THUSSAT2023年3月诊断性测试数学答案_2024年2月_01每日更新_16号_2023届THUSSAT中学生标准学术能力高三下学期3月月诊断性测试_THUSSAT2023年3月诊断性测试数学试卷及答案

中学生标准学术能力诊断性测试 2023 年 3 月测试 数学参考答案 一、单项选择题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一 项是符合题目要求的． 1 2 3 4 5 6 7 8 A D B D C C A B 二、多项选择题：本题共4小题，每小题5分，共20分．在每小题给出的四个选项中，有多项 符合题目要求．全部选对的得5分，部分选对但不全的得2分，有错选的得0分． 9 10 11 12 AC BCD ABD AD 三、填空题：本题共4小题，每小题5分，共20分． 13． 第1页 共7页 5 − 2 14． ( −  , − 1 ) 15．108 16．1 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．（10分） （1） ( 3 n + 9 )  ( n + 1 ) 2 a n + 1 = ( n + 2 ) 3 a n  3 ( n ( n + + 3 ) 2 a 2 ) n + 1 = ( n ( n + + 2 1 ) a 2 ) n 即 ( n ( + n 3 + ) 2 a ) n2 + 1 = 1 3  ( n ( n + + 2 1 ) a 2 ) n ····································································· 2分 又 ( 1 ( + 1 + 2 1 ) ) a 2 1 = 1 3   (n+2)a   ，所以数列 n 是首项为  (n+1)2  1 3 ，公比为 1 3 的等比数列 从而 ( n ( n + + 2 1 ) a 2 ) n =  1 3  n (n+1)2 ，则a = ·················································· 5分 n (n+2)3n （2） a n = ( ( n n + + 2 1 ) )  2 3 n = n n + + 1 2  n 3 + n 1  n 3 + n 1 ··························································· 6分  S n  2 3 1 + 3 3 2 + + n 3 + n 1 2 3 n+1 1 2 3 n+1 设T = + + + ，则 T = + + + n 31 32 3n 3 n 32 33 3n+1两式相减得： 1 1 n−1 1−    2 2 1 1 n+1 2 9 3  n+1 T = + + + − = + − 3 n 31 32 3n 3n+1 3 1 3n+1 1− 3 第2页 共7页 = 2 3 + 1 6  1 −  1 3  n − 1  − n 3 + n + 1 1 = 5 6 − 2 2 n  3 + n 5 + 1 ················································ 9分 5 2n+5 5 2n+5 从而T = − ，故S  − ····················································· 10分 n 4 43n n 4 43n 18．（12分） （1）由已知及正弦定理得：2sinBcosA−sinA=2sinC 又在  A B C 中， s i n C = s i n ( A + B ) = s i n A c o s B + c o s A s i n B ························· 2分  2 s i n B c o s A − s i n A = 2 s i n A c o s B + 2 c o s A s i n B 即2sinAcosB=−sinA 又 s i n A  0 ，  c o s B = − 1 2 ········································································ 4分 又 0 B    ， B = 2 3   ，即角 B 的大小为 2 3  ·············································· 5分 （2） S  A B C = 1 2 a c s i n B = 4 3 a c ····································································· 6分 B D 是  A B C 的角平分线，而 S  A B C = S  A B D + S  B C D 3 1 1  ac= ABBDsin60+ BDBCsin60 4 2 2 3 3 即 ac= BD(a+c)， 4 4  B D = a a + c c ················································· 8分 B D = 2 ，ac=2(a+c) a + c  2 a c ，ac4 ac ，即ac16 ··············································· 10分 当且仅当 a = c = 4 1 3 时取等号，则S = acsinB 16=4 3 ABC 2 4 即ABC的面积的最小值为4 3 ······························································· 12分19．（12分） （1）在 第3页 共7页 B E 上取一点 N ，使得 B N = 1 2 N E ，连接 F N ， N M B D = 6 ，  B N = 1 6 B D = 1 ， N E = 2 ， E D = 3 AF 1 BN AF 1 = ， = = FE 2 NE FE 2 则FN AB ··························································································· 2分 F N  面 A B C ， A B  面 A B C ，  F N 面 A B C B N N D = C M M D = 1 5 ，  N M B C ······························································· 4分 N M  面 A B C ， B C  面 A B C ，  N M 面 A B C F N N M = N ，  面 F N M 面 A B C F M  面 F N M ，  F M 面 A B C ·························································· 5分 （2） A E ⊥ B D ，CE ⊥BD 2 所以二面角A−BD−C的平面角为AEC = ············································ 6分 3 又 AE CE = E，BD⊥面 A E C B D  面 A B D ，  面 A B D ⊥ 面 A E C 面ABD 面 A E C = A E ，过点C作 C H ⊥ A E ，则 C H ⊥ 面 A B D 则 C H C E s i n 3 2 3 C E  =  = C E = ( 3 3 ) 2 − 3 2 = 3 2 3 3 6 ，CH = 3 2 = ······························· 8分 2 2 即 C 到面 A B D 的距离为 3 2 6 M D = 5 6 C D 5 3 6 5 6 ，M到面ABD的距离为  = ······························ 9分 6 2 4 计算EM： c o s  C D B = 3 3 3 = 1 3 5 3 在DME中，DM = ，DE =3 2第4页 共7页  1 3 =  5 2 3 2   2 5 + 2 3 3 2 −  E 3 M 2  E M = 5 2 1 ··············································· 11分  E M 与面 A B D 所成角的正弦值为 5 45 2 6 1 = 5 3 3 4 4 ········································· 12分 （其他方法酌情给分） 20．