数学答案青岛

青岛市2023年高三年级第一次适应性检测数学参考答案及评分标准一、单项选择题：本题共8小题，每小题5分，共40分。1--8：ACBBDCCA二、多项选择题：本题共4小题，每小题5分，共20分。9．AC10．AB11．ABD12．BCD三、填空题：本题共4个小题，每小题5分，共20分。3113．(1,2)答案不唯一；14．；15．23；16．0，．31011四、解答题：本题共6小题，共70分。解答应写出文字说明，证明过程或演算步骤。17.（本小题满分10分）解：（1）由题意得f(x)2cos2xsin2x1cos2xsin2xπ2sin(2x)1························································2分42π因为T2π，所以21······································································3分Tπ所以f(x)2sin(x)14πππ令xkπ得，xkπ（kZ）424π所以函数f(x)图象的对称轴方程为xkπ（kZ）····································5分41π2（2）由f()得sin()···························································6分343244所以sincos，所以(sincos)2，即1sin23995所以sin2·······················································································10分918．（本小题满分12分）解：（1）因为S2，S4，S54成等差数列，a2，a4，a8成等比数列2S4S2S54所以2·················································································2分a4a2a82(4a16d)(2a1d)(5a110d)4a1d4所以2，整理得2(a13d)(a1d)(a17d)a1dd因为d0，解得：a1d2·······································································5分n(n1)所以S2n2n2n·································································6分n2数学评分标准第1页（共8页）n2n3（2）由（1）得2bT，2bT，·····················7分nnn(n1)n1n1(n1)(n2)n2n3所以两式相减得：2b2b+TT························8分nn1n1nn(n1)(n1)(n2)21整理得：2bbnn1n(n1)(n1)(n2)11所以2[b]bnn(n1)n1(n1)(n2)11即bn12(bn)············································································10分Sn1Sn31因为b,b10，121121所以{bn}是以1为首项，2为公比的等比数列············································11分Sn11所以b2n1，所以b2n1···········································12分nn(n1)nn(n1)19．（本小题满分12分）π解：（1）当D为圆弧BC的中点，即CAD时，BCPD···························1分3证明如下：因为D为圆弧BC的中点，π所以CADBAD，即AD为CAB的平分线3因为ACAB，所以AD为等腰△CAB的高线，即ADBC····························2分因为PAAB,PAAC,ABACA,AB,AC平面ABDC所以PA平面ABDC，所以PABC··························································3分因为PAADA，所以BC面PAD所以BCPD····························································································4分（2）由（1）得，PA为四棱锥PABDC的高，因为PA4，所以，当底面积SABDC取最大值时，四棱锥PABDC体积最大·······5分2π2π设CAD，则BAD,(0,)33112SSS22sin22sin()ABDCCADBAD2232ππ2[sinsin()]23sin()362πππ5π因为(0,),(,)3666ππ所以时，sin()1,S取最大值2336ABDC数学评分标准第2页（共8页）π所以，当四棱锥PABDC体积最大时，CADBAD····························7分3过A在平面ABDC内作直线AEAB，交圆弧BC于点E，由题，AE,AB,AP两两垂直，以A为原点，分别以AE,AB,AP所在直线为x轴，y轴，z轴建立如图所示空间直角坐标系，································································8分则A(0,0,0),P(0,0,4),B(0,2,0),D(3,1,0),C(3,1,0)因为PD(3,1,4),CD(0,2