2024-2025二调参考答案（数学）０(1)_2025年1月_250121吉林省吉林市普通中学2024-2025学年高三上学期二模（全科）_吉林省吉林市普通中学2024-2025学年高三上学期二模试题数学Word版含答案

令ab1，则 f(1)0，A选项正确； 吉林地区普通高中 2024—2025 学年度高三年级第二次模拟考试 令ab1，则 f(1)0； 数学学科参考答案 令a1,b x，则 f(x)f(x)，B选项不正确； 1 1 1 令a x,b ，则 f(1) xf( ) f(x)0， 一、单选题：本大题共8题，每小题5分，共40分。 x x x 1 1 当x1时，0 1， f(x)0， f( )0，即当0 x1时， f(x)0. 1 2 3 4 5 6 7 8 x x 又 f(x)是奇函数，当1 x0时， f(x)0；当x1时， f(x)0. C B A C B C A D  f(0) f(1) f(1)0，C选项正确，D选项正确. 8. 教学提示 lnt （法二）当x0时， f(x)0. 令t [x](tN*)，则k  . t f(ab) f(a) f(b) 当x0时，   . lnt ab a b 原不等式的解集对应区间的长度为1，不等式k  的正整数解有且只有一个. t f(x) 根据等式特点，构造函数 log |x| (c0且c1)， lnx x c 易知 f(x) 在(0,e)上单调递增，(e,)上单调递减， x 即 f(x) xlog |x|(c0且c1). c ln3 ln2 又 f(3) ，f(2) f(4) ， 3 2 又当x1时， f(x)0，c1，即 f(x) xlog |x|(c1). c ln2 ln3  k  . 2 3 0,x0, 综上 f(x) 易得ACD选项正确，B选项错误. 二、多选题：本大题共3题，每小题6分，共18分。 xlog |x|,x0. c 9 10 11 三、填空题：本大题共3小题，每小题5分，共15分。 AC BCD ACD 3 2 6 12. 1 13. 14. (2分)； [0, ] （3分） 10. 教学提示 3 3 4 D.易证 sinx  x . 14. 教学提示 （法一）由CP 2PE 得，点P轨迹是以A为球心，1为半径的球面， 1 1  f (x)  sinx sin2x sinnx 2 n 2 n 又点P在平面SAB内，点P在以A为圆心，1为半径， 为圆 3 1 1 2  sinx  sin2x  sinnx 心角的圆弧上，因此点P的轨迹长度为 . 2 n 3 1 1 2  x  2x  nx 建系如图，设P(cos,0,sin)([0, ])，则AB(2,0,0),CP (cos1, 3,sin). 2 n 3 nx ABCP 2(cos1) cos1 11. 教学提示 cos AB,CP     . AB CP 2 (cos1)2 3sin2 52cos （法一）已知 f(ab)af(b)bf(a)， 令ab0，则 f(0)0； 第 1 页 共 6 页5t2 令t  52cos,t[ 3, 6],cos ， 3 2 即ABC 的面积S为 . ·························································································6分 4 3 t 3t2 t 6 π π cos AB,CP    [0, ]. （Ⅱ）因为CD为角C 平分线，C  ，所以ACDBCD . 2t 2 4 3 6 故直线CP与直线AB所成角的余弦值的取值范围为[0, 6 ]. 在ABC 中，S ABC  S ACD  S BCD ， 4 1 π 1 π 1 π 所以 absin  aCDsin  bCDsin ，··························································8分 （法二）设直线CP与直线AB所成角为， 2 3 2 6 2 6 取AB的中点M, CPM  ,PMA ， 1 2 2 3 2 2 2 6 由CD ，得 ab a b (ab)，所以ab (ab).·························10分 根据三余弦定理可知，coscoscos 2 4 8 8 8 6 1 2 CM tan  ，易知P 从点M运动至N 处，tan 逐渐减小，则cos 逐渐增大， ab 6 ab 1 PM 1 1 因为a0，b0，所以由基本不等式 ab  ，得 (ab)( )2， 2 6 2 由图可知，P从点M运动至N 处cos 逐渐增大， 2 2 6 6 则P在点M处时，cos取得最小值，此时cos0， 所以ab ，当且仅当ab 时取等号. 3 3 2 3 6 则P在点N 处时，cos取得最大值，此时coscoscos    ， 1 2 2 2 4 2 6 所以ab的最小值为 .·······················································································13分 3 6 故直线CP与直线AB所成角的余弦值的取值范围为[0, ]. 4 16．【解析】 （Ⅰ） f(x)的定义域为{x|x1}.············································································2分 四、解答题：本大题共5小题，共77分。 xex 15．【解析】 f(x) . ····································································································4分 (x1)2 （Ⅰ）由sin2 AsinAsinBcos2Bcos2C (1sin2B)(1sin2C)sin2C sin2B， 令 f(x)0，得x1或1 x0， 得sin2 Asin2Bsin2C sinAsinB.  