中学生标准学术能力诊断性测试2024届高三下学期3月测试数学试卷答案_2024年3月_013月合集_2024届中学生标准学术能力诊断性测试高三3月联考

中学生标准学术能力诊断性测试 2024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 D C B D A C B C 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 CD ABD BCD ABD 三、填空题：本题共4小题，每小题5分，共20分． 185 13．16 14．− 8 15． 645 16．a+b 451 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） 2tanC 3 （1） tan2C = =− ········································································· 2分 1−tan2C 4 10 解得tanC=3，故cosC = ·································································· 4分 10 10 （2） a2 +b2 =c2 +2abcosC =16+ ab2ab ·············································· 6分 5 80+8 10 解得ab ·················································································· 8分 9 3 由（1）知sinC = 10 1 4+4 10 故S = absinC  ································································ 10分 ABC 2 3 4+4 10 故ABC面积的最大值为 ． 3 第1页 共5页 {#{QQABbYAEggCAQBIAAQhCAwGYCAGQkACCCKoOABAEsAAAyRNABCA=}#}18．（12分） （1）对于n2时，a +4(n+1)=2a +8n=2(a +4n) ······································ 2分 n+1 n n a =2,a +41=6， 1 1 a +4n=32n,a =32n −4n ··································································· 3分 n n 经验算，a 符合上述结果，故a =32n −4n ················································· 4分 1 n （2）设b =n+3na =n+36n −4n3n， n n n(n+1) 18 ( 1−6n) 则S = + −4n3n ························································· 6分 n 2 1−6 设T =4n3n， n T =431+832 +1233+ +4n3n， n 3T =432 +833+1234+ +4n3n+1 ····················································· 7分 n 作差得到2T =−431−432 −433− −43n +4n3n+1 ······························ 8分 n −6 ( 1−3n) 故T = +2n3n+1 =(2n−1)3n+1+3 ·········································· 10分 n 1−3 n(n+1) 18 ( 1−6n) S = + −T n 2 1−6 n n2 n 18 33 故S = + + 6n −(2n−1)3n+1− ················································ 12分 n 2 2 5 5 19．（12分） （1）圆C 的方程可变形为(x−2)2 + y2 =4 ···························································· 2分 1 故C 的圆心坐标为 (2,0) ，半径为2 ······························································· 4分 1 （2）设M (x ,y ) ，因为点M是AB的中点，CM ⊥ AB， M M 1 k k =−1 ························································································ 6分 CM AB 1 y y 故 M  M =−1 ················································································ 8分 x −2 x −1 M M 第2页 共5页 {#{QQABbYAEggCAQBIAAQhCAwGYCAGQkACCCKoOABAEsAAAyRNABCA=}#}由此可得x2 −3x +y2 +2=0 ··································································· 10分 M M M  3 2 1 3  1 故轨迹方程为  x −  + y2 = ，轨迹是以圆心为 ,0 ，半径为 的圆 ······· 12分  M 2 M 4 2  2 20．（12分） （1）解：10x+100.005+100.01+10y+100.019+100.02+100.027=1， 故x+ y=0.019 ························································································· 1分 20010y−20010x=22， 故y−x=0.011 ························································································· 2分 解得x=0.004,y=0.015 ············································································ 3分 （2）①学院毕业生年薪在 30,80) 区间的人数比例为： (0.02+0.027+0.015+0.01+0.005) 10=77%， 故同类型合作办学高校毕业生平均年薪最高为30万元 ········································ 5分 ②对于单个毕业生，其年薪高于50万的概率P=(0.005+0.01+0.015)10=0.3，  3  故随机变量~ B  4, ，  10 故P(=0)=(1−0.3)4 =0.2401 ·································································· 6分 P(=1)=C1(1−0.3)3 0.3=0.4116 ························································ 7分 4 P(=2)=C2(1−0.3)2 0.32 =0.2646 ······················································ 8分 4 P(=3)=C3(1−0.3)0.33 =0.0756 ······················································· 9分 4 P(=4)=0.34 =0.0081 ········································································· 10分 的分布列为：  0 1 2 3 4 P 0.2401 0.4116 0.2646 0.0756 0.0081 第3页 共5页 {#{QQABbYAEggCAQBIAAQhCAwGYCAGQkACCCKoOABAEsAAAyRNABCA=}#}的数学期望E()=0.34=1.2 ······························································· 12分 21．（12分） (4x−a)ex −ex( 2x2 −ax+a ) −2x2 +(a+4)x−2a （1） f(x)= = ······················· 2分 ( ex)2 ex −2x2 +5x−2 当a=1时， f(x)= ， f(0)=−2 ············································ 4分 ex 又 f (0)=1，故曲线y = f (x) 在 ( 0, f (0)) 处的切线方程为y=−2x+1 ············ 5分 −2x2 +(a+4)x−2a (−2x+a)(x−2) （2） f(x)= = =0， ex ex a 解得知x =2,x = ··················································································· 7分 1 2 2 a   a 若a>4, f (x) 在 (−,2),  ,+ 递减， 2, 递增 ······································· 8分 2   2 a a f = >0 极大值   ··············································································· 9分 2 a e2 若a=4，函数单调递减，无极大值 ····························································· 10分  a a  若a<4, f (x) 在 −,  ,(2,+) 递减， ,2 递增 ····································· 11分  2 2  8−a 极大值 f (2)= >0 ············································································ 12分 e2 综上， f (x) 的极大值恒为正数． 22．（12分） x2 ( ) ( ) （1）椭圆E: + y2 =1的左，右焦点分别为F − 3,0 ,F 3,0 ， 1 2 4 设A(m,n),AF ⊥ AF ，故AF AF = ( − 3−m,−n )( 3−m,−n ) =0 ··············· 1分 1 2 1 2 即m2 +n2 =3 ··························································································· 2分 m2 2 6 3 又 +n2 =1，解得m= ,n= ························································· 3分 4 3 3 2 6 3 A点坐标为 ,  ············································································ 4分   3 3   第4页 共5页 {#{QQABbYAEggCAQBIAAQhCAwGYCAGQkACCCKoOABAEsAAAyRNABCA=}#}（2）设P点坐标为 (p,0) ，则可得Q点坐标为 ( 2 6−2p, 3 ) ································· 5分 2 ( )  6 OPOQ=(p,0) 2 6−2p, 3 =−2p2 +2 6p=−2p−  +3 ················· 7分   2   6 当 p = 时，OPOQ取最大值，最大值为3 ················································ 8分 2 2 6 3  2 6 3 （3）A点坐标为 , ，B点坐标为− , ，     3 3 3 3     3 点O到线段AB的距离h = ····································································· 9分 1 3 3 若S =2S ，则点M到线段AB的距离应为h = ， 1 2 2 6 3 3 故M点的纵坐标为 或 ，代入椭圆方程， 6 2 33 解得M点的横坐标为 或1 ······························································· 11分 3  33 3  3 故M点的坐标为： , 或1,  ·············································· 12分     3 6 2     第5页 共5页 {#{QQABbYAEggCAQBIAAQhCAwGYCAGQkACCCKoOABAEsAAAyRNABCA=}#}