2018年全國卷3高考數(shù)學(xué)文科真題及詳細(xì)解析.pdf

2018cccppp?)?III 000 555 1.c)7gC?6OyW?3Kk 2.JKz?KY?Y)rKkAK8?YI“XIU“Z ?2YIJKY?3Kk“?3? 3.(?Kk?. !JJJKKK?KKK?12?KKKzzz?KKK5?60333zzz?KKK?oookkkKKK888. 1. 8A = x|x 1 0,B = 0,1,2KA B A 0B 1C 1,2D 0,1,2 2. (1 + i)(2 i) = A 3 iB 3 + iC 3 iD 3 + i 3. I?/-i7?5?i7?m? NeX?7?,i?7?NKi?7?: 4. esin = 1 3Kcos2 = A 8 9 B 7 9 C 7 9 D 8 9 5. e,+N?y7|G?V0.45Qy7|Gy7|G?V0.15Ky7| G?V A 0.3B 0.4C 0.6D 0.7 6. f(x) = tanx 1 + tan2x? A 4 B 2 C D 2 7. e?”y = lnx?”ux = 1? A y = ln(1 x)B y = ln(2 x)C y = ln(1 + x)D y = ln(2 + x) 56 8. x + y + 2 = 0Oxy?uA,B:P3?(x 2)2+ y2= 2K4ABP? A 2,6B 4,8C 2,32D 22,32 9. y = x4+ x2+ 2?” 10. V-C : x2 a2 y2 b2 = 1(a 0,b 0)?l%2K:(4,0)?C?C?l A 2 B 2C 32 2 D 22 11. 4ABC?S?A,B,C?Oa,b,ce4ABC?a 2 + b2 c2 4 KC = A 2 B 3 C 4 D 6 12. ?A,B,C,D4?o:4ABC?n?/93Knc ID ABC N? A 123B 183C 243D 543 ?!WWWKKK?KKK?4?KKKzzz?KKK5?20. 13. a = (1,2),b = (2,2),c = (1,)ec k (2a + b)K = 14. ,ikr#r?dk?)r?dTiO?1 ?N?J?k?!?X?K? 15. eCx,yv? 2x + y + 3 0 x 2y + 4 0 x 2 0 Kz = x + 1 3y? 16. f(x) = ln(1 x2 x) + 1,f(a) = 4Kf(a) = nnn!)KKK?70.)AAA?iii!yyyLLL.1111721 KKK777KKKzzzKKK) 777LLL11122!23KKKKKK). 777KKK?60 17. 12?ana1= 1,a5= 4a3 1an? 2PSnan?cneSm= 63m 57 18. 12,Jp)?mEM#J?,)?#?)? ?)?40 0) 1yk ? NeX?7?,i?7?NKi?7?: YYYA 4. esin = 1 3Kcos2 = A 8 9 B 7 9 C 7 9 D 8 9 YYYB 5. e,+N?y7|G?V0.45Qy7|Gy7|G?V0.15Ky7| G?V A 0.3B 0.4C 0.6D 0.7 YYYB 6. f(x) = tanx 1 + tan2x? 60 2018cccppp?)?III 000YYY A 4 B 2 C D 2 YYYC 7. e?”y = lnx?”ux = 1? A y = ln(1 x)B y = ln(2 x)C y = ln(1 + x)D y = ln(2 + x) YYYB 8. x + y + 2 = 0Oxy?uA,B:P3?(x 2)2+ y2= 2K4ABP? A 2,6B 4,8C 2,32D 22,32 YYYA 9. y = x4+ x2+ 2?” YYYD 10. V-C : x2 a2 y2 b2 = 1(a 0,b 0)?l%2K:(4,0)?C?C?l A 2 B 2C 32 2 D 22 YYYD 11. 4ABC?S?A,B,C?Oa,b,ce4ABC?a 2 + b2 c2 4 KC = A 2 B 3 C 4 D 6 YYYC 12. ?A,B,C,D4?o:4ABC?n?