2024届安徽省皖江名校联盟高三二模数学答案

数学参考答案题号1234567891011答案DCDACACCBDBDBCD2i1.【解析】z2ii12i，故z12i，故选D.i【解析】由已知，所以，又，所以，故选2.ABARABRRBxxaa3C.3.【解析】选D.31014【解析】由已知.a2,n2时,aSS2n1,ba3,ba9,b310，故2m1310,m,1nnn11225102故选A.k2k【解析】由已知n故故通项为k8kkk82k（）故奇数项5.2256,n8,Tk1C8x1C82xk0,1,,8,xkkk1k1的系数为正数，偶数项的系数为负数，由CC8282，解得k5.所以选C.【解析】问题等价于函数x的图像与直线有两个公共点当时由图象得6.ya1y2a,,0a1,102a1,故0a;当a1时,由图象得2a2,不符合条件.所以选A.2yyy=2ay=2aOxxO1247.【解析】由已知得a2c2b2ac,故cosB,所以B,由a2c22ac,得ac,ABC面积的最大23333值为ac,故选C.438.【解析】令x1,y1,则f11;令x1,y1,则f11.令y1,得fxfx,x2故yfxx0为偶函数.任取x1,x20,,x1x2,则1,x1x2则fx1fx2f1fx2，故yfxx0在0,上为减函数.由已知f2x11，x111可得f2x1f1，故2x11，解得1x0，且x.若f2，则22f1024f210f29f211029f4，故选C.数学参考答案第1页（共7页）{#{QQABYQAAggCoAJJAABhCAQ2QCECQkAGCACoGRBAAsAABCRFABAA=}#}729.【解析】由图可知fx的周期为：T2，又T，所以2；1212由f0，sin0，且，所以；1262263由f0，所以A3，故A错误；所以fx3sin2x26因为fx2sin2x2cos2x为偶函数，B正确；32x0,，则2x0,，故fx在0,上单调递增，C错误；42345411因为f3,f3,f3,f3，故D正确。所以选BD。363610.【解析】由已知得PQPFQF，由双曲线定义知：PFQFQF2a,22112yQPF2PF12a4a，故QF24a,QF16a，MNP2221在QF1F2中，由余弦定理得：16a36a224a167，2F1OF2xx2y2解得：a24，所以b224，方程为1，A错误。424123PF1F2的面积为S8a83，B正确。22111取的中点，两圆内切，故错误。QF1MOMQF2QF12aQF1a,C222111取的中点，则，两圆外切，故正确。QF2NONQF1QF22aQF2aD222211.【解析】因为ABP的面积为，Q到平面ABP的距离不是定值，故A错误；2221因为ADQ的面积为，P到平面ADQ的距离为，体积为，故B正确；12126323D因为QAC的最大值为，P到平面ACQ的最大距离为，1PC123A113231B1故四面体PQAC的体积最大值为=，故C正确。3233OG过点Q作，，设，QGH//BCPMDGPNCD,CGt,PD1sDHNMC1122则t,s0,1，SADDG1t，PMDG1t,ABADQ22tt1stOMDMsinCDGts，OP1t2st，1t21t21t2数学参考答案第2页（共7页）{#{QQABYQAAggCoAJJAABhCAQ2QCECQkAGCACoGRBAAsAABCRFABAA=}#}111st11故四面体PQAD的体积为V1t21st，其最大值为，故D正确.321t266012.【答案】37.5【解析】从小到大排序为：10,12,14,16,20,24,30,35,40,43;108008,故第80百分位数是37.5.51213.【答案】3x4y40,【解析】由已知得a抛物线的方程为x4y，所以F0,1，243直线l的方程为yx1，与x24y联立整理得x23x40，415故x14,x21，故OAB的面积为SOFx1x2.222【答案】【解析】14.11fxsinxx1cosxxxcos1sinxxxxcossin,4当x0,时,fx0,fx递增;当x,时,fx0,fx递减;442fxf,f01,f1,fx1,max44min2故最大值与最小值的和为：11.43f115.【解析】（1）由已知fx2x10，···················································2分x所以f183f1，解得f14故fxx210x12lnx，f19···············4分,所求切线方程为：y94x1，即y4x13·····················································6分（2）由已知函数fxx210x12lnx，定义域为0,2122x5x6fx2x10，xx由fx0，解得x2或x3············································································8分随x的变化fx和fx的变化如下x0,222,333,fx+00+fx单调递增极大值单调递减极小值单调递增函数fx单调递增区间为0,2和3,，单调递减区间为2,3································11分数学参考答案第3页（共7页）{#{QQABYQAAggCoAJJAABhCAQ2QCECQkAGCACoGRBAAsAABCRFABAA=}#}当x2时，fx取得极大值f21612ln2，当x3时，fx取得极小值f32112ln3·····················································13分16.【解析】（1）记A,B,C参赛获胜事件分别记为A,B,C表示，参赛失败分别记为ABC,,,所以PAPBPC0.4,0.6,0.8，PAPBPC0.6,0.4,0.2······················2分则甲班至少获胜2场事件记为M,则MABCABCABCABCPMPABCPABCPABCPABCPAPBPCPAPBPCPAPBPCPAPBPC······················4分0.40.60.80.40.60.20.40.40.80.60.60.80.656所以甲班至少获胜2场的概率为0.656························································6分（2）由已知X取值为0,4,5,6,9,10,11,15，PX0PABC0.60.40.20.048，PX4PABC0.60.40.80.192,·······················································8分PX5PABC0.60.60.20.072，PX6PABC0.40.40.20.032PX9PABC0.60.60.80.288，························································10分PX10PABC0.40.40.80.128，PX11PABC0.40.60.20.048，PX15PABC0.40.60.80.192·······························································12分所以EX00.04840.19250.07260.03290.288100.128110.048150.1920.7680.3600.1922.5921.2800.5282.8808.600…………………15分17.【解析】（1）由已知得平面ABCD平面ABEF,AFAB,所以AF平面ABCD因为BD平面ABCD，故BDAF··································································2分因为ABCD是正方形，所以BDACAC,AF平面ACF，ACAFABD平面ACF又BD平面BDE，所以平面ACF平面BDE。……………………5分（2）又（1）知：AD,,AFAB两两垂直，数学参考答案第4页（共7页）{#{QQABYQAAggCoAJJAABhCAQ2QCECQkAGCACoGRBAAsAABCRFABAA=}#}以AD,,AFAB所在直线分别为x,,yz轴，建立空间直角坐标系如图。设MAD，AB1，则AM0,0,0,cos,sin,0，CE1,0,1,0,1,1，故AMcos,sin,0，AC1,0,1,AE0,1,1············································7分设平面AMC的法向量为mx1,,y1z1，则mAC0,mAM0x1z10故，取x1sin，则y1cos，z1sinx1cosy1sin0所以msin,cos,sin·········································································9分设平面AME的法向量为nx2,,y2z2，nAE0,nAM0y2z20故，取x2sin，则y2cos，z2cosx2cosy2sin0所以nsin,cos,cos······································································11分1sincos所以cosm,n，1sin21cos21sincos1由已知得················