数学参考答案_2024年3月_013月合集_2024届山东省青岛市高三年级第一次适应性检测_青岛市2024年高三年级第一次适应性检测（青岛一摸）数学

2024 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：ADBA CCBA 二、多项选择题：本题共3小题，每小题6分，共18分． 9．AB 10．AC 11．BCD 三、填空题：本题共3个小题，每小题5分，共15分． 51 12．0； 13． ； 14．1 7． 2 四、解答题：本题共5小题，共77分．解答应写出文字说明，证明过程或演算步骤． 15. （13分） 解：（1）由题知：各组频率分别为：0.15，0.25，0.3，0.2，0.1······················3分 日均阅读时间的平均数为： 300.15500.25700.3900.21100.167（分钟）······················· 6分 （2）由题意，在[60,80),[80,100),[100,120]三组分别抽取3,2,1人·······················7分 的可能取值为：0,1,2·················································································8分 C3C0 1 则P(0) 4 2  ················································································9分 C3 5 6 C2C1 3 P(1) 4 2  ···············································································10分 C3 5 6 C1C2 1 P(2) 4 2  ·············································································· 11分 C3 5 6 所以的分布列为：  0 1 2 1 3 1 P 5 5 5 1 3 1 E()0 1 2 1······································································13分 5 5 5 16．（15分） x2  x1 解：（1）当a1时， f (x)  ， f (x ) 1解得x 1························3分 x 0 0 1 3 又因为 f(1) ，所以切线方程为：x y 0···········································5分 2 2 x2 ax1 （2） f(x)的定义域为(0,)， f (x)  ········································6分 x 当a0时，得 f(x)0恒成立， f(x)在(0,)单调递增·································8分 数学评分标准 第1页（共5页） {#{QQABYYCEggigAhBAAQgCEwEICEKQkAACACoOQEAAoAAAyQNABCA=}#}当a 0时，令g(x) x2 ax1,  a2 4····················································9分 （ⅰ）当0即0a2时， f(x)0恒成立， f(x)在(0,)单调递增·········································· 11分 a a24 a a24 (x )(x ) （ⅱ）当 0即a 2时， f(x) 2 2 ···············12分 x a a2 4 a a2 4 由 f(x)0得，0 x 或x ， 2 2 a a2 4 a a2 4 由 f(x)0得，  x 2 2 a a2 4 a a2 4 所以 f(x)在(0, ),( ,)单调递增， 2 2 a a2 4 a a2 4 在( , )单调递减···········································14分 2 2 综上：当a2时， f(x)在(0,)单调递增； a a2 4 a a2 4 当a 2时， f(x)在(0, ),( ,)单调递增； 2 2 a a2 4 a a2 4 f(x)在( , )单调递减·······················15分 2 2 17．（15分） 解：(1) 取棱AA中点D，连接BD，因为AB  AB ，所以BD  AA ·················1分 1 1 1 因为三棱柱ABC ABC ，所以AA //BB ······················································2分 1 1 1 1 1 所以BD BB ，所以BD  3·····································································3分 1 因为AB2，所以AD1，AA 2； 1 因为AC 2，AC 2 2 ，所以AC2  AA2  AC2 ，所以AC  AA ，············4分 1 1 1 1 同理AC  AB，·························································································5分 因为AA AB A,且AA,AB平面AABB ，所以AC 平面AABB ， 1 1 1 1 1 1 因为AC 平面ABC， z 所以平面AABB 平面ABC ··············6分 A B 1 1 1 1 （2）取AB中点O，连接A 1 O， C 1 取BC中点P，连接OP，则OP// AC， D 由（1）知AC 平面AABB ， 1 1 O 所以OP平面AABB ·······················7分 A 1 1 B y 因为AO平面AABB ，AB平面AABB ， 1 1 1 1 1 M P 所以OP AO，OP AB， C x 1 因为AB  AA AB ，则AO AB······························································8分 1 1 1 数学评分标准 第2页（共5页） {#{QQABYYCEggigAhBAAQgCEwEICEKQkAACACoOQEAAoAAAyQNABCA=}#}以O为坐标原点，OP，OB，OA 所在的直线为x轴、 y 轴、z轴，建立如图所示的空 1 间直角坐标系Oxyz，则A(0,1,0)，A(0,0, 3)，B (0,2, 3),C(2,1,0)， 1 1 设点N(a,0, 3)，(0a2)···········································································9分    AB (0,2,0)，AC (2,1, 3)，AN (a,1, 3)， 1 1 1 1     nAB 0 2y 0 设面ABC的法向量为n(x,y,z)，得  1  1 ，得 ， 1 1 nAC 0 2x y 3z 0 1  取x 3，则 y 0，z 2，所以n( 3,0,2)·············································10分 设直线AN 与平面ABC所成角为， 1 1     |nAN | 3 a2 则sin|cosn,AN |     |n|| AN | 7 a2 4 3 (a2)2  7 a2 4 3 a2 4a4  ································11分 7 a2 4 21 若a 0，则sin ，·········································································12分 7 3 4 3 4 42 若a0，则sin 1  1  ，·······························13分 7 4 7 4 7 a a 4 当且仅当a ，即a 2时，等号成立，·······················································14分 a 42 所以直线AN 与平面AMB所成角的正弦值的最大值 ·································15分 1 7 18．