f(x)的单调递减区间为(,1)，(1,0). ·····························································7分 由正弦定理得a2 b2 c2 ab.···················································································2分 a2 b2 c2 ab 1 ex ex 所以cosC    ， （Ⅱ）x(1,), m(x1)，x10m . ································9分 2ab 2ab 2 x1 (x1)2 π 因为C(0,π)，所以C  . ····················································································· 4分 3 ex ex(x1) 设g(x) ，x(1,)，g(x) . ···················································11分 在ABC 中，c 3，ab 6 ，由余弦定理c2 a2 b2 ab(ab)2 3ab， (x1)2 (x1)3 令g(x)0，则x1. 得( 3)2 ( 6)2 3ab，解得ab1. 当1 x1时，g(x)0，g(x)在(1,1)上单调递减； 1 π 1 3 3 所以S  absin  1  . 2 3 2 2 4 第 2 页 共 6 页当x1时，g(x)0，g(x)在(1,)上单调递增. ······················································13分 取 y 1，则z h，n (0,1,h).········································································13分 2 2 2 e e e 则g(x)  g(1) ，m ，即m的取值范围是(，). ····································15分 设平面BB C C 与平面PBB 夹角为， min 4 4 4 1 1 1 17.【解析】 |n n | |h1| 10 1 则cos|cosn ,n | 1 2   ，解得h2或 . （Ⅰ）直三棱柱ABC  A 1 B 1 C 1 中，AA 1 平面A 1 B 1 C 1 ， 1 2 |n 1 ||n 2 | 2 1h2 10 2 1 又A B 平面A B C ，AA  A B .·····································································3分 所以正四棱锥P ABB A 的高为2或 .·····································································15分 1 1 1 1 1 1 1 1 1 1 2 【此处也可直接证明：正四棱锥P ABB A 中，A B  AA .】 （建系方式二）以A为原点，AC，AA ，AB所在直线 1 1 1 1 1 1 分别为x轴、 y轴、z轴建立如图所示的空间直角坐标系， 又A B  AC ，AA AC  A ,AA ,AC 平面ACC A ， 1 1 1 1 1 1 1 1 1 1 1 1 1 此时可得平面BB C C 的一个法向量为n (1,0,1)， 1 1 1 A B 平面ACC A .····························································································5分 1 1 1 1 平面PBB 的一个法向量为n (1,0,h). 1 2 又A B 平面PA B ， 1 1 1 1 （法二）取BB 中点M,CC 中点N ，连接PM ，MN . 1 1 平面PA B 平面ACC A .·················································································7分 1 1 1 1 正四棱锥P ABB A ， 1 1 （Ⅱ）（法一）以A为原点，AA ，CA，AB所在直线分别为x轴、 y轴、z轴建立如图所示的空间 1 PB PB ，PM⊥BB . 1 1 直角坐标系，则B(0,0,2), B (2,0,2) ,C(0,2,0). 1 又直三棱柱ABC  A B C 中,四边形BB C C是矩形， 1 1 1 1 1 设正四棱锥P ABB A 的高为h，则P(1,h,1)， 1 1 M，N 分别为BB ，CC 的中点， 1 1 BB (2,0,0)，BC (0,2,2)，BP (1,h,1)，·························································9分 1 MN //B C ，B C  BB ，MN⊥BB ， 1 1 1 1 1 1 设平面BB C C 的一个法向量为n (x ,y ,z ). 1 1 1 1 1 1 ∠PMN 或其补角即为平面BB C C 与平面PBB 的夹角， 1 1 1  BB n 0  2x 0  x 0 则 1 1   1   1 ，  BCn 1 0 2y 1 2z 1 0 y 1 z 1 即cosPMN  10 或 10 .···············································································11分 10 10 取z 1，则 y 1， n (0,1,1).