辽宁省大连市2020届高三双基测试卷 数学（理）试题（扫描版含答案）

辽宁省大连市2020届高三双基测试卷 数学（理）试题（扫描版含答案）

. . 13 2 14 1 (15) 16 9,4 2 . . 17 ( 12 ) (I) AC CE ACE 3 ABCD AD AB PA AB AD PA E PD AE PD CE PD AE CE E AE CE PD . 6 (II) BD AC O PO ABCD AC BD O AC BD PA PC AC PO BD PO AC BD O AC BD ABCD PO ABCD O OB x OC y OP z ················································································· 7 2PA PC AB 60ABC 2AC (0, 1,0)A E D CB A P z yx O E D CB A P PD ·················· 8 PAB ( , , )x y zn 0 0 AB AP 1x 1z ········································································ 10 PAB PAB ······················· 12 (18)( 12 ) 2 2 n n a 1 1 2 2 8 a a na n 2 2 n n a 1 2n na n 1 2 2 8 a a 1 2 3 1 2 3 1 2 2 2 3 2 2n n nS a a a a n 2 3 4 12 1 2 2 2 3 2 2n nS n 1( 1) 2 2n nS n 12 (2 2) 2 (2 4) 2 ( )n n n na n n n n N 1(2 4) 2n nb n 1n n na b b 1 2 2 1 3 2 1 1 1( ) ( ) ( ) (2 2) 2 2n n n n n nS a a a b b b b b b b b n 1( 1) 2 2n nS n (19)( 12 ) (I) A ··························································· 1 B ··························································· 2 C 39 64 . ··················· 4 (II) X 3 4 5 6 ······························· 5 ················· 9 X X 3 4 5 6 P 15 56 5 28 ····························································································· 10 ··················· 12 (20) 12 4a ( ) 0g x ( )g x (0, ) 4a 2( ) ( 2) 1m x x a x ( ) 0m x ( )g x 4a (0, ) 4a 1( ) ( ) ( 1) ln( 1)xh x f x f x ax e x ax (0, ) (0) 0h a 0 (0,ln( 1))x a 0( ) 0h x 0(0, )x x ( ) 0h x 0( , )x x ( ) 0h x ( )h x 0(0, )x 0( , )x ( )h x (0, ) 0x 0 0 ( ) 0 ( ) 0 h x h x (0,ln( 1))x a ( )u x (0,ln( 1))x a 1( ,1)2 (21) 12 E 1c E ··································· 3 l y kx m 1 1 2 2( , ), ( , )C x y D x y 2 2 2(4 3) 8 4 12 0k x kmx m 0 2 23 4k m ······ 4 2 2 1 2 1 2(4 9) (4 18)( ) 4 36 0k x x km x x m ····················· 6 2 22 3 0k km m k m 2k m 0 ··· 7 l y kx k l ( 1,0) ······························································ 8 l x my n 1 1 2 2( , ), ( , )C x y D x y 2 2 2(3 4) 6 3 12 0m y mny n 0 2 23 4m n ······ 4 2 2 1 2 1 2(9 4) 9 ( 2)( ) 9( 2) 0m y y m n y y n ······················· 6 2 3 2 0n n 1n 2n 0 ······· 7 l ( 1,0) ······························································ 8 2x x y y l 1mx ny 1m : 1l x ny 2 1x ny 1x ny l ( 1,0) ······························································ 8 l y kx m 1 1 2 2( , ), ( , )C x y D x y 2 2 2(4 3) 8 4 12 0k x kmx m 2 2=48(4 3) 0k m 0OC OD OB 1 2 1 2( , )B x x y y B E 10 B l 11 O l 11 12 0OC OD OB B E 10 11 12 (22) 10 4 4 (I) 2sin 4cos 2 2sin 4 cos C 2 4y x ············································· 3 l 2 cos sin x t y t t ······························ 5 (II) 2 cos sin x t y t 2 4y x 2 2sin 4 cos 8 0t t 0 1 2 2 4cos sint t 1 2 2 8 sint t 2 2 4 2 1 2 1 2 2 2 2 2 2 21 2 1 2 2 16cos 16 ( ) 21 1 1 1 16 1sin sin 8| | | | ( ) 64 4( )sin t t t t MA MB t t t t ·································································································· 10 (23) 10 4 5 3 3 3 3 3 2+ 2 2( ) 2 23 3 3 3+( ) 8 42 a bab 5 (II) 3 3 2 2+ ( )( )a b a b a ab b 2( )[( ) 3 ]a b a b ab + 4a b ··········································································· 10 3 3 3+2 +2 3 4a a 3 3 3+2 +2 3 4b b 48 12( )a b + 4a b ··········································································· 10