东北三省精准教学2024年12月高三联考数学-简版答案及考点细目表_2024-2025高三（6-6月题库）_2024年12月试卷_1204东北三省精准教学2025届高三上学期12月联考

东北三省精准教学 2024 年 12 月高三联考 数学  4 1 AK  AB AC 所以 9 3 ， ·································································································11分 参考答案  4 3  9 AD AB AC AD AK 又 7 7 ，所以 7 ·················································································12分 ， 1 2 3 4 5 6 7 8 9 10 11 所以AD过点K，即AD，BE和CF三线交于一点K. ································································13分 （备注：按考点给分，漏步骤相应扣分；可使用建系方法解题，过程酌情给分） C A B D B B C C ABC BCD ACD 2 22 16．【答案】（1）证明见解析（6分） （2） （9分） 11 12. 21(5分） 【解析】（1）由勾股定理，AD2＝AC2﹣CD2=4，··········································································1分 13. 5/6/7（写出一个即可得满分5分） 满足PA2+PD2＝AD2，所以PA⊥PD．························································································2分 14.（0,1）（5分） 因为平面PAD⊥平面ABCD，平面PAD∩平面ABCD＝AD，CD⊂平面ABCD，CD⊥AD， 所以CD⊥平面PAD，············································································································4分 15．【答案】（1）a7，b5，c6（6分） （2）证明见解析（7分） 又PA⊂平面PAD，所以PA⊥CD．····························································································5分 【解析】（1）因为△ABC的周长为18，所以abc18， ···········································1分 因为PD，CD⊂平面PCD，且PD∩CD＝D，所以PA⊥平面PCD.···················································6分 由于b，c，a是递增的等差数列，故ab2c，·························································2分 （备注：证明PA⊥PD得2分，证明PA⊥CD得3分，证明PA⊥平面PCD得1分） 所以c6， ab12(ab)① ， ········································································3分 1 b2 c2 a2 （2）方法一：取AD的中点O，作OM//CD交BC于M，连接OP，则OP⊥AD， 又cosA  ②， ··········································································4分 5 2bc 因为平面PAD⊥平面ABCD，平面PAD∩平面ABCD＝AD，OP⊂平面PAD，所以OP⊥平面ABCD，以O为原点， 由①②，解得a7，b5，c6.··············································································6分 OA，OM，OP所在直线分别为x轴，y轴，z轴建立空间直角坐标系Oxyz，······································7分 （备注：推导出①得3分，否则按考点给分；推导出②得1分；联立①②得出a,b,c得2分） 所以A  1,0,0  ,P  0,0,1  ,C  1, 5,0  ,E     1 , 5 , 1   ，设B  a,b,0  ，a＞0，b＞0 ，·················8分 （2）由题意可得BD AE 3,CD AF 4,BF CE 2，  2 2 2  2  3   1 5 1        所以AF  AB，AE AC，······························································································7分 则EB   a ,b ,   ，AB  a1,b,0 ，CB  a1,b 5,0 ，CP  1, 5,1 .······9分 3 5  2 2 2     3  设BE和CF交于点K，由B，K，E三点共线，得AK t AB1t AE t AB 1t AC，··········8分  1 1 1 5 1 易知平面PAD的一个法向量为e  0,1,0 ，因为BE//平面PAD，    2   由C，K，F三点共线，得AK t 2 AF 1t 2 AC  3 t 2 AB1t 2 AC ， ·······································9分 所以  E  B  e  0，解得b  5 . ·························································································10分 2  2  4 t  t， t  ，  1 3 2  1 9   3 3  5 5  所以 3 1t 1t， 解得  t  2 ， ···················································································10分 因为AB⊥BC，所以ABCB 0，解得a  2 或 a  2 （舍），即CB   2 , 2 ,0   . ···········11分 5 1 2  2 3  设平面PBC的一个法向量为m x,y,z ， 第 1 页 共 4 页 {#{QQABbYSEggCgABBAAAgCUwEQCgEQkgACCagOwBAIoAAAyRFABAA=}#} m  ·C  P  0,   x 5yz 0,    因为AD⊥CD，所以BT⊥CD，且CT＝DT＝ 5 ．