·······································································11分 1 1 1 设正四棱锥P ABB 1 A 1 的高为h,则 PM  h2 1 , 设平面PBB 的一个法向量为n (x ,y ,z )， 1 2 2 2 2 A B  AC 且A B  AC 2, 1 1 1 1 1 1 1 1  BB n 0 2x 0 x 0 则 1 2  2  2 ，  BPn 2 0 x 2 hy 2 z 2 0 x 2 hy 2  z 2 B 1 C 1  MN  2 2 . 第 3 页 共 6 页作PH⊥平面AA 1 C 1 C ，垂足为H ，连接NH ， MQ 2MP  2 1h2 ， 则NH 2h，PN  (2h)21 . 在MNQ中，由余弦定理得，NQ2  MN2 MQ2  2MN MQcosNMQ , 在PMN 中，由余弦定理得，PN2  PM2 MN2  2PMMN cosPMN , 即（22h）2 844h2 22 22 1h2 cosNMQ ,············································13分 即( (2h)2 1)2 ( h2 1)2 (2 2)2 2 h2 12 2cosPMN ,······························13分 10 1 当cosNMQ  时，h ， 10 2 10 1 当cosPMN  时，h ， 10 2 10 当cosNMQ   时，h 2 . 10 10 当cosPMN   时，h 2 . 1 10 所以正四棱锥P ABB 1 A 1 的高为2或 .·····································································15分 2 1 18.【解析】 所以正四棱锥P ABB A 的高为2或 .·····································································15分 1 1 2 1 1 （Ⅰ）（法一）x (1234)2.5， y (9096100108)98.5，················2分 4 4 （法三）取AA 中点F ,BB 中点M,CC 中点N ,连接NF、MN、MP、MF ,并延长NF、MP交于 1 1 1 4 点Q，取MN 中点E ,连接EP,EP、MF 交于点G.  (x  x)(y  y)(1.5)(8.5)(0.5)(2.5)0.51.51.59.529， i i i1 正四棱锥P ABB A 中，PB PB 1 1 1 4  (x  x)2 (1.5)2 (0.5)2 0.52 1.52 5， PM⊥BB . i 1 i1 又直三棱柱ABC  A B C 中,四边形BB C C是矩形， 4 1 1 1 1 1  (x  x)(y  y)  i i 29   所以b i1  5.8，a ybx98.55.82.584， M，N 分别为BB ，CC 的中点， 4 5 1 1  (x  x)2 i i1 MN//B C ，B C  BB ，MN⊥BB ， 1 1 1 1 1 1  所以国家自然科学技术基金项目支出 y关于年份序号x的经验回归方程为 y5.8x84.······· 5分 NMQ或其补角即为平面BB C C 与平面PBB 的夹角， 1 1 1  当x6时， y5.8684118.8（百亿元）, 10 10 即cosNMQ 或 .··················································································11分 预测2025年的国家自然科学技术基金项目支出为118.8百亿元.·········································6分 10 10 1 1 （法二）x (1234)2.5， y (9096100108)98.5，························2分 4 4 设正四棱锥P ABB A 的高为h，则PG 即为四棱锥P ABB A 的高， 1 1 1 1 4 4 x y 190296310041081014，x2 1491630， MP  MG2GP2  1h2 . MFQ中，GP  MF ，MF  FQ， i1 i i i1 i GP//FQ，又G为MF 的中点，GP 为MFQ的中位线，FQ2h. 第 4 页 共 6 页4 即抛物线方程为x2  y. ····························································································1分 x y 4xy  i i 101442.598.5   b i1  5.8，a ybx98.55.82.584，  4 x2 4x 2 3042.52 A n (x n ,x n 2),B n (x n ,x n 2),A n1 (x n1 ,x n 2 1 ). i i1 x2  x2 所以国家自然科学技术基金项目支出 y关于年份序号x的经验回归方程为  y 5.8x84.······5分 即k n  x n n   1 1  x n n  x n1  x n . ···················································································3分 当x6时，  y5.8684118.8（百亿元）, k n1 2k n ，x n2  x n1 2(x n1  x n ). 预测2025年的国家自然科学技术基金项目支出为118.8百亿元.·········································6分 x  x k  x  x 20，x  x 0， n2 n1 2. 