惠州市2025届高三第二次调研考试试题高三数学参考答案与评分细则一、单项选择题：本题共8小题，每小题满分5分，共40分．题号12345678答案BDCBAADC1.【解析】因为Bx0x4,所以ABx2x4.故选：B.22.【解析】因为z210，即z21，所以zi，所以z11i1212.故选：D.9a36d273.【解析】设等差数列的公差为，由已知得：1解得，and,a11，d1a19d8所以．故选：C．a100a199d199984.【解析】连接FB，在正方体ABCDA1B1C1D1中，BC平面ABB1A1，棱BC的中点为E，则BE平面ABB1A1，而BF平面ABB1A1，故BEBF，则EFB即为直线EF与平面ABB1A1所成角，设正方体棱长为2，则22，BE1,BFB1FB1B145BE16则EFBF2BE26，故sinEFB．故选：B.EF66255．【解析】由b2，(2ab)b3，得2abb2ab23，即ab,25abaab535由已知得，所以向量在向量上的投影向量为2，，．a2ba2a(31)()aaa488故选：A．a116.【解析】若函数f(x)在(1,)上单调递增，则，解得a，12a02所以“a0”是“函数f(x)在(1,)上单调递增”的充分不必要条件．故选：A．高三数学答案第1页，共11页{#{QQABLQAAogAgAgAAAQgCQQkwCgCQkhCCAQgOwFAIIAIAyBFABAA=}#}7.【解析】设优弧BC所在圆的圆心为O,半径为,连接OA,OB,OC.易知“水滴”的“竖直高度”�OAR45为OAR,“水平宽度”为2R,由题意知,解得OAR.因为AB与圆弧相切于点B,2R33OBR3所以OBAB.在RtABO中,sinBAO，又BAO(0,),5OAR5234所以cosBAO1sin2BAO，由对称性知,BAOCAO,则BAC2BAO,53424所以sinBAC2sinBAOcosBAO2.故选：D.55258.【解析】根据已知条件设理科女生有x1人，理科男生有x2人；文科女生有y1人，文科男生有y2人；根据题意可知：x1x2y1y2，x1y1x2y2，根据同向不等式可加的性质有：x1x2x1y1y1y2x2y2，即x1y2，所以理科女生多于文科男生，C正确.其他选项没有足够证据论证.故选：C.二、多项选择题：本题共3小题，每小题满分6分，共18分。在每小题给出的四个选项中，有多项符合题目要求。全部选对得6分，部分选对得部分分，有选错的得0分。题号91011全部正确选项ADABDACD9.【解析】数据从小到大排列为:1，1，2，3，3，3，3，4，5，5．对于A，该组数据的极差为514，故A正确；12234452对于B，众数为3，平均数为3，两者相等，故B错误；101对于C，方差为(13)22(23)21(33)24(43)21(53)221.8，故C错误；10对于D，1080%8，这组数据的第80百分位数为第8个数和第9个数的平均数4.5，故D正确．故选：AD．10.【解析】由图像可知：，；fxmax2A2高三数学答案第2页，共11页{#{QQABLQAAogAgAgAAAQgCQQkwCgCQkhCCAQgOwFAIIAIAyBFABAA=}#}1又f02sin1，故sin，又，所以，所以A项正确；226777已知f2sin0，由五点作图法可知：，解得：2，所以B项正确；12126126故fx2sin2x；则xf(x)2xsin2x，设h(x)xf(x)2xsin2x，61212则h(x)2(x)sin(2x)2xsin2xh(x)，所以函数yxf(x)是偶函数，故C项错误；12g(x)f(x)2sin2(x)2sin(2x)2cos(2x)6666262cos(2x)2cos(2x)，所以D项正确；33故选：ABD.22211.【解析】A选项，经验算，点(,0)和(1,1)的坐标满足曲线L的方程：x(yx)1，所以点22(,0)和(1,1)均在L上，故A项正确；222B选项，OPx2y2，因为曲线L：x(yx)1关于y轴对称，当x0时，x2(yx)21，设xcos,yxsin，,，2221cos2所以OPx2y2cos2(cossin)21sin2231351sin2cos2sin(2)，其中tan，2222235513551所以OP，OP，所以OP的最大值和最小值之和min222max222为5，故B项错误；C选项，因为曲线L：x2(yx)21关于y轴对称，当x0时，x2(yx)21，则(yx)21x2，所以yx1x2，因求点P的纵坐标的最大值，故取yx1x2，1又y2（x1x2）212x1x212x2(1x2)1x2(1x2)2(当且仅当x2时2等号成立)，所以y2，故C项正确；高三数学答案第3页，共11页{#{QQABLQAAogAgAgAAAQgCQQkwCgCQkhCCAQgOwFAIIAIAyBFABAA=}#}y2x2D选项，PAPB23等价于点P在椭圆1内（包含椭圆），由B项可知，即满足：323(1cos2)2(cossin)23cos26，即2(1sin2)6，整理得：234sin23cos25，即5sin(2)5，其中其中tan，即sin(2)1恒成立，则故4D项正确.