山东省青岛市2025年高三年级第三次适应性检测数学答案_2025年5月_250529山东省青岛市2025年高三年级第三次适应性检测（青岛三模）（全科）

2025 年高三年级第三次适应性检测 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：DABC CACA 二、多项选择题：本题共3小题，每小题6分，共18分． 9．BD 10．ACD 11．BCD 三、填空题：本题共3个小题，每小题5分，共15分． 5 7 3 12． ； 13． ； 14． 5100 8 4 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)销售额不少于60万元的有3天，X 的所有可能取值为1,2,3························1分 C1 3 C1C2 3 C3 1 P(X 1) 3  ,P(X 2) 2 3  ,P(X 3) 3  ······················ 4分 C3 10 C3 5 C3 10 5 5 5 所以X 的分布列为： X 1 2 3 3 3 1 P 10 5 10 3 3 1 9 所以E(X)1 2 3  ································································6分 10 5 10 5 1 1 (2)x  （24568）5， y  （2030606080）50，·······················7分 5 5 5 x y 2204305606608801460··········································8分 i i i1 5 x2 5x2 41625366452514512520····································9分 i i1 14605550 b ˆ 10.5，aˆ  y b ˆ x 5010.552.5······························ 11分 20 所以经验回归方程为yˆ 10.5x2.5，··························································· 12分 当x10时，yˆ 102.5 答：所求方程为yˆ 10.5x2.5，预测销售额为102.5万元································· 13分 16．（15分） ur 1 解：（1）由直线过原点且方向向量为(3,1)可知l的方程为 y  x·················· 1分 3 1 因此对任意正整数n，a  a 即a 3a ·················································2分 n 3 n1 n1 n 数学评分标准 第1页（共6页）因为a 2a 6，所以9a 6a 6，a 20················································3分 3 2 1 1 1 a 所以 n1 3，{a }是首项为2，公比为3的等比数列，···································5分 a n n 所以a 23n1 ·························································································7分 n （2）因为数列{b }是以4为首项，2为公差的等差数列， n 所以b 2n2··························································································9分 n 23n1 2 因为a  23n1 ，所以2m2 23n1 ，即m 3n1 1·········· 12分 n 2 3n 1 所以c 3n1 1，解得S  n·························································15分 n n 2 17（15分） 解：（1）EA,CD均垂直于平面ABC， \EA//CD ··································································································1分 取AB中点M ，连接FM,CM ， FE = FB,AM = BM ，\FM // AE 且FM =1， 又CD// AE且CD =1，故四边形FMCD是平行四边形·····································3分 \FD//CM ， 又FDË平面ABC，CM Ì平面ABC， \FD//平面ABC·························································································6分 （2）令DC =a(a>0)，取AC中点为O，连接BO，过O作OZ // AE，且交DE于Z， OZ // AE，AE ^平面ABC，OZ ^平面ABC， ΔABC 是正三角形，所以AO=CO， \OB^ AC ，·····························································································7分    以O为坐标原点，OB,OC,OZ 方向为x,y,z 轴正方向，建立空间直角坐标系O-xyz， 则B( 3,0,0)，E(0,-1,2)，D(0,1,a)，C(0,1,0)，   所以EB=( 3,1,-2)，ED=(0,2,a-2)，··························································9分    ì