&#

48、183;··········································· 11 分因此，僅從最終用時(shí)考慮，乙選手水平更高. ··

49、·········································· 12 分解法二：（1）同解法一·····&#

50、183;·················································&#

51、183;·······················5 分?jǐn)?shù)學(xué)參考答案及評(píng)分細(xì)則 第 3頁（共 16頁）x æ ö4 4（2）設(shè)甲在一次射擊中命中目標(biāo)的子彈數(shù)為x ，則 Ex = ´ = ，所，所以 5 4è 5 ø 5 以甲在四次射擊中命中目標(biāo)的子彈數(shù)的期望為 4Ex =16 ，···

52、····························7 分設(shè)乙在一次射擊中命中目標(biāo)的子彈數(shù)為h ，則h æ 3 ö，所以 Eh = 5´ = ，所3 15 è 4 ø 4 4以乙在四次射擊中命中目標(biāo)的子彈數(shù)的期望為 4Eh =15 ，··

53、3;····························9 分 所以在四次射擊中，甲命中目標(biāo)的子彈數(shù)的期望比乙多 1，所以乙被罰時(shí)間的期望比 甲多 1 分鐘，又因?yàn)樵谫惖郎霞椎幕袝r(shí)間比乙慢 3 分鐘，所以甲最終用時(shí)的期望比 乙多 2 分鐘，······&

54、#183;·················································&

55、#183;····························· 11 分 因此，僅從最終用時(shí)考慮，乙選手水平更高.················

56、;······························ 12 分19本小題主要考查直線與直線、直線與平面、平面與平面的位置關(guān)系，直線與平面所成角、二面角等基礎(chǔ)知識(shí)；考查空間想象能力，邏輯推理能力，運(yùn)算求解能力等；考查化歸與轉(zhuǎn)化思想，數(shù)形結(jié)合思想，函數(shù)與方程思想等；考查直觀想象，邏輯推理，數(shù)學(xué)

57、運(yùn)算等核心素養(yǎng)；體現(xiàn)基礎(chǔ)性和綜合性滿分 12 分 解法一：（1）如圖，在 VAC 內(nèi)過 P 作 PM VC ，垂足為 M ，V在 VBC 內(nèi)過 M 作 MN VC 交VB 于 N ，M連結(jié) PN ，則直線 PN 即為直線l . ··························· 2 分理由如下：因?yàn)?PM VC ， MN V

58、C ， PM M ，lPN所以VC 平面 PMN ，AC由于過空間一點(diǎn)與已知直線垂直的平面有且只有一個(gè)，所以平面 PMN 與平面a 重合，B因?yàn)槠矫?PMN 平面VAB = PN ，所以直線 PN 即為直線l . ····························4 分（2）因?yàn)?VAB 和 ABC 均為等邊三角

59、形，所以VA =VB, AC = BC ，又因?yàn)閂C =VC ,所以 VACVBC ，所以 ÐPVM = ÐNVM ，又VM =VM ，所以 RtVPMRtVNM ，所以VP =VN ，所以 2 VN = VB . ·······5 分3 如圖，設(shè) AB 的中點(diǎn)為 D ，連結(jié)VD,CD ，V因?yàn)?VAB 和 ABC 均為等邊三角形，M所以VA =VB, AC = BC ，所以 AB VD, AB CD ，P又因?yàn)閂D ，所 以 AB 平面VCD ，因?yàn)?AB Ì 平面 ABC ，所以平面 ABC

60、 平面VCD .AN COD在 VCD中，作VO CD，垂足為O，B因?yàn)槠矫?ABC 平面VCD = CD，VO Ì 平面VCD ，所以VO 平面 ABC ，所以 ÐVCD 是直線VC 與平面 ABC 所成的角，所以 ÐVCD = .························7 分3因?yàn)?VAB 和 ABC 均是邊長(zhǎng)為 4