1 2 1 n1 n x  x （Ⅱ）（ⅰ）零假设为 n1 n H :申报天元基金者的所在地区与性别无关联.   x  x 是以2为首项，2为公比的等比数列，即x  x 2n. ································5分 0 n1 n n1 n 根据列联表中的数据，经计算得到 12n x  x (x  x )(x  x )(x  x ) 121 22 2n1  2n 1(n2) n 1 2 1 3 2 n n1 12 200(65554535)2 800 2   8.0817.879.······················································9分 10010011090 99 x 1符合上式， 1 依据小概率值0.005的独立性检验，我们推断H 不成立，即认为申报天元基金者的所在地区与性 数列 x 的通项公式是x 2n 1.···········································································7分 0 n n 别有关联，此推断犯错误的概率不大于0.005.·······························································11分 （Ⅱ）（法一）A (x ,x2),A (x ,x2 ),A (x ,x2 ). n n n n1 n1 n1 n2 n2 n2 （ⅱ）把男生样本的满意度平均数记为x，方差记为s2； x x2  x2 直线A A 的斜率为 n1 n  x  x . 女生样本的满意度平均数记为 y，方差记为s2；总样本的满意度平均数记为z，方差记为s2 . n n1 x  x n1 n y n1 n 则x9，s2 7.19， y7，s2 6.79， 直线A A 的方程为： y x2 (x  x )(x x )，即(x  x )x y x x 0. x y n n1 n n n1 n n n1 n n1 根据男、女样本量按比例分配的分层随机抽样总样本平均数与各层样本平均数的关系， A 到直线A A 的距离为 n2 n n1 11 9 可得总样本的满意度平均数为z 9 78.1，·················································14分 20 20 (x  x )x  x2  x x (x  x )(x  x ) 322n1 11 9 d  n n1 n2 n2 n n1  n2 n1 n2 n  ， s2  [7.19(98.1)2] [6.79(78.1)2]8. n (x  x )2 1 (x  x )2 1 (x  x )2 1 20 20 n n1 n n1 n n1 总样本的满意度的平均数为8.1，方差为8.··································································16分  A A  (x  x )2 (x2  x2)2  (x  x )2[1(x  x )2]2n 1(x  x )2 . n n1 n1 n n1 n n1 n n1 n n1 n 并据此估计全体参加论坛活动的天元基金者的满意度的平均数为8.1，方差为8.·················17分 1 1 322n1 S n  2 A n A n1 d n  2 2n 1(x n1  x n )2  (x  x )2 1 323n. ·······················12分 19.【解析】 n n1 1 （Ⅰ）证明：点A (1,1)在抛物线x2 2py(p0)上，2p1， （法二）证明：在ΔABC 中，AB(x，y )，AC (x ，y )，则ΔABC 的面积S  x y  x y . 1 1 1 2 2 2 1 2 2 1 第 5 页 共 6 页ABAC 证明如下：cosBAC  ， AB AC 1 1 1 S  AB ACsinBAC  AB AC 1cos2BAC  (AB AC )2 (ABAC )2 2 2 2 1  (x 2  y 2)(x 2  y 2)(x x  y y )2 2 1 1 2 2 1 2 1 2 1 1  2 x 1 2y 2 2  x 2 2y 1 2 2x 1 x 2 y 1 y 2  2 x 1 y 2  x 2 y 1 . ·························································9分 下面求ΔA A A 的面积S . n n1 n2 n A A (x  x ，x2  x2 )(2n，2n(32n 2))， n1 n n n1 n n1 A A  (x  x ，x2  x2 ) (2n1,2n1(32n1 2)). n1 n2 n2 n1 n2 n1 1 S  2n2n1(32n1 2)2n1[2n(32n 2)] 323n.···································12分 n 2 n(n1) （Ⅲ）a n log 2 (x n 1)n，T n  2 . ·····························································14分 a2 2T 16，(r1)2 k(k1)16. r1 k 4(r1)2 4k(k1)64(2k1)2 63. (2r2k3)(2r2k1)6363121397.r,kN*， 2r2k363 2r2k321 2r2k39  或 或 .  2r2k11  2r2k13 2r2k17 r 15 r 5 r 3 解得 或 或 （舍）. k 15 k 4 k 0 r 15，k 15或r 5，k 4.················································································ 17分 第 6 页 共 6 页