（17分） 解: （1）设M(x,y)，切点为N ，则|MN |2|MW |2|OM |2 |OW |2 ， 所以|x2|2 x2  y2 4··············································································3分 化简得 y2 4x，所以C的方程为： y2 4x···················································· 4分     （2）（ⅰ）因为l //l ，所以可设GAGA,GBGB ， 1 2  1       又因为GE  (GAGB) (GAGB)GF， 2 2 所以G,E,F 三点共线，同理，H,E,F三点共线， 所以G,E,H 三点共线···················································································6分 （ⅱ）设A(x ,y ),B(x ,y ),A(x ,y ),B(x ,y ) ，AB中点为E，AB中点为F ， 1 1 2 2 3 3 4 4 数学评分标准 第3页（共5页） {#{QQABYYCEggigAhBAAQgCEwEICEKQkAACACoOQEAAoAAAyQNABCA=}#}将x ym代入 y2 4x得： y2 4y4m0，所以 y  y 4,y y 4m， 1 2 1 2 y  y 所以 y  1 2 2，同理 y 2（G,E,H,F 均在定直线y 2上）················8分 E 2 F 因为TT//l ，所以EAT 与EAH 面积相等，EBT与EBH 面积相等； 1 所以四边形GTET面积等于四边形GAHB 面积··············································· 10分 设G(x ,2),H(x ,2)， G H y  y y  y y2 直线AA: y y  1 3 (xx ) ，即 y y  1 3 (x 1 )， 1 x x 1 1 y2 y2 4 1 3 1  3 4 4 4x y y 2(y  y ) y y 整理得：直线AA: y  1 3 ，又因为 y 2，所以x  1 3 1 3 ， y  y G G 4 1 3 4x y y 2(y  y ) y y 同理，直线BA: y  2 3 ， y 2，所以x  2 3 2 3 ·············12分 y  y H H 4 2 3 y  y (y  y )( 3 4  y ) (y y )(2y ) 1 2 2 3 所以|GH ||x x | | 1 2 3 | | | G H 4 4 |(y  y )(y  y )|  1 2 3 4 ····························14分 8 1 (y  y )2| y  y | 所以四边形GAHB 面积S  |GH || y  y | 1 2 3 4 2 1 2 16 [(y  y )24y y ] (y  y )24y y  1 2 1 2 3 4 3 4 16 (1616m) 1616n  16 4 (1m)2(1n) (1m)2 1n 4[ ]2(2m2 2mn)16············16分 2 m2 2mn m1 当且仅当(1m)2 1n，即 ，即 时取等号， nm2 2m6 n3 所以GAT 面积的最大值为16······································································17分 数学评分标准 第4页（共5页） {#{QQABYYCEggigAhBAAQgCEwEICEKQkAACACoOQEAAoAAAyQNABCA=}#}19．（17分） 解：（1）因为m x3 (ab a b a b a b )x3 ， 4 1 4 2 3 3 2 4 1 所以m ab a b a b a b ··································································· 4分 4 1 4 2 3 3 2 4 1 （2）因为 f({a }{b }) f({a }) f({b }) ， n n n n 所以 f({a }) f({b }) f({c }) f({a }{b }) f({c }) f(({a }{b }){c }) ， n n n n n n n n n 又因为 f({a }) f({b }) f({c }) f({a })[f({b }) f({c })] n n n n n n  f({a }) f({b }{c }) n n n  f({a }({b }{c })) n n n 所以 f(({a }{b }) f{c }) f({a }({b } f{c })) n n n n n n 所以({a }{b }){c }{a }({b }{c }) ···············································10分 n n n n n n （3）对于{a },{b }S ， n n 因为(a a xa xn1)(b b xb xn1)d d xd xn1 1 2 n 1 2 n 1 2 n 所以d xn1 a (b xn1)a xk1(b xnk)a xn2(b x)a xn1b ， n 1 n k n1k n1 2 n 1 所以d ab a b a b a b a b ， n 1 n 2 n1 k n1k n1 2 n 1 n 所以{a }{b }{d }{a b }····························································13分 n n n k n1k k1 200 100 200 100 100 (k1)2 1 d a b a b   a b a b  ·········14分 200 k 201k k 201k k 201k k 201k k(k 1)2k2 k1 k1 k101 k1 k1 100 1 2 1 所以d  (1  ) 200 2k2 k k 1 k1 100 1 100 1 1  [  ] 2k2 k2k1 (k1)2k2 k1 k1 1 102 1    ········································································17分 2 1012102 2 数学评分标准 第5页（共5页） {#{QQABYYCEggigAhBAAQgCEwEICEKQkAACACoOQEAAoAAAyQNABCA=}#}