2020年全国卷Ⅲ高考理科数学试题及答案

1、 5 , )| , , ,B(x,y)|xy8A BAx y x y N y x*11 3 1 1 341 , , , p p p p p1234ii1 ppp pp pp p p 14231423 p pp3 pp pp1421423 K(t) tI （ ItK1e0.23(t53)It)0.95K * O ： 2 ( E y pC2 1( 41( 2b|a5 |b6 6,a b a,ab =2ABC= =3191312236+4 24+4 2 3 3+ =41l=和xy= l2 2x5111 1= += 222 2： x22y221FF 5 P是 C F121abFPFF =21 25 5

2、44558ccab45 x y ，2x y ， z 3x2y则x，2(x )26 x1xx= 2aa3 4 an1annn1aaa23nanSnnn256778201或3或22, a b c d) ac bd2k ABCD ABC D , E FDD BB,2 2 DE ED BF FB11111111 CAEF12,1,AA3 A EF A11 4 : x2y21(0 5), , A B CCm25 m2 C 在 Q | , P Cx6| 11 f(x) x bxcy f(x) , ( 3fy22 ( )1 ( )f x f x 2 xt t2xOy Ct CBy 2tt2| |；xAB设

3、,0 abc1a b cab bc ca ；0 , , c , , 34 a b ca b c23 5, 7, 2 1,aaan23na (2n3)a (2n1),n1na (2n1)a (2n1),nn1a 5a 3).21 3 2 1.aan1n2an nnnS 325272 (2n1)2n . 32n2S 32 52 72 (2n1)2 234n1n 得S 3222 22 22 (2n1)2 ,23nn1n Snn1n12341 (100228837)2K23.841 , ,AB a AD b c x C DC1111C 系 121c F b c11b c3,则 , (a,c), ( )

4、 , )b c C F , ) , , )（1, ,得C F1C1E a3331EAC F1 C F A E F C , , , CAEF111A E F,A ,AE(0,1,1),AF(2,0,2) ,(0,1,2) ,AE11AF (2,0,1)1n1x y z设 ( , , )AEFAE0, yz0,n即 (1,1,1)n112x2z 0,nAF 0, 1设 nA120,nAE 1 ( n212 2nAF 0,21n n 77n ,n A A12|n |n |71211225m2 m2,54x2y21C.P(x ,y ),Q(6, y )y 0Qy 0P,PPQ1B(5,0)y (x|

5、| y y1|BQ 1 y,BP BP,22yPQQQ|BP|BQ|y 1y 1Cx 3 3或 .PPPy 2QBP或,QQ P Q .(3,1), (6,2); (3,1), (6,8)P P11221102|PQ 10 x(5,0) APQ PQ yPQ 31 11111111 10 105为.2 227 130|PQ 130 APQ PQ yx PQ A9322222226221 13052 130.2 265 .APQ2 ( )3 2 f xx b13 ( )0 0.fb243故 b4343f x x ( ) , ( )3 .x c f x3x24令 f(x)0 x1或x1.22f(x)与 f(x)111x222f(x)f(x)+00+c 1c144111 (1) ( ) f(x)ffcc214141 ( ( ) ffcc24411 ,c4411当 = f(x) c4121当 = f(x) .c421111 11当 f(x) , , , x x x( )( , )x, ( 2cx1x24422 21233 f(x) f(x)2 0得 2C与yt tt2由23 0得C与xt t(2故|AB 4 10 AB x