ïn×EB=0 ì ï 3x+y-2z=0 设平面BED法向量n=(x,y,z)，则í  ，所以í ， ï în×ED=0 ïî2y+(a-2)z=0  可取n=(a+2, 3(2-a),2 3)，····································································12分  又因为CE =(0,-2,2)，设CE与平面BED所成角为θ ，   |CE×n| 6 a 6 1 sinθ =   = ´ = ·······································14分 |CE|×|n| 4 a2 -2a+7 4 1+ 7 - 2 a2 a 7 所以，当a=7时，sinθ最大值为 4 7 综上，直线CE与平面EBD所成角正弦值的最大值为 ···································15分 4 数学评分标准 第2页（共6页）18．（17分） 解：（1）因为 A(1,1)在E 上，所以12p，···················································· 2分 所以E 的方程为 y2  x················································································ 3分 （2）若l与x轴平行，不合题意； 若l不与x轴平行，则l与E 相切于 A， y2  x 设l:x myn，由 ，得 y2 myn 0， x myn  m2 4n 0 m 2 所以 ，解得： ， 1mn n1 x1 所以l: y  ··························································································6分 2 1 1 因为B(1,0),F( ,0),A(1,1),D(0, )，故D为AB中点， 4 2 1 3 所以，AB中垂线的方程为y2x ，BF 中垂线的方程为x ， 2 8 3 5 因此VABF 的外接圆圆心坐标为( , )， 8 4 3 5 5 5 VABF 的外接圆半径等于 ( 1)2 ( 0)2  8 4 8 3 5 125 所以，VABF 外接圆方程为(x )2 (y )2  ········································10分 8 4 64     (3)（解法1）设CPCA(01)，CQCB(01)       CPCA(CPPA)(01)，故(1)CPPA   |PA| 1 |QB| 1 所以   ，同理   ， |CP|  |CQ|  |PA| |QB| 1 1 1 1 因为  1，所以  1，所以  3， |CP| |CQ|        1 1 设CG mCD(0m1)，因为CD CA CB， 2 2  m m m  m  所以CG  CA CB CP CQ 2 2 2 2 m m 2 因为P,G,Q三点共线，所以  1，所以m ··············14分 2 2 3  2 CG  CD，设C(x ,y ),y 2  x ,y 1,G(x,y)， 3 0 0 0 0 0  x x 0   1   3 因为CG (xx ,y y )，CD(x ,  y )，所以 0 0 0 2 0  y 1 y 0  3 数学评分标准 第3页（共6页）所以(3y1)2 3x，····················································································16分 2 因为A,C 不重合，所以y 1，y ， 0 3 2 综上，点G 的轨迹方程为(3y1)2 3x(y )·················································17分 3 （解法2）因为D为AB中点， |CD| |CA| |AP| |CB| |QB| 记 ,t  1 ,t  1 ， |CG| 1 |CP| |CP| 2 |CQ| |CQ| 则t t 3因为CD为△ABC 中线， 1 2 S |CQ||CP| 1 S S S 1 1 1 t t △CQP   且 △CQP  △CGP  △CGQ  (  ) 1 2 ， S |CB||CA| tt S 2S 2S 2 t t 2tt  △CAB 1 2 △CAB △CAD △CBD 1 2 1 2 t t 1 3 故 1 2  ，所以 ，所以G是△ABC 的重心·······································14分 2tt  tt 2 1 2 1 2  x x 0  3   y 1 设C(x ,y ),y 2  x ,y 1,G(x,y)，所以y 0 ，解得(3y1)2 3x············· 16分 0 0 0 0 0  3 y2  x  0 0  2 因为A,C 不重合，所以y 1，y ， 0 3 2 综上，点G 的轨迹方程为(3y1)2 3x(y )·················································17分 3 （解法3）设G(x,y),P(x ,y ),Q(x ,y ),C(y2,y )(y 1),G(x,y)， 1 1 2 2 0 0 0 |PA|   设 ，则APPC ，所以(x 1,y 1)(y2x ,y y )， |CP| 1 1 0 1 0 1 1y2 1y 所以x  0 ,y  0 ， 1 1 1 1 |QB| 1y2 y 设 ，同理可得：x  0 ,y  0 ， |CQ| 2 1 2 1 1y 1y2 y 0 x 0 1 1 直线PQ的方程为：  ， y 1y 1y2 1y2 0  0 0  0 1 1 1 1 又因为1，化简得：[()y (1)]x[()y23]y1 y y2①， 0 0 0 0 1 y  直线CD的方程为：y 0 2 x 1  (2y 0 1)x y 0 2 ②，·································· 14分 y2 2 2y2 0 0 数学评分标准 第4页（共6页） y2 x 0 由①②解得： 3 ，消去y 得：(3y1)2 3x，······································16分 1 y 0 y 0  3 2 因为A,C 不重合，所以 y 1， y ， 0 3 2 综上，点G的轨迹方程为(3y1)2 3x(y )·················································17分 3 19．