（12分） （1）列联表如下： 感兴趣 不感兴趣 合计 男生 12 4 16 女生 9 5 14 合计 21 9 30 ·········································· 2分 K 2 = 3 0  1 ( 6 1  2 1  4 5  − 2 4 1   9 9 ) 2  0 . 4 0 8 2  2 . 0 7 2 所以没有85%的把握认为学生对“数学建模”选修课的兴趣度与性别有关 ············ 5分 （2）由题意可知X 的取值可能为0，1，2，3 C3 5 则P(X =0)= 5 = ··········································································· 6分 C3 42 9 P ( X = 1 ) = C 14 C C 39 25 = 1 2 0 1 ············································································ 7分 C2C1 5 P(X =2)= 4 5 = ············································································ 8分 C3 14 9 C3 1 P(X =3)= 4 = ················································································ 9分 C3 21 9 故X 的分布列为 X 0 1 2 3 5 10 5 1 P 42 21 14 21第5页 共7页 E ( X ) = 0  5 4 2 + 1  1 2 0 1 + 2  1 5 4 + 3  1 2 1 = 4 3 ··············································· 12分 21．（12分） （1）因为双曲线C以2x 5y =0为渐近线 设双曲线方程为 ( 2 x 5 y ) ( 2 x 5 y )  + − = ，即 4 x 2 5 y 2  − = ························· 1分 F ( 0 , 3 ) ， 0    ，即： y 2 5 x 2 4 1   − − − =   − − =9， 5 4 9 2 0 9   − = ，即 2 0  = − ····················································· 3分 所以双曲线 C 的方程为： y 4 2 − x 5 2 = 1 ··························································· 4分 （2）设直线 l : y = k x + 3 ， P ( x 1 , y 1 ) ， Q ( x 2 , y 2 )  5 y y = 2 − k x 4 + x 2 3 = 2 0  5 ( k x + 3 ) 2 − 4 x 2 = 2 0 化简得： ( 5 k 2 − 4 ) x 2 + 3 0 k x + 2 5 = 0 ··························································· 6分 此方程的两根为 x 1 , x 2  30k x +x =−   1 2 5k2 −4 ，则 25  x x =  1 2 5k2 −4 900k2 100 9k2 − ( 5k2 −4 )  PQ = 1+k2  − =10 1+k2  ( 5k2 −4 )2 5k2 −4 ( 5k2 −4 )2 4k2 +4 20 ( k2 +1 ) =10 1+k2  = ·············································· 8分 ( 5k2 −4 )2 5k2 −4  15k −12  PQ中点M 坐标为 − ,  ······················································· 9分  5k2 −4 5k2 −4  P Q 12 1 15k  中垂线方程为：y+ =−  x+  5k2 −4 k 5k2 −4 −27  −27  令x=0，y= ，T  0,  5k2 −4  5k2 −4则 第6页 共7页 T F = 3 + 5 k 2 2 7 − 4 = 1 5 5 k k 2 2 + − 1 4 5 ···························································· 11分  T P F Q = 1 2 5 5 0 5 k k( k 2 2 k 2 + − 2 − 1 4 + 4 5 1 ) = 3 4 ········································································ 12分 22．（12分） （1）令 f  ( x ) = e x ( − e x x ) e 2 x = 1 − e x x  0 ··································································· 1分  x  1 ，即函数 f ( x ) 的单调递增区间为 ( −  , 1 ) ，函数 f ( x ) 的单调递减区间为 ( 1 , +  ) ·········································· 2分 当 x = 1 2 时， f  1 2  = 1 2 e 12 = 2 1 e ，  切点为  1 2 , 2 1 e  1 又 f 1 = 2 = 1 ，   2 1 2 e e2  f ( x ) 在 x = 1 2 处的切线方程为： y − 2 1 e = 2 1 e  x − 1 2   y = 2 1 e x + 4 1 e ················································· 4分 （2） e x e 2 x  k  ( l n x + 1 ) ， x e x  k  l n x e x + 1 = k  l n e x ln + x + 1 1 x(1,+) ，  l n e x ln + x + 1 1  0 ，  k  l n e x x ex + ln x + 1 1 ··············································· 6分 由（1）可知 f ( x ) = x e x 在 (1,+) 上单调递减，下证： x  l n x + 1 即证：x−lnx1在x(1,+) 恒成立 令 g ( x ) = x − l n x ，则 g  ( x ) = 1 − 1 x = x − x 1  0 g(x) 在 (1,+) 上单调递增 又 x1，g(x)g(1)=1−ln1=1 ························································ 9分 xlnx+11第7页 共7页 f ( x ) 在 x  ( 1 , +  ) 上单调递减  f ( x )  f ( l n x + 1 ) ，即 x e x  l n e x ln + x + 1 1 ，  l n e x x ex + ln x + 1 1  1 ································· 11分  k  1 ·································································································· 12分