故选：ABD.三、填空题：本题共3小题，每小题5分，共15分．512.3213.14.e2512.【解析】当x1时，二项式展开式各项的系数和为2532.故答案为：32.13.【解析】由题意知，且三者成等比数列，则2AF1ac,F1F22c,F1BcaF1F2AF1F1B155即4c2(ca)(ca)c2a2，所以e2，所以e.故答案为：.555b21b2114.【解析】设方程ln(ax)x的实根为x0，则ln(ax)x，2402041122bx0bx0所以axe4，即axe40.020212yx0设点P(a,b),则点P在直线xxe40上.0221x0142eyx0设点O(0,0)到直线4的距离为d，则d，x0xe012x2041et1et(t1)设tx2，f(t)(t)，则f'(t)，04t2t21求得在，上单调递减，在，上单调递增，所以，f(t)11f(t)minf(1)e222则df(t)e，又a2b2OP,由几何意义可知OPd，所以a2b2OPe2.高三数学答案第4页，共11页{#{QQABLQAAogAgAgAAAQgCQQkwCgCQkhCCAQgOwFAIIAIAyBFABAA=}#}33baeae0检验：当时，3，由，解得2；t1x0222222eabeb233baeae02222由22，解得，所以则ab可以取到最小值e.222eabeb2故答案为：e2.四、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．（本小题满分13分，其中第一小问6分，第二小问7分。）111【解析】（1）因为：f110，所以切点坐标为：1,，···························1分2222又fxx1，·····························································································2分x所以f12，································································································3分即所求切线的斜率为2.1所以切线方程为：y2x1，·······································································5分2化简得：4x2y30，所以曲线yfx在点(1，f1)处的切线方程为4x2y30．··························································································································6分3【注】切线方程写成y2x不扣分22x2x2x2x1（2）fxx1，（x0）········································8分xxx由f'(x)0得x2；由f'(x)0得0x2.············································10分所以fx在1，2上单调递减，在2,e上单调递增.·················································11分所以函数fx在区间1,e上的极小值为f22ln2，也是最小值.所以函数fx在区间1,e上的最小值为2ln2.························································································································13分高三数学答案第5页，共11页{#{QQABLQAAogAgAgAAAQgCQQkwCgCQkhCCAQgOwFAIIAIAyBFABAA=}#}16．（本小题满分15分，其中第一小问5分，第二小问10分。）【解析】（1）证明：已知PA底面ABCD，且BC底面ABCD，所以PABC.···································································································2分由ACB90,可得BCAC.············································································3分又PAACA,PA,AC平面PAC,·································································4分所以BC平面PAC．······················································································5分（2）取CD的中点E．由AB∥CD,BAD120，可得ADC60，又因为ADCD1，所以三角形ADC