61、 的等邊三角形，所以VD = DC = 2 3 ,所以 VCD 是等邊三角形，所以VO = 3， DO = OC = 3 .以 O為原點(diǎn)，分別以O(shè)C,OV 的方向?yàn)?y 軸和 z 軸正方向建立如圖所示的空間直角坐標(biāo)數(shù)學(xué)參考答案及評(píng)分細(xì)則 第 4頁（共 16頁）系 O - xyz ，則 A(-2,- 3,0),B(2,- 3,0),C (0, 3,0),V (0, 0,3)，··················

62、3;····················8 分所以CV = (0,- 3,3),CA = (-2,-2 3,0), AB = (4, 0, 0),ö4 5 3 öCP CV CA = = ç ÷= + = ç- - ÷, ,1 , PN AB ,0,0 .3 3 3 3 3 è 3 øè ø過

63、l 及點(diǎn)C 的平面為平面CPN ，設(shè)平面CPN 的法向量為 n = (x, y, z)，PzVìïCP則 íïPNîn× n =0,0,ì4 5 3- x - y + z = 3 3ïï即íï =8x 0.ïî30,取 n = (0, 3,5) ，ADNBOxCy即平面CPN 的一個(gè)法向量為 n = (0, 3,5) .············&#

64、183;····································· 10 分易知，平面 ABC 的一個(gè)法向量為 m = (0, 0,1) ， ······

65、····································· 11 分m × n 5 5 7所以 cos < m,n >= = =m n× 2 7 14，5 7所以過l 及點(diǎn)C 的平面

66、與平面 ABC 所成的銳二面角的余弦值為 ············ 12 分14 解法二：（1）如圖，在 VAB 內(nèi)過 P 作 PNAB ，交VB 于 N ，則直線 PN 即為直線l .························

67、83;·················································

68、83;··2 分 理由如下：V取VC 的中點(diǎn)Q ，連結(jié) AQ,BQ ，因?yàn)?VAB 和 ABC 均為等邊三角形，所以VA = AC,VB = BC ，所以VC AQ,VC BQ ，PlQ又因?yàn)?AQ ，所以VC 平面 ABQ ，NA C又因?yàn)閂C 平面a ，所以平面a平面 ABQ ， 又因?yàn)槠矫鎍 平面VAB = l ，平面 ABQ 平面VAB = AB ，B所以 ABl ，所以直線 PN 即為直線l .··············&#

69、183;·········································4 分VP = 2VA，所以 2（2）由（1）知， PNAB ，因?yàn)?VN = VB . ·

70、;····························5 分3 3設(shè) AB 的中點(diǎn)為 D ，連結(jié)VD ，交 PN 于 G ，連結(jié)CG ，V因?yàn)?VAB 和 ABC 均為等邊三角形，所以VA =VB, AC = BC ，所以 AB VD, AB CD ，P又因?yàn)閂D ，GN所以 AB 平面VCD ， AB Ì

71、; 平面 ABC ，A CO所以平面 ABC 平面VCD . DB在 VCD中，作VO CD，垂足為O，因?yàn)槠矫?ABC 平面VCD = CD，VO Ì 平面VCD ，所以VO 平面 ABC ，數(shù)學(xué)參考答案及評(píng)分細(xì)則 第 5頁（共 16頁）所以 ÐVCD 是直線VC 與平面 ABC 所成的角，所以 ÐVCD = ，·····················&

72、#183;7 分3因?yàn)?VAB 和 ABC 均是邊長(zhǎng)為 4 的等邊三角形，所以VD = DC = 2 3 , ÐVDC = ，31 2 3因?yàn)?ABPN ，所以 DG = DV = 由（1）知，過l 及點(diǎn)C 的平面為平面CPN ，3 3因?yàn)?AB Ë 平面CPN ， PN Ì 平面CPN ，所以 AB平面CPN ， ·····················

73、8 分設(shè)平面CPN 平面 ABC = l' ，因?yàn)?AB Ì 平面 ABC ，所以 ABl'，因?yàn)?AB 平面VCD ， CG Ì平面VCD，CD Ì平面VCD ，所以 AB CG, AB CD ，所以CG l ',CD l '，又因?yàn)镃G Ì平面CPN ， CD Ì平面 ABC ，所以 ÐGCD 為平面CPN 與平面 ABC 所成的銳二面角的平面角，·············

