理数答案一、选择题题号123456答案ADCCAC题号789101112答案CDBDAA二、填空题题号1314151613532答案e35241三、解答题417.解：（1）男生10500人，女生4500人，3抽取女生占总人数的比例为∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(1分)10又分层抽样收集300位学生3女生样本数据应收集为30090∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(2分)10（2）由频率分布直方图可知，学生每周平均体育运动时间超过4个小时的概率为(0.150.1250.0750.025)20.75∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(4分)33（3）由（2）可知运动时间超过4小时的概率p，则X~B(4,)∙∙∙∙∙∙∙∙∙∙(5分)44131P(x0)C0()4()0，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(6分)444256133P(x1)C1()3()1，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(7分)444641327P(x2)C2()2()2，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(8分)4441281327P(x3)C3()1()3，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(9分)444641381P(x4)C4()0()4，则X的分布列为：∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(10分)444256X0123413272781P25664128642563且E(X)43.∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(12分)418.解：（1）证明：方法一：因为E、F、G分别是棱AB、AP、PD的中点，所以EF//BP，GF//AD，又EF平面PBC，BP平面PBC，所以EF//平面PBC，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(1分)又因为底面ABCD为平行四边形，所以AD//BC，则GF//BC，又GF平面PBC，BC平面PBC，所以GF//平面PBC，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(2分)因为GFEFF，GF,EF平面EFG，所以平面EFG//平面PBC，∙∙∙∙∙∙∙∙∙∙∙∙∙∙(3分)又PC平面PBC，所以PC//平面EFG，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(4分)方法二：取AC的中点H，连接FH，HE，取AC的中点H，连接FH，HE，GH∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(1分)因为E、H分别是棱AB、AC的中点，所以EH//BP分=∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(2)同理可知FG//AD=所以EH//FG=所以E、H、G、F四点共面∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(3分)同理可知FH//PC又PC平面EFG，FH平面EFG，所以PC//平面EFG∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(4分)（2）因为PCPDCD22,ACADAP2,所以APAD,APAC又，平面，平面，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(6分)所�以�∩????平=面�????⊂．以�为�原��点，��⊂，�，���所在直线分别为轴、轴、轴建立��空⊥间直角????坐�标�系，∙�∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙�∙∙∙�∙∙∙∙∙∙�∙∙�∙∙∙∙∙∙�∙∙∙�∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙�∙∙∙∙∙∙∙∙∙�∙∙(7分�)则，，，，，所�以(0,0,0)�(1,−1，,0)�(0,0,1)，∙�∙∙∙(∙0∙∙,∙1∙∙,∙1∙∙)∙∙∙∙∙�∙∙(∙0∙∙,∙0∙∙,∙2∙∙)∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(9分)��=(−1,1,1)��=(0,1,0)，设平面的一个法向量为，��=(0,2,−2)����=(�0,�0,�0)所以，即取，则，∙∙∙∙∙∙(10分)000�⋅��=0,−�+�+�=0,00�=1�=(−1,0,1)设与�⋅平F面G=0,所成的�角=为0,，则，−21�>|=|8⋅2|=2则直��线与平��面�的夹角为�．∙∙∙s∙i∙n∙∙�∙∙∙=∙∙∙∙|∙c∙∙o∙∙s∙∙<∙∙∙∙�∙∙�∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(12分)π19.解�：����6（1）选①，设递增等差数列an的公差为dd0，由a22d，a322d，2a625d，有22d22d25d4，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(2分)化简得d24．则d2，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(4分)ana1(n1)d2n，所以an的通项公式为an2n．∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(6分)选②，设递增等差数列an的公差为dd0，由S24d，a625d，S486d，2有25d4d86d4，化简得19d216d440，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(2分)即19d22d20，解得d2，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(4分)则ana1(n1)d2n，所以an的通项公式为an2n．∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(6分)111选③，设递增等差数列an的公差为dd0，由是与的等差中项，得a5S2S5112，S2S5a5即S2S5a52S2S5，则有4d1010d24d24d1010d，化简得12d211d260，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(2分)即d212d130，解得d2，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(4分)则ana1(n1)d2n，所以an的通项公式为an2n．∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(6分)bnbnn1（2）由是以1为首项，2为公比的等比数列，得2，由（1）知an2n，anann1n即有bn22nn2，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(8分)123n则Tn122232n2，234n1于是2Tn122232n2，两式相减得：2(12n)T2122232nn2n1n2n1(1n)2n12，∙∙∙∙∙∙∙(11分)n12因此n1Tn(n1)22∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(12分)20.解：（1）由已知c2，得a2b24①，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(1分)x2y231设椭圆方程1，代入点M(3,1)得1②，联立①②解得a2b2a2b2a26,b22，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(3分)x2y2所以椭圆方程为1∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(4分)62（2）方法一：由题可知直线PQ斜率存在，设直线PQ的方程为ykxm.设点P(x1,y1)，Q(x2,y2)∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(5分)x2y21联立62得，(3k21)x26kmx3m260，满足Δ0时，∙∙∙∙∙∙∙∙∙∙(7分)ykxm6km3m26有xx,xx，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(8分)123k21123k21由AMBM可得kMPkMQ0,∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(9分)y1y1kxm1kxm1120，120，即x13x23即x13x23，化简得2kx1x2(m13k)(x1x2)23(m1)02代入韦达定理，可得33k3k(m2)3(m1)0，又点M(3,1)不在直线PQ上，因此3km10，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(11分)33所以3k30，即k，故直线PQ的斜率为定值.∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙(12分)33xx'3(x'3)2(y'1)2方法二：令,则M'(0,0)，则椭圆C'方程1∙∙∙∙∙∙∙∙∙∙(5分)yy'162即椭圆C'：x23y223x6y0，设直线PQ的方程为mxny1，∙∙∙∙∙∙∙∙∙∙(6分)y1y2m则kPM,kQM,kPQ，联立可得x1x2nx23y223x(mxny