···································································12分 则    5 5 令x1，得 y  5,z  4，可得m 1, 5,4 ，··························12分 2 m·CB 0,  x y  0, 2 2 30 由CT2+BT2＝BC2，可知BC  ，  2 易知平面ABC的一个法向量n0,0,1， ··············································································13分   QN CT 6   mn 2 22 由△BTC∽△BNQ得  ，所以QN  ，····································································13分 则cos m,n     ，····························································································14分 BQ BC 3 m  n 11 1 1 EQ 6 EQ PO ，因此tanENQ  ， 2 22 2 2 QN 4 因为二面角PBCA为锐二面角，所以二面角PBCA的余弦值为 . ······························15分 11 2 22 cosENQ ··········································································································14分 （备注：建系给1分，写出点的坐标给1分，计算出平面PBC的法向量给4分，求出二面角的余弦值给3分， 11 ， 过程酌情给分） 2 22 所以二面角P﹣BC﹣A的余弦值为 ．·············································································15分 方法二：取AD的中点O，连接OP和OC，再取OC的中点Q，连接QE， 11 （备注：分析出∠ENQ为二面角给3分，分析出TB∥AD给2分，计算出∠ENQ的余弦值给3分，过程酌情给 在平面ABCD内过点Q作BC的垂线，垂足为点N，连接EN， 分） 因为PA＝PD，且O是AD的中点，所以PO⊥AD． 5 ln2 17．【答案】（1）m ，n （6分） （2）14小时（9分） 又平面PAD⊥平面ABCD，平面PAD∩平面ABCD＝AD，PO⊂平面PAD， 2 6 所以PO⊥平面ABCD．···········································································································7分 【解析】（1）由正午12点的城市活力度为20，知M(12)20，···················································1分 因为EQ是△POC的中位线，则EQ∥PO， 5 代入数据得12m56m20，解得m ，··········································································3分 所以EQ⊥平面ABCD． 2 24点到次日早上6点期间的城市活力度均为工作日内活力度的最低值， 因为BC⊂平面ABCD，所以BC⊥EQ．·······················································································8分 ln2 因为BC⊥QN，QN，EQ⊂平面ENQ，且QN∩EQ＝Q， 故M(24) M(6)5，代入数据得520e12n，解得n .······················································6分 6 所以BC⊥平面ENQ．又EN⊂平面ENQ，所以BC⊥EN， 5  t10,6„t„12, 由二面角的平面角的定义可知，∠ENQ即为二面角P﹣BC﹣A的平面角．·······································9分 （2）由（1）知M  t  2   ln2 (t12) 20e 6 ,12 t„24. 连接BQ，并延长BQ交CD于点T． 由EQ∥PO，PO⊂平面PAD，EQ⊄ 平面PAD，所以EQ∥平面PAD．···········································10分 当6t 12时，令M(t)10，解得t[6,8]，···········································································8分 当EB∥平面PAD时，EQ，EB⊂平面EBQ，且EQ∩EB＝E，  ln2 (t12) 当12t 24时，令 M(t)10 ，则 1020e 6 , ··································································10分 所以平面EBQ∥平面PAD．  ln2 (t12) 1 ln 1 ln2 1 由平面与平面平行的性质定理可知TB∥AD．············································································11分 e 6  e 2, (t12)ln ，·················································································11分 2 6 2 记AC交BT于F，因为点Q是OC的中点，TB∥AD，所以F是AC的中点， 解得t[18,24]，故一日内只有8＜t＜18时活力度大于10，···························································13分 由此可知BT＝FT+BF= AD+ AC= ，BQ2， 即该工作日内有14个小时活力度不大于10.·················································································15分 1 1 5 （备注：（1）问求得m给3分，求得n给3分；（2）问求出t[6,8]给2分，求出t[18,24]给5分，答案给2 2 2 2 第 2 页 共 4 页 {#{QQABbYSEggCgABBAAAgCUwEQCgEQkgACCagOwBAIoAAAyRFABAA=}#}分） （备注：求n1时a 的值得2分，求出公比得2分，写出a 的通项公式得3分） 1 n 1 18．【答案】（1）y (e1)x （6分） （2）证明见解析（11分） 2 n1 1 n 3 n （2）方法一：（i）由sin  sin  cos ·······························································9分 【解析】（1）对g(x)求导可得 f(x)ex ax，当a1时， f(x)ex x，········································1分 3 2 3 2 3 ， 1 n 3  n1 n 故 f(1)e1，g(1)e ，···································································································3分 得2ncos  2n1sin 2nsin  ··········································································12分 2 3 3  3 3  ， 故曲线 y  g(x)在点 1,g(1) 处的切线方程为 yg(1)= f(1)(x1)，················································4分 3 n1 故T  2n1sin 1.