/93Knc ID ABC N? A 123B 183C 243D 543 YYYB ?!WWWKKK?KKK?4?KKKzzz?KKK5?20. 13. a = (1,2),b = (2,2),c = (1,)ec k (2a + b)K = YYY 1 2 61 14. ,ikr#r?dk?)r?dTiO?1 ?N?J?k?!?X?K? YYY? 15. eCx,yv? 2x + y + 3 0 x 2y + 4 0 x 2 0 Kz = x + 1 3y? YYY3 16. f(x) = ln(1 x2 x) + 1,f(a) = 4Kf(a) = YYY2 nnn!)KKK?70.)AAA?iii!yyyLLL.1111721 KKK777KKKzzzKKK) 777LLL11122!23KKKKKK). 777KKK?60 17. 12?ana1= 1,a5= 4a3 1an? 2PSnan?cneSm= 63m ) 1?an?q, q2= a5 a3 = 4, q = 2 an= 2n1an= (2)n1 2d1Sn= 1 2n 1 2 = 2n 1Sn= 1 + (2)n 1 + 2 = 1 31 (2) n Sm= 2m 1 = 63Sm= 1 31 (2) m = 63? m = 6. 18. 12,Jp)?mEM#J?,)?#?)? ?)?40 x21 )?mu1?1?)?p 62 2d“?m = 80?L LmLmO 1)?15520 1?)?51520 O202040 3K2= n(ad bc)2 (a + b)(c + d)(a + c)(b + d) = 40(15 15 5 5)2 20 20 20 20 = 10 6.635k99%?r)? ?k?. 19. 12X/ABCD3?l?CD3RM?CDuC,D ?: 1yAMDBMC 23AM3:P?MC kPBDnd ) 1?/ABCD?CMD AD?CMD ADMCD. CM3MCDS ADCM q M?lCDuC,D?: CMMD. q AD DM = D, CM ADM CM3BCM S BCMADM 2AM3:PPAM:yXe ?BD,AC?u:O?PD,PB,PO 3/ABCDOAC:PAM?: OP k MC OP3PDBSMC3PDBS MC k PDB. 20. 12?k?l?C : x2 4 + y2 3 = 1?uA,B:AB?:M(1,m)(m 0) 1yk 0?4k2+ 3 t2 1 ? x1+ x2= 8kt 3 + 4k2 = 2,y1+ y2= k(x1+ x2) + 2t = 6t 3 + 4k2 = 2m m 0, t 0k (3 + 4k2)2 16k2 k 1 2k 0 g0(x)3(,+)N4O g0(x)3(,+)3x0g0(x0) = 0 ex0+1+ 2ax0+ 1 = 0=ex0+1= 2ax0 1g(x)3(,x0)N43(x0,+)N4O g(x) g(x0).qg(x0) = ex0+1+ax2 0+x01 = ax20+(12a)x02 = (ax0+1)(x02)g0 ? 1 a ? = e1 1 a1, a 1, 0 e1 1 a 1 1 tan 1tan 1 45 9090 135n (45,135) 2:PI(x,y) ? = 90:PI(0,0) ? 6= 90?l?y = kx 2,A(x 1,y1),B(x2,y2) x2+ y2= 1 1 ? y = kx 2 2 ? kx2+ (kx 2)2 = 1 ?n? (1 + k2)x2 22kx + 1 = 0 x1+ x2= 22k 1 + k2 ,y1+ y2= 22 1 + k2 , x = 2k 1 + k2 3 ? y = 2 1 + k2 4 ? ?k = x y “ 4 ?x2+ y2+ 2y = 0. ?:P(0,0) vx2 + y2+ 2y = 0 AB:?P?;, x2+ y2+ 2y=0=x2 + y + 2 2 !2 = 1 2 dA 2 2 , 2 2 ! ,B 2 2 , 2 2 ! K 2 2 y 0 ?:P? x = 2 2