74、;······· 10 分2 21在 GCD 中，由余弦定理得，CG2 = DG2 + DC2 - 2DG × DC ×cosÐGDC ，CG = ,3 ·····························

75、;··················································

76、;························· 11 分CG2 + DC2 - DG2 5 7所以 cosÐGCD = =2CG × DC 14，5 7所以過l 及點(diǎn)C 的平面與平面 ABC 所成的銳二面角的余弦值為 ·········

77、;··· 12 分1420本小題主要考查正弦定理、余弦定理及三角恒等變換等基礎(chǔ)知識(shí)，考查邏輯推理能力、運(yùn)算求解能力等，考查化歸與轉(zhuǎn)化思想、函數(shù)與方程思想、數(shù)形結(jié)合思想等，考查數(shù)學(xué)運(yùn)算、邏輯推理等核心素養(yǎng)，體現(xiàn)基礎(chǔ)性和綜合性滿分 12 分a + c - b2 2 2解法一：（1）在 ABC 中，由余弦定理得 cos B = ，····················&#

78、183;········1 分 2ac又因?yàn)?a = 6 ，b +12cos B = 2c ， a2 c2 b212 2 2ac+ -所以b + × = c ，·····························&

79、#183;·····································2 分整理得b2 + c2 - 36 = bc .········

80、··················································

81、·················3 分在 ABC 中，由余弦定理得b2 + c2 - 36 = 2bccos A ，1所以bc = 2bccos A ，即 A = .······················

82、;·········································4 分cos 2p又因?yàn)?AÎ(0,p) ，所以 = ···

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

84、83;··········5 分A3（2）選 ······································

85、;··················································

86、;·······6 分 1 p因?yàn)?M 為 ABC 的內(nèi)心，所以 ÐBAD = ÐCAD= ÐBAC = ， 2 6由SD = SD + SD ，·····························

87、83;··················· 7 分ABC ABD ACD得1 p = 1 × p + 1 × p bcsin c ADsin b ADsin ，2 3 2 6 2 6因?yàn)?AD=3 3 ，所以3 1 bc bc = 3 3 ´ (b + c) ，即b + c = . 8 分2 2 3數(shù)學(xué)參考答案及評(píng)分細(xì)則 第 6頁（共 16頁）由（1）可得b2 + c2 -

88、36 = bc ，即 (b + c)2 -3bc = 36 ，·····································9 分(bc) bc2所以 3bc 36 0- - = ，即 (bc + 9)( - 4) =

89、 0 ，········································ 10 分9 9又因?yàn)閎c > 0，所以bc = 36，····&

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

91、#183;·········· 11 分1 p 1 3所以 S bc ····································

92、;············ 12 分= sin = ´36´ = 9 3 DABC2 3 2 2 解法二：（1）因?yàn)?a = 6 ， b +12cos B = 2c ，所以b + 2acosB = 2c ，·····················&

93、#183;·················································&

94、#183;··· 1 分在 ABC 中，由正弦定理得sin B + 2sin AcosB = 2sinC ，即 sin B + 2sin Acos B = 2sin(A + B) ， ································&

95、#183;························ 2 分即 sin B + 2sin AcosB = 2sin AcosB + 2 cos Asin B ，即 sin B = 2 cos Asin B ， ············

96、··················································

97、············ 3 分因?yàn)?BÎ(0,p)，所以sin B ¹ 0，故 cos 1A = . ·····························

98、3;·······································4 分2p又因?yàn)?AÎ(0,p) ，所以 = ······

99、··················································

100、········5 分A3（2）選 ········································&#

101、183;·················································&#

102、183;····6 分 因?yàn)?M 為 ABC 的垂心，p p æ p öÐBMD = - ÐMBD = - ç - ÐACB÷ = ÐACB ，又 MD = 3 ，·················7 分 所以2 2 è 2 ø所以在 MBD 中， BD = MD × tan&#

103、208;BMD = 3 tanÐACB ，同理可得CD = 3 tanÐABC ， ····································· 8 分又因?yàn)?BD + CD = 6 ，所以 3

