山东省青岛市2022-2023学年高三下学期第一次适应性检测（一模）数学试题（PDF版含答案）

青岛市 2023 年高三年级第一次适应性检测 数学参考答案及评分标准 一、单项选择题：本题共 8小题，每小题 5分，共 40分。 1--8：AC B B D C CA 二、多项选择题：本题共 4小题，每小题 5分，共 20分。 9．AC 10．AB 11．ABD 12．BCD 三、填空题：本题共 4个小题，每小题 5分，共 20分。 13． ( 1, 2) 3 1答案不唯一； 14． ； 15． 2 3 ； 16．0 ， ． 3 1011 四、解答题：本题共 6小题，共 70分。解答应写出文字说明，证明过程或演算步骤。 17. （本小题满分 10分） 解：（1）由题意得 f (x) 2cos2 x sin 2 x 1 cos 2 x sin 2 x 2 sin(2 x π ) 1 ························································ 2 分 4 T 2π 2 2π因为 ，所以 1 ······································································3 分 T 所以 f (x) 2 sin(x π ) 1 4 令 x π kπ π π 得， x kπ （ k Z） 4 2 4 所以函数 f (x) π图象的对称轴方程为 x kπ （ k Z）····································5 分 4 1 π 2 （2）由 f ( ) 得 sin( ) ··························································· 6 分 3 4 3 所以 sin cos 2 4 4 ，所以 (sin cos )2 ，即1 sin 2 3 9 9 所以 sin 2 5 ······················································································· 10 分 9 18．（本小题满分 12分） 解：（1）因为 S2 ，S4 ，S5 4成等差数列， a2 ，a4 ，a8 成等比数列 2S 所以 4 S2 S5 4 2 ················································································· 2 分 a4 a2 a8 2(4a1 6d) (2a所以 1 d) (5a1 10d) 4 a d 4 2 ，整理得 1 (a1 3d ) (a1 d )(a1 7d ) a d d 2 1 因为 d 0，解得： a1 d 2 ······································································· 5 分 所以 S 2n n(n 1) 2 n 2n n ································································· 6 分2 数学评分标准 第 1页（共 8页） n 2 n 3 （2）由（1）得 2bn Tn ， 2b T ，····················· 7 分n(n 1) n 1 n 1 (n 1)(n 2) 2b n 2 n 3所以两式相减得： n 2bn 1+Tn 1 Tn ························8 分n(n 1) (n 1)(n 2) 2 1 整理得： 2bn bn 1 n(n 1) (n 1)(n 2) 2[b 1 1所以 n ] b n(n 1) n 1 (n 1)(n 2) b 1 1即 n 1 2(bn ) ············································································10 分Sn 1 Sn b 3因为 1 ,b 1 1 1 0 ，2 1 2 {b 1所以 n }是以1为首项， 2 为公比的等比数列············································ 11 分Sn 1 1 所以b n 1 n 1n 2 ，所以bn 2 ··········································· 12 分n(n 1) n(n 1) 19．（本小题满分 12分） 解：（1）当D π为圆弧BC ... ...