四川省元三维大联考·高2022级第三次诊断性测试数学答案

高中2022级第三次诊断性考试数学参考答案及评分标准一、选择题：本大题共8小题，每小题5分，共40分．CBADBCAD二、选择题：本题共3小题，每小题6分，共18分．9．ABD10．ABD11．AD三、填空题：本大题共3小题，每小题5分，共15分．512．13．314．66四、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．sinAa715．解：（1）由正弦定理知：，·············································2分sinBb8∴8sinA7sinB，·········································································3分又8sinA7sinB83，··································································4分3∴16sinA83，则sinA，·······················································5分2又a0，1∴f(x)x3x23(x4)，····························································7分31设g(x)(x4)exx3x23x12，3∴g(x)(x3)exx22x3(x3)(exx1)，································9分易知ex≥x1，···········································································10分由g(x)0得0x3，由g(x)0得x3，·····································11分所以g(x)在(0，3)上单调递减，在(3，+∞)上单调递增，························12分所以≥3，分g(x)gmin(x)g(3)21e0·············································14∴当x>0时，f(x)(x4)ex．························································15分17．解：（1）连接B1D1交A1C1于O1，连接DO1，在平行六面体ABCD-A1B1C1D1中，BB1=DD1且BB1//DD1，∴四边形BDD1B1是平行四边形，······················································2分∴BD=B1D1，且BD//B1D1，且O，O1分别为BD，B1D1的中点，∴OD=O1B1，OD//O1B1，可得四边形ODO1B1是平行四边形，·················3分∴OB1//O1D，·················································································4分由平面ACE//平面A1C1D，平面ACE∩平面BDD1B1=OE，平面A1C1D∩平面BDD1B1=O1D，∴OE//O1D，··················································································5分∴OB1//OE，且都经过点O，∴O，E，B1三点共线，·····························6分又因为△OBE∽△B1D1E，BEBD∴1112，所以BE2EO；·················································7分EOOB1（2）记CD的中点为点F，连接BF，FD1，∵DC=BC=BD=2，则BF⊥DC，且BF3，······································8分参考答案及评分标准第2页共8页∵平面CC1D1D⊥平面ABCD，且平面CC1D1D∩平面ABCD=CD，BF平面ABCD，BF⊥DC，∴BF⊥平面CC1D1D，·····································································9分∴⊥，即222，BFFD1BD1BFD1F6∴，则222，∴，分D1F3D1FDFDD14FD1DC······················10以F为原点，以FD，FB，FD1分别为x轴，y轴，z轴，建立如图所示的空间直角坐标系O-xyz．13则A(2，3，0)，C(−1，0，0)，A(1，3，3)，B(1，3，3)，O(，，0)，1122B(0，3，0)，·····················································································11分113333∴OEOB(，，)，BEBOOE(0，，)，·········12分3126333，，，，，，分AC(330)AA1(103)···········································13不妨设平面ACC1A1的法向量为n=(x，y，z)，直线BE与平面ACC1A1所成角为θ，ACn03x3y0，即，不妨取n=(3，3，−1)，·················14分AA1n0x3z0nBE226则sin|cosn，BE|，|n||BE|13226∴直线BE与平面ACC1A1所成角的正弦值为．···························15分13b1a318．解：（1）由解得：，···················································2分c2b1参考答案及评分标准第3页共8页x2∴双曲线C的方程为y21；······················································3分3（）当轴时，必然与轴重合，由曲线对称性知中点在轴上，2l1xl2xABMxDE的中点N也在x轴上，故MN经过的定点P1(x1，y1)在x轴上．···········4分设为P1(n，0)，直线AB方程为：xty3，设A(x1，y1)，B(x2，y2)，xty3联立，得：22，分22(t3)y6ty60······································5x3y36t18由韦达定理可得：yy，xxt(yy)6，············6分12t231212t2393t故AB中点M(,)，···························································7分t23t2313()19同理：直线DE方程为：xy3，DE中点N(，t)，t1212()3()3tt9t23t即：N(，)，···································································8分13t213t23t3t3tt2313t2t23由P1MN三点共线可得：kk，即：，········9分P1MMN99t29nt2313t2t2399解得：n，即P1(，0)；···························································10分229（3）由（2）可知，P(3，0)时，定点P1(，0)，2同理可知，，也一定在轴上．P2(x2y2)x考虑一般情况，假设点P(m，0)时，设P1(n，0)，并设直线AB方程为：xtym，A(x1，y1)，B(x2，y2)，xtym联立得：222，22(t3)y2tmym303xy32tm6m由韦达定理可得：yy，xxt(yy)2m，12t231212t233mtm故AB中点M(，)，························································11分t23t23参考答案及评分标准第4页共8页1()m13m同理：直线DE方程为：xym，DE中点N(，t)，t1212()3()3tt3t2mtm即：N(，)，··································································12分13t213t2tmmttmt2313t2t23由P1MN三点共线可得：kk，即：，·······13分P1MMN3m3mt23mnt2313t2t233m3m解得：n，即：点P(m，0)时，P1(，0)，223323n故当点P(3，0)时，可得P1(3，0)，P2(3()，0)，……，P(3()，0)，22n21