104、tan ÐABC+ 3 tan ÐACB = 6 ，即 tanÐABC+tanÐACB = 2 3 ， ······································

105、 9 分又因?yàn)樵?ABC 中， tan(ÐABC + ÐACB)= -tanÐBAC = - 3 ，tan ABC tan ACBÐ + Ð = - 所以 3 ，1- tan ÐABC tan ÐACB因此 tanÐABC tanÐACB=3 ······················&

106、#183;··········································· 10 分故 tanÐABC，tanÐACB 為方程

107、x2 - 2 3x + 3 = 0 兩根，即 tan ÐABC= tan ÐACB= 3 ，因?yàn)?ÐABC，ÐACB Î(0,p)，p所以 ÐABC=ÐACB= ，所以 ABC 為等邊三角形， ····························

108、······· 11 分3所以 1 62 3 9 3SD = ´ ´ = ····································&#

109、183;························· 12 分ABC2 2解法三：（1）同解法一. ····················&#

110、183;·················································&#

111、183;··········5 分（2）選 ······································

112、··················································

113、·······6 分 因?yàn)?M 為 ABC 的垂心，數(shù)學(xué)參考答案及評(píng)分細(xì)則 第 7頁（共 16頁）p p所以 ÐAMB = p - ÐACB ， Ð = - Ð = ， ··························

114、3;··········8 分ABM BAC2 6 AM AB所以在 ABM 中，由正弦定理得 = sinÐABM sinÐAMB，AM AB=即 ···························

115、3;·············· 9 分p Ðsin ACBsin 6又因?yàn)樵?ABC 中，AB BC由正弦定理得 = ，··························10 分sin&

116、#208;ACB sinp3AM = BC 所以，因?yàn)?a = 6 ，所以 AM = 2 3 . ········································ 11 分p psin

117、 sin6 3又因?yàn)?MD = 3 ，所以 1 1 6 (2 3 3) 9 3SD = a × AD = ´ ´ + = .··················· 12 分ABC2 2解法四：（1）同解法一. ··············&

118、#183;·················································&

119、#183;················5 分（2）選 ································

120、;··················································

121、;·············6 分 1 p因?yàn)?M 為 ABC 的內(nèi)心，所以 ÐBAD = ÐCAD= ÐBAC = . 2 6在 ABD 中，由正弦定理得BD AD=psin Bsin6，因?yàn)?AD = 3 3 ，所以 2BD3 3= ， sin B同理可得 2CD3 3 3 3= =psinC sin(B )+3.··········

122、··················································

123、·7 分3 3 3 3又因?yàn)?BD + CD = 6 ，所以 + =12，psin B sin(B )+3p é p ù即 4sin sin( + ) = 3 êsin + sin( + )úB B B B3 ë 3 û，p 1 3即 4sin Bsin(B + ) = 3(sin B + sin B + cos B)，·················&#

124、183;·····················8 分3 2 2 p p即 4sin Bsin(B + ) = 3sin(B + ) ， 3 6p p p p p即 4sin(B + ) - sin(B + ) + = 3sin(B + ) ，············

125、·····························9 分6 6 6 6 6é p p ù é p p ù p3 1 3 1即 4 sin(B + ) - cos(B + ) sin(B + ) + cos(B + ) = 3sin(B + )ê 

126、50; ê ú2 6 2 6 2 6 2 6 6ë û ë û，數(shù)學(xué)參考答案及評(píng)分細(xì)則 第 8頁（共 16頁）即 4é3 sin2 ( + p) - 1 cos2 ( + p)ù = 3sin( + p) B B Bê ú，ë4 6 4 6 û 6p p é + p - ù é + p + ù =即 4sin2 (B + ) - 3sin(B + ) -1 = 0 ，即 sin( ) 1 4sin( ) 1 0B Bê 

127、50; ê ú6 6 ë 6 û ë 6 û，···· 10 分因?yàn)閜A = ,所以 03< < 2 ，所以 ( , 5 )B B + Î ，p p p pB + Î ，3 6 6 6p p所以sin( + ) > 0，故sin(B + ) =1，B6 6 p p pB + = ，即 B = ，所以 ABC 為等邊三角形，···········

128、83;······················ 11 分 即6 2 3所以 1 62 3 9 3SD = ´ ´ = .···················

129、3;············································· 12 分ABC2 2解法五：（1）同解法一. 

