数学参考答案题号123456789101112答案CDBADCDAABCADBCDBD1.【解析】Bxx2或x4，所以AB1,5,6，故选C.222.【解析】z1i，故选D.1i2cos223.【解析】因为2，所以cos2sin2，cos4sincos4sin4，1sin34sincos3cos2，tan或tan0（舍去），故选B.41l4.【解析】设底面半径为r，由已知得r3，高为h3.故体积为Vr2h3.3R=1225.【解析】由已知fxln2x4xaln2x4xalna，因为rfxfx，故lna0，所以a1，故fxln2x4x21.f2ln2232ln12.25【解析】一次分形长度为4，二次分形长度为4，次分形长度为452故选6.a1a25a54,C.333243【解析】设，则，，7.PF23mQF22mPF12a3mQF12a2m.22264417在PQF中得：2a3m25m22a2m，即ma.在PFF中得：a2a24c2，故e.1151225255D1PC18.【解析】过点Q作GH//BC，PMDG，PNCD,设CGt,PD1s，A1B11122则，，Gt,s0,1SADQADDG1tPMDG1t,O22QDHtt1stMCOMDMsinCDGts，OP1t2st，N1t21t21t2AB111st11故四面体PQAD的体积为V1t21st，当st0时，其最大值为，故A正确.321t2669.【解析】可求甲乙平均数为x1x27，中位数均为7，故A，B正确；甲的极差为6，乙的极差为4，故140112C正确；甲的标准差为：1449499，乙的标准差为：441111，故D7777错误.10.【解析】由已知A,B为单位圆上任意两点，OAOB1，PAPBAB2，A正确；设D为AB的第1页共6页{#{QQABLYCEggCIAAJAABhCQQ0QCEKQkAECCIgOQEAIMAIBCQNABAA=}#}中点，则PA+PB=2PD6，故B错误；当P,A,B共线时，PAPBAB2，故C错误；当P,A,B共线时，PA+PB2PD4，故D正确.1111.【解析】fx2x3，由f10，故lna1，所以a,fxx23xlnx此时xlnae，2x23x12x1x111fx，fx在0,,1,上单调递增，在,1单调递减.fx极小值xx223111为f12，故ff1.又fln42f1，故BCD正确.5416212.【解析】可求D5,2，由ABBC,BCCD,M的方程为：x1y24，22N的方程为：x3y24，MN22，P,Q两点之间的距离的最大值y为，故错误由已知PQ//MN，故直线PQ的斜率为，所以正确AD224A.1B.N当PQ斜率为0时，直线PQ被M截得的弦长为4，当PQ斜率不为0时，直线PQBOMCx被M截得的弦长不为4，故C错误.当DP与M相切时，tanADP最大，此时14tanADM，故tanADP，所以D正确.23【答案】【解析】展开式第项为：r3r6，由，得，故常数项为413.15r1Tr1C6x3r120r4C615.333314.【答案】,，,或0,，0,等【解析】因为fxfx，所以fx为偶函444433数，由x0,，fx2sinx，故fx在0,上单调递增，在,上单调递减，由4443对称性可知在,上单调递增.4515【.答案】【解析】由已知得F0,1，设直线l的方程为ykx1代入x24y整理得x24kx40，2设，故，①②，又，故③。由①Ax1,y1,Bx2,y2x1x24kx1x24AF4FBx14x21②③解得29此时，2，到距离为：k.ABy1y22kx1x244k1OABd.161k215故OAB的面积为SABd2k21.22lnx1216.【答案】【解析】fx2x2xlna1lnx在0,上有两个不等的1,elnalnalna实数根，故0lna1，解得a1,e.第2页共6页{#{QQABLYCEggCIAAJAABhCQQ0QCEKQkAECCIgOQEAIMAIBCQNABAA=}#}17.【解析】（1）由已知及正弦定理得：ababcbc，即b2c2a2bc.·······2分b2c2a212由余弦定理得：cosA，又A0,，所以A.···················3分2bc232故C,341262所以sinCsinsincoscossin····················································5分343434427（2）由（1）知A，又ADAC，所以BAD，ADB3612762sinADBsinsin·······································································7分12344在ABD中，由正弦定理得：7ADsinABAD326，所以AB12···············································10分sinADBsinBsin24【解析】（）由已知，由已知，又18.1b1a23b12a225当时，分n2bna2na2n122a2n222bn12··························································2故，bn22bn12所以数列是首项为，公比为的等比数列分bn252···························································4（）由（）知：n1分21bn522···········································································6分T2na1a2a2na1a3a2n1a2a4a2n2b1b2bn2n·······8设Sbbb5122n12n52n2n5············································10分n12n故n1分T2n2Sn2n526n10··············································································1219.【解析】（1）记A,B,C,D,E参赛获胜事件分别用A,B,C,D,E表示，5场全胜的概率为：PABCDE0.40.50.60.70.80.0672·······································2分甲班至多获胜4场与5场全胜为对立事件，故甲班至多获胜4场的概率为P1PABCDE10.06720.9328甲班至多获胜4场的概率为0.9328·················································································4分（2）记BCD三人组合班级得分为Y，Y的取值分别为0,7,6,5,11,12,13,18，由已知得PY00.50.40.30.06，PY70.50.40.30.06第3页共6页{#{QQABLYCEggCIAAJAABhCQQ0QCEKQkAECCIgOQEAIMAIBCQNABAA=}#}PY60.50.60.30.09，PY50.50.40.70.14PY110.50.60.70.21，PY120.50.40.70.14PY130.50.60.30.09，PY180.50.60.70.21·········································10分EY70.0660.0950.14110.21120.14130.09180.2110.6因为EY10.610，所以BCD三人组合具有参赛资格····················································12分20.【解析】（1）由已知可得：BC2，PB2，PCCDAB22在PBC中，PB2BC2PC2，故PBBC······························································2分又ABBC，且PBABB，BC平面PAB因为BCABC，所以平面PAB平面PBC····························································4分（2）取AB、CD的中点O、E，连接OP,OE.z因为PAPB，所以POAB，P由（）知：平面平面，………………分y1PABABC6DC所以PO平面ABC.AOBx以OB,OE,OP所在直线分别为x,y,z轴，建立空间直角坐标系。则A2,0,0,B2,0,0,C2,2,0,P0,0,2···················································8分设平面的法向量为，APCmx1,y1,z1AC22,2,0,AP2,0,222x12y10mAC0,mAP0，故，2x12z10取，则分x11,y12,z11m1,2,1······················································10mn1又平面APC的法向量为n0,0,1，cosm,n。mn23所以二面角PACB的正弦值为··········································································12分2c163621.【解析】（1）由已知得：e2，c2a，所以b23a2，又1，················2分aa2b2x2y2解得a24，b212，故双曲线C的方程为1····························