湖北省武汉市七彩联盟2024-2025学年度下学期高一年级6月联考数学答案_2024-2025高一（7-7月题库）_2025年7月_250702湖北省武汉市七彩联盟2024-2025学年度下学期高一年级6月联考

七彩联盟高一期末考试数学答案 2025.06.25 1 2 3 4 5 6 7 8 单选 A C C B B D A B 9 10 11 多选 BCD ACD AC 3  填空 12.【答案】100 13.【答案】 13.【答案】2 2 2 2 1 15.【答案】（1）a ,b1； （2）a4,b3 2 【详解】（1）解法一：由已知有，方程az2bz10的两根为1i，··························2分  b  1i1i2   a 由根与系数的关系得 ····························································5分  1 1i1i2  a 1 a ,b1··································································································6分 2 解法二：由已知有a1i2 b1i10，····························································2分  b10 b12abi0由复数的相等得 ，···················································5分 2ab0 1 a ,b1··································································································6分 2 （2）设zabia,bR，················································································7分 3a a2 b2 7 3abi a2 b2 79i，由复数的相等有 ，································9分  3b9 解得a 4,b3.······························································································13分 5 （注：a 是增根，未舍去的扣2分） 4 16.【答案】（1）证明见解析； （2）2 2 5 【详解】（1）证明：在ABC中，∵AB2AC2 BC2 ，∴AC  AB，·························2分 在直三棱柱ABCABC 中，AA 平面ABC，∴AA AC， 1 1 1 1 1 ∵AA ,AB是平面AABB内两条相交直线， 1 1 1 ∴AC 平面AABB，∴AC  AB. ····································································5分 1 1 1 又知四边形AABB是正方形，∴AB AB. 1 1 1 1 ∵AC,AB 是平面ABC内两条相交直线，∴AB 平面ABC. 得证.·····························8分 1 1 1 1（2）在直三棱柱ABCABC 中，BC //BC， 1 1 1 1 1 ∴BC 与平面ABC所成的角等于BC平面ABC所成的角，设大小为，·······················10分 1 1 1 1 设AB ABO，连接OC. 1 1 由（1）知，BO平面ABC，∴BC平面ABC所成的角为BCO，·····························13分 1 1 BO 2 2 在RtABC中，sinsinBCO   . ·························································15分 BC 5 17.【答案】（1）a0.012； （2）93；（3）792 【详解】（1）解法一：设成绩在90,150的频率为p，则成绩在30,90的频率为1 p， 根据题目x 65，x 115，平均成绩 x97 ， 1 2 有p1151 p6597，解得p0.64；···························································· 2分 则根据频率分布直方图有0.014a0.006200.64， 解得a0.012. ································································································4分 解法二：根据频率分布直方图以及x 115，得 2 0.014 a 0.006 x  100 120 140115，···············2分 2 0.014a0.006 0.014a0.006 0.014a0.006 解得a0.012. ································································································4分 （2）设获得优胜奖的成绩为Y 分， 易计算得频率分布直方图成绩在90,110,110,130,130,150的频率分别为0.28、0.24、0.12； 则优胜奖成绩Y 位于90,110中，·········································································6分 Y 90 0.640.60 0.640.60 20 由此有  ，解得Y 90 90 93， 11090 0.01420 0.014 7 故以样本估计总体，估计获得优胜奖的成绩为93分. ···············································9分 （3）样本方差s2 10.64 x x 2 s2  0.64  x x 2 s2  ，·······························12分  1 1  2 2 代入有s2 0.36 65972200 0.64 115972225 351.36440.64792，     则样本的方差s2 792. ·····················································································15分 （注：为降低难度，减小运算量，本题增加了条件）    18.【答案】（1）A ,C ； （2） AD 2 31 3 4  1 S  bcsinA3 3 【详解】（1）由已知得 ABC 2 ，················································2分     ABAC bccosA6两式相除得tanA 3，  又0 A，∴A ；···················································································· 4分 3 cosA cosB cosC 又已知3  2 ，（下用两种方法求C，任一种都算对） a b c 解法一：根据余弦定理有 b2c2a2 a2c2b2 a2b2c2 3  2 ，化简得2a2 3c2，∴ 2a  3c················· 7分 2abc 2abc 2abc 2 2 由正弦定理得，sinC  sinA ， 3 2 2  又∵0C  ，∴C . ················································································9分 3 4 cosA cosB cosC 解法二：由正弦定理得，3  2 ()， sinA sinB sinC  ∵A ， 3 3tanC ∴tanB tan  AC  tan AC  ，代入()， 1 3tanC 1 3tanC 1 得 3 2 ，化简得tan2C1.·······················································7分 3tanC tanC 2  又0C  ，∴C . ···················································································9分 3 4   5 （2）∵A ，C  ，∴BAB ， 3 4 12 sinB 31 ∴b c c； sinC 2   又由（1）得bc12，∴c2 12 31 ，······························································· 11分 2 ∴b2    31 c   ，化简得4b2   31 2 c2 24  31  ，即b2 6  31  ；·············13分    2       ∵BD 2DC，∴3AD 2AC AB，·································································15分 两边同时平方有        2  2 2 2 9 AD  2ACAB 4 AC  4ABAC AB     4b2 46c2  24 31 2412 31 ，   36 31  化简得， AD 2 31.················································································· 17分  m     1  19.【详解】（1）证明：∵OP mPM  OM ，由NRtRO知OR ON . 1m 1t    m  1 又P,Q,R三点共线，∴OQOP1OR OM  ON①， 1m 1t    1  n  由MQ nQN 知OQ OM  ON ②， 1n 1n  m 1 1 m       1m 1n 1nm ∵OM,ON 不共线，由①②及平面向量基本定理有 ， 1 n 1tn  1  1t 1n 1n 1m 1tn ∴1  ，化简得mnt 1. 得证. ·····················································5分 1nm 1n （2）①证明：∵AC//，平面ACB经过AC且与平面相交于EG， ∴AC//EG. 又E为AB的中点， ∴G为BC中点.······························································································ 7分 ∵F 为CD中点，∴FG//BD. ∵GF ,BD ∴BD//. 得证.····························································································· 9分 ②由已知有AC,EG,TF三线相交于点K. AE BG CK CK AT DF 由（1）有   1，   1, EB GC KA KA TD FC ∵E,F分别为AB，CD中点， BG AT ∴  .···································································································11分 GC TD CG DT 记  ，四面体ABCD的体积为V ，多面体ETBGFD的体积为V'. CB DA V BE S 1 S  1  连接EF,ED，则有 EBGFD   四边形BGFD  1 CGF  1 ，···························14分 V AB S 2 S  2 2 BCD BCD V AE S 1 DTDF  EFDT   FDT    ，··································································· 16分 V AB S 2 DADC 4 DAC V' V V 1 ∴  EBGFD EFDT  . V V 2 即平面平分四面体ABCD的体积. 得证. ·····························································17分