2024届山东省青岛市高三一模数学参考答案

2024年高三年级第一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：ADBACCBA二、多项选择题：本题共3小题，每小题6分，共18分．9．AB10．AC11．BCD三、填空题：本题共3个小题，每小题5分，共15分．5112．0；13．；14．17．2四、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）由题知：各组频率分别为：0.15，0.25，0.3，0.2，0.1······················3分日均阅读时间的平均数为：300.15500.25700.3900.21100.167（分钟）·······················6分（2）由题意，在[60,80),[80,100),[100,120]三组分别抽取3,2,1人·······················7分的可能取值为：0,1,2·················································································8分C3C0142则P(0)3················································································9分C65C2C1342P(1)3···············································································10分C65C1C2142P(2)3··············································································11分C65所以的分布列为：012131P555131E()0121······································································13分55516．（15分）x2x1解：（1）当a1时，f(x)，f(x)1解得x1························3分x0013又因为f(1)，所以切线方程为：xy0···········································5分22x2ax1（2）f(x)的定义域为(0,)，f(x)········································6分x当a0时，得f(x)0恒成立，f(x)在(0,)单调递增·································8分数学评分标准第1页（共5页）{#{QQABYYCEggigAhBAAQgCEwEICEKQkAACACoOQEAAoAAAyQNABCA=}#}当a0时，令g(x)x2ax1,a24····················································9分（ⅰ）当0即0a2时，f(x)0恒成立，f(x)在(0,)单调递增··········································11分aa24aa24(x)(x)（ⅱ）当0即a2时，f(x)22···············12分xaa24aa24由f(x)0得，0x或x，22aa24aa24由f(x)0得，x22aa24aa24所以f(x)在(0,),(,)单调递增，22aa24aa24在(,)单调递减···········································14分22综上：当a2时，f(x)在(0,)单调递增；aa24aa24当a2时，f(x)在(0,),(,)单调递增；22aa24aa24f(x)在(,)单调递减·······················15分2217．（15分）解：(1)取棱A1A中点D，连接BD，因为ABA1B，所以BDAA1·················1分因为三棱柱ABCA1B1C1，所以AA1//BB1······················································2分所以BDBB1，所以BD3·····································································3分因为AB2，所以AD1，AA12；222因为AC2，A1C22，所以ACAA1A1C，所以ACAA1，············4分同理ACAB，·························································································5分因为AA1ABA,且AA1,AB平面A1ABB1，所以AC平面A1ABB1，因为AC平面ABC，zA所以平面A1ABB1平面ABC··············6分1B1C（2）取AB中点O，连接A1O，1取BC中点P，连接OP，则OP//AC，D由（1）知AC平面A1ABB1，O所以OP平面A1ABB1·······················7分ABy因为AO平面AABB，AB平面AABB，11111MP所以OPA1O，OPAB，Cx因为ABA1AA1B，则A1OAB······························································8分数学评分标准第2页（共5页）{#{QQABYYCEggigAhBAAQgCEwEICEKQkAACACoOQEAAoAAAyQNABCA=}#}以O为坐标原点，OP，OB，OA1所在的直线为x轴、y轴、z轴，建立如图所示的空间直角坐标系Oxyz，则A(0,1,0)，A1(0,0,3)，B1(0,2,3),C(2,1,0)，设点N(a,0,3)，(0a2)···········································································9分A1B1(0,2,0)，A1C(2,1,3)，A1N(a,1,3)，nA1B102y0设面A1B1C的法向量为n(x,y,z)，得，得，nA1C02xy3z0取x3，则y0，z2，所以n(3,0,2)·············································10分设直线AN与平面A1B1C所成角为，|nAN|3a2则sin|cosn,AN||n||AN|7a243(a2)27a243a24a4································11分7a2421若a0，则sin，·········································································12分7343442若a0，则sin11，·······························13分47a747a4当且仅当a，即a2时，等号成立，·······················································14分a42所以直线AN与平面AMB所成角的正弦值的最大值·································15分1718．（17分）解:（1）设M(x,y)，切点为N，则|MN|2|MW|2|OM|2|OW|2，所以|x2|2x2y24··············································································3分化简得y24x，所以C的方程为：y24x····················································4分（2）（ⅰ）因为l1//l2，所以可设GAGA,GBGB，1又因为GE(GAGB)(GAGB)GF，22所以G,E,F三点共线，同理，H,E,F三点共线，所以G,E,H三点共线···················································································6分（ⅱ）设A(x1,y1),B(x2,y2),A(x3,y3),B(x4,y4)，AB中点为E，AB中点为F，数学评分标准第3页（共5页）{#{QQABYYCEggigAhBAAQgCEwEICEKQkAACACoOQEAAoAAAyQNABCA=}#}22将xym代入y4x得：y4y4m0，所以y1y24,y1y24m，yy所以y122，同理y2（G,E,H,F均在定直线y2上）················8分E2F因为TT//l1，所以EAT与EAH面积相等，EBT与EBH面积相等；所以四边形GTET面积等于四边形GAHB面积···············································10分设G(xG,2),H(xH,2)，yyyyy2直线AA:yy13(xx)，即yy13(x1)，1xx11y2y241313444xyy2(yy)yy131313整理得：直线AA:y，又因为yG2，所以xG，y1y344xy2y32(y2y3)y2y3同理，直线BA:y，yH2，所以xH·············12分y2y34yy(yy)(34y)(yy)(2y)123所以|GH||xx||123||2|GH44|(yy)(yy)|1234····························14分81(yy)2|yy|所以四边形GAHB面积S|GH||yy|123421216[(yy)24yy](yy)24yy1212343416(1616m)1616n164(1m)2(1n)(1m)21n4[]2(2m22mn)16············16分2m22mnm1当且仅当(1m)21n，即，即时取等号，2nm2m6n3所以GAT面积的最大值为16······································································17分数学评分标准第4页（共5页）{#{QQABYYCEggigAhBAAQgCEwEICEKQkAACACoOQEAAoAAAyQNABCA=}#}19．（17分）33解：（1）因为m4x(a1b4a2b3a3b2a4b1)x，所以m4a1b4a2b3a3b2a4b1···························································