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Bx0x4,所以ABx2x4.故选:B.22.【解析】因为z210,即z21,所以zi,所以z11i1212.故选:D.9a36d273.【解析】设等差数列的公差为,由已知得:1解得,and,a11,d1a19d8所以.故选:C.a100a199d199984.【解析】连接FB,在正方体ABCDA1B1C1D1中,BC平面ABB1A1,棱BC的中点为E,则BE平面ABB1A1,而BF平面ABB1A1,故BEBF,则EFB即为直线EF与平面ABB1A1所成角,设正方体棱长为2,则22,BE1,BFB1FB1B145BE16则EFBF2BE26,故sinEFB.故选:B.EF66255.【解析】由b2,(2ab)b3,得2abb2ab23,即ab,25abaab535由已知得,所以向量在向量上的投影向量为2,,.a2ba2a(31)()aaa488故选:A.a116.【解析】若函数f(x)在(1,)上单调递增,则,解得a,12a02所以“a0”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”�OAR45为OAR,“水平宽度”为2R,由题意知,解得OAR.因为AB与圆弧相切于点B,2R33OBR3所以OBAB.在RtABO中,sinBAO,又BAO(0,),5OAR5234所以cosBAO1sin2BAO,由对称性知,BAOCAO,则BAC2BAO,53424所以sinBAC2sinBAOcosBAO2.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:x1x2y1y2,x1y1x2y2,根据同向不等式可加的性质有:x1x2x1y1y1y2x2y2,即x1y2,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514,故A正确;12234452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(13)22(23)21(33)24(43)21(53)221.8,故C错误;10对于D,1080%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:,;fxmax2A2高三数学答案第2页,共11页{#{QQABLQAAogAgAgAAAQgCQQkwCgCQkhCCAQgOwFAIIAIAyBFABAA=}#}1又f02sin1,故sin,又,所以,所以A项正确;226777已知f2sin0,由五点作图法可知:,解得:2,所以B项正确;12126126故fx2sin2x;则xf(x)2xsin2x,设h(x)xf(x)2xsin2x,61212则h(x)2(x)sin(2x)2xsin2xh(x),所以函数yxf(x)是偶函数,故C项错误;12g(x)f(x)2sin2(x)2sin(2x)2cos(2x)6666262cos(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(yx)1,所以点22(,0)和(1,1)均在L上,故A项正确;222B选项,OPx2y2,因为曲线L:x(yx)1关于y轴对称,当x0时,x2(yx)21,设xcos,yxsin,,,2221cos2所以OPx2y2cos2(cossin)21sin2231351sin2cos2sin(2),其中tan,2222235513551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(yx)21关于y轴对称,当x0时,x2(yx)21,则(yx)21x2,所以yx1x2,因求点P的纵坐标的最大值,故取yx1x2,1又y2(x1x2)212x1x212x2(1x2)1x2(1x2)2(当且仅当x2时2等号成立),所以y2,故C项正确;高三数学答案第3页,共11页{#{QQABLQAAogAgAgAAAQgCQQkwCgCQkhCCAQgOwFAIIAIAyBFABAA=}#}y2x2D选项,PAPB23等价于点P在椭圆1内(包含椭圆),由B项可知,即满足:323(1cos2)2(cossin)23cos26,即2(1sin2)6,整理得:234sin23cos25,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x1时,二项式展开式各项的系数和为2532.故答案为:32.13.【解析】由题意知,且三者成等比数列,则2AF1ac,F1F22c,F1BcaF1F2AF1F1B155即4c2(ca)(ca)c2a2,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,2402041122bx0bx0所以axe4,即axe40.020212yx0设点P(a,b),则点P在直线xxe40上.0221x0142eyx0设点O(0,0)到直线4的距离为d,则d,x0xe012x2041et1et(t1)设tx2,f(t)(t),则f'(t),04t2t21求得在,上单调递减,在,上单调递增,所以,f(t)11f(t)minf(1)e222则df(t)e,又a2b2OP,由几何意义可知OPd,所以a2b2OPe2.高三数学答案第4页,共11页{#{QQABLQAAogAgAgAAAQgCQQkwCgCQkhCCAQgOwFAIIAIAyBFABAA=}#}33baeae0检验:当时,3,由,解得2;t1x0222222eabeb233baeae02222由22,解得,所以则ab可以取到最小值e.222eabeb2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。)111【解析】(1)因为:f110,所以切点坐标为:1,,···························1分2222又fxx1,·····························································································2分x所以f12,································································································3分即所求切线的斜率为2.1所以切线方程为:y2x1,·······································································5分2化简得:4x2y30,所以曲线yfx在点(1,f1)处的切线方程为4x2y30.··························································································································6分3【注】切线方程写成y2x不扣分22x2x2x2x1(2)fxx1,(x0)········································8分xxx由f'(x)0得x2;由f'(x)0得0x2.············································10分所以fx在1,2上单调递减,在2,e上单调递增.·················································11分所以函数fx在区间1,e上的极小值为f22ln2,也是最小值.所以函数fx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PABC.···································································································2分由ACB90,可得BCAC.············································································3分又PAACA,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,BAD120,可得ADC60,又因为ADCD1,所以三角形ADC
广东省惠州市2025届高三第二次调研考试数学答案
2024-10-30
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