·······························································································14分 n 3 3 1 化简得 y (e1)x .············································································································6分 2 n1  3 2 3 n1 （ii）因为 sin 0, ，故 T 1  sin a a .·············································17分 n n n （备注：求导给1分，求出切线方程给3分，化简给2分） 3  2  3 3 （2）由第（1）问得 f  x ex ax，求导得 f' x ex a，···························································7分 （备注：（i）三角恒等变换拆分给2分，移项构造裂项相消得到T n 给5分；（ii）证明给3分）  n n 当a0时， f(x)单调递增，故 f(x)不存在极值，·········································································8分 方法二：（i）设数列a n cos 的通项公式为b n ，则b n 2ncos ，············································9分  3  3 当a0时，存在x R，使得 f' x 0，且x ln(a)，···························································9分 0 0 0 当n为6的倍数时， 此时在(,x 0 )上， f'(x)0， f(x)单调递减，在(x 0 ,)上， f'(x)0， f(x)单调递增，此时 f(x)存在极值 T n (b 1 b 2 ...b 6 )(b 7 b 8 ...b 12 )...(b n 5 b n4 ...b n )63(1 26 212  2n6) 2n 1 f(x 0 )，·······························································································································11分 ·········································································································································12分 由计算得 f(x ) f(ln(a))eln(a) aln(a)aaln(a)，·························································12分 0 当n除6余1时，T n T n1 b n  2n1， a 设h(a)aaln(a)，a0，则h'(a)1ln(a) (1)ln(a)，··········································13分 a 当n除6余2时，T n T n1 b n 1， 当a1时，a1，h'(a)ln(a)0， 当n除6余3时，T n T n1 b n 2n1， 当1a0时，h'(a)ln(a)0，·····················································································15分 当n除6余4时，T n T n1 b n 2n1， 所以当a1时，h(a)取得最大值，·····················································································16分 故h(a)h(1)1ln11，即 f(x)的极值小于或等于1.··························································17分 当n除6余5时，T n T n1 b n 1， （备注：讨论极值 f(x )存在情况得5分，未讨论扣2分；构造函数给1分，求最大值给3分，答案给1分） 综上所述： 0 -1，n6n1或6n4，  3 n1 T n 2n 1,n6n2或6n3，·······························································································14分 19．【答案】（1）a 2n，n∈N*（7分） （2）（i）T  2n1sin 1，n∈N*（7分）；（ii）证明见  n n 3 3 2n 1,n6n或6n5. 解析（3分） 0，n6n1或6n4， （ii）|Tn1| ···································································16分 【解析】（1）当n1时，S 2a 1，解得a 2 ····································································2分 2n,n6n2或6n3或6n或6n5. 1 1 1 ， |Tn1|a 当n2时，S 2a 1，所以S S 2a 12a 1，···············································4分 n.·······················································································································17分 n1 n1 n n1 n n1 （备注：过程酌情给分） 即a 2a ，a 是首项和公比均为2的等比数列，a 2n n∈N*.··············································7分 n n1 n n ， 第 3 页 共 4 页 {#{QQABbYSEggCgABBAAAgCUwEQCgEQkgACCagOwBAIoAAAyRFABAA=}#}题号 题型 分值 考查的主要内容及知识点 难度 1 单选 5 复数的运算，复数的模 易 2 单选 5 充分条件与必要条件，诱导公式 易 3 单选 5 向量的平行，向量的运算 易 4 单选 5 三角函数的图象 中 5 单选 5 正四棱台，向量运算，数学文化 中 6 单选 5 等比数列与等差数列的综合 中 7 单选 5 双曲线的离心率 中 8 单选 5 抽象函数，函数的周期性 中难 9 多选 6 不等式，指数函数的单调性 易 点到平面距离，外接球，异面直线所成角，线面 10 多选 6 中 角 11 多选 6 函数的图象，函数的周期性、对称性与奇偶性 中难 12 填空 5 等差数列的前n项和 易 13 填空 5 直线和圆，集合 中 14 填空 5 通过导数解决不等式恒成立问题 中难 15 解答 13 等差数列，解三角形 易 16 解答 15 证明线面垂直，求二面角余弦值 中 17 解答 15 函数的实际问题，指数函数 中 18 解答 17 切线方程，函数的极值 中 19 解答 17 数列的通项公式与前n项和，数列的证明 难 第 4 页 共 4 页 {#{QQABbYSEggCgABBAAAgCUwEQCgEQkgACCagOwBAIoAAAyRFABAA=}#}