130、3;·················································

131、3;······························5 分（2）選 ··················&#

132、183;·················································&#

133、183;··························6 分 1 p因?yàn)?M 為 ABC 的內(nèi)心，所以 ÐBAD = ÐCAD= ÐBAC = . 又因?yàn)?AD = 3 3 ， 2 63在 ABD 中，由余弦定理得 BD2 =c2 + 27 - 2´3 3 ´ c = c2 - 9c

134、 + 27 ，2同理可得CD2 = b2 - 9b + 27 ······························ 7 分又因?yàn)? × AB × AD ×sinBD S AB2 6= DABD = =CD S AC AD AC1 × × &#

135、215;sinDACD2 6，所以c 9c 27 c2 - + = 2b - 9b + 27 b2 2，即 (b - c)3(b + c) - bc = 0 ，bc故 b = c 或b + c = . ·································&

136、#183;·············································8 分3（i）當(dāng)b = c 時(shí)， ABC 為等邊三

137、角形，所以 1 62 3 9 3SD = ´ ´ = .ABC2 2 bc（ii）當(dāng)b + c = 時(shí)，由（1）知b2 + c2 - 36 = bc ，即 (b + c)2 -3bc = 36 ，··········9 分3(bc) bc2所以 - 3bc - 36 = 0 ，即 (bc + 9)( - 4) = 0 ，·············&

138、#183;·························· 10 分9 9因?yàn)閎c > 0，所以bc = 36. ··················

139、··················································

140、·· 11 分又因?yàn)锳p= ，所以31 p 1 3SD bcsin 36 9 3= = ´ ´ = ABC2 3 2 2綜上所述， SD = 9 3 ··································&

141、#183;··································· 12 分ABC說 明 ： 設(shè) ABC 的 外 接 圓 半 徑 為 R ， 則 在 ABC 中 ， 由 正 弦 定 理 得數(shù)學(xué)參考答案及評(píng)分細(xì)則 第 9頁（共 16頁）B

142、C 62R = = = 4 3psin sinA3，即 R = 2 3 ，因 為 M 為外心，所以 AM = 2 3 ，與 AM = 4 矛盾，故不能選21本小題主要考查橢圓的標(biāo)準(zhǔn)方程及簡(jiǎn)單幾何性質(zhì)，直線與圓、橢圓的位置關(guān)系，平面向量等基礎(chǔ)知識(shí)；考查運(yùn)算求解能力，邏輯推理能力，直觀想象能力和創(chuàng)新能力等；考查數(shù)形結(jié)合思想，函數(shù)與方程思想，化歸與轉(zhuǎn)化思想等；考查直觀想象，邏輯推理， 數(shù)學(xué)運(yùn)算等核心素養(yǎng)；體現(xiàn)基礎(chǔ)性，綜合性與創(chuàng)新性滿分 12 分解法一：（1）以O(shè) 為坐標(biāo)原點(diǎn)，橢圓C 的長(zhǎng)軸、短軸所在直線分別為 x 軸、 y 軸，建立平 面直角坐標(biāo)系，如圖. ···

143、83;·················································

144、83;························1 分 設(shè)橢圓的長(zhǎng)半軸為 a ,短半軸為 b ,半焦距為 c ，ìc 2ï ,=a 2ïïï - = -6 6íb 1 ,依題意得3 3ïï = +a b c ,2 2 2ïïîì =a

145、 2,ïíb =1,解得ï =c 1,î·········· 3 分yOAxB所以C 的方程為x22+ y2 =1 ····························

146、3;······································4 分（2）因?yàn)橹本€ AB與以O(shè)A為直徑的圓的一個(gè)交點(diǎn)在圓O 上，所以直線 AB 與圓O 相切 ····································