（17分） sinx sin1 1 解：（1）m(x)= 在[0,1]上单调递增，且m(0)=0，m(1)= < <1， 2 2 2 所以0m(x)1···························································································1分 sinx -sinx 对任意的x ,x Î[0,1]，|m(x )-m(x )|=| 1 2 |， 1 2 1 2 2 令 f(x)sinxx(0x1)， f ¢ (x)=cosx-1£0 ， f(x)在[0,1]上单调递减， 不妨设0 x  x 1，则 f(x )³ f(x )，sinx -x ³sinx -x ， 1 2 1 2 1 1 2 2 所以x -x £sinx -sinx £0，故|sinx -sinx |£|x -x |， 1 2 1 2 1 2 1 2 1 1 所以|m(x )-m(x )|£ |x -x |，故λ取 即可． 1 2 2 1 2 2 所以m(x)ÎP·····························································································4分 1 （2）对任意的x[0, 1 ](nN*)，nxÎ[0,1]，m ¢ (x)= 1 ( 1 -x),xÎ[0,1]， n 2n+1 x+1 5-1 - 5-1 令m ¢ (x)=0解得：x= 或x= （舍）···········································5分 2 2 5-1 5-1 所以m(x)在(0, )上单调递增，在( ，1)上单调递减； 2 2 1 1 又因为m(0)=0，m(1)= (ln2- )>0 ， 2n+1 2 所以m(nx)³0． 1 令g(x)ln(x1)x(0x1)，g ¢ (x)= -1£0，所以g(x)在[0,1]上单调递减， x+1 x2 x2 所以g(x)£g(0)=0，ln(x+1)£ x，ln(x+1)- £ x- £1， 2 2 1 1 所以0m(nx)  ， 2n1 n 1 所以0m(nx) ························································································7分 n 1 对任意的x ，x Î[0, ]， 1 2 n 数学评分标准 第5页（共6页）1 n2x2 n2x 2 因为|m(nx )-m(nx )|= |ln(nx +1)-ln(nx +1)- 1 + 2 | ， 1 2 2n+1 1 2 2 2 n2x2 n2x 2 又因为|ln(nx +1)-ln(nx +1)- 1 + 2 | 1 2 2 2 n2x2 n2x 2 £|ln(nx +1)-ln(nx +1)|+| 1 - 2 | 1 2 2 2 n2x2 n2x2 n 其中| 1 - 2 |= | nx +nx || x - x |£ n| x - x | ， 2 2 2 1 2 1 2 1 2 1 令h(x)=lnx-x(x³1)，h(x) 10，h(x)在[1,+¥)上单调递减， x 1 不妨设0 x  x  ，1£nx +1£nx +1£2，则h(nx +1)³h(nx +1)， 1 2 n 1 2 1 2 所以ln(nx +1)-(nx +1)³ln(nx +1)-(nx +1) ， 1 1 2 2 所以0³ln(nx +1)-ln(nx +1)³nx -nx ， 1 2 1 2 所以|ln(nx +1)-ln(nx +1)|£|nx -nx |=n|x -x | ， 1 2 1 2 1 2 2n 2n 所以|m(nx )-m(nx )|£ x -x ，故λ取[ ,1)内常数即可． 1 2 2n+1 1 2 2n+1 所以m(x)ÎP ··························································································· 11分 n （3）因为|a -a |=|m(ta )-m(ta )|£ λ| a - a | ， n+1 n n n-1 n n-1 所以|a a ||a a |n1|a a | ·················································13分 n1 n n n1 2 1 又因为|a -a |=|(a -a )+(a -a )++(a -a )| k+l k k+l k+l-1 k+l-1 k+l-2 k+1 k £|a -a |+|a -a |++|a -a | k+l k+l-1 k+l-1 k+l-2 k+1 k £ λk+l-2 |a -a |+λk+l-3 |a - a |++ λk-1| a - a | 2 1 2 1 2 1 =(λk+l-2+λk+l-3++ λk-1)| a - a | 2 1 λk-1-λk+l-1 = |a -a | 1-λ 2 1 λk-1 £ |a -a |···································································17分 1-λ 2 1 数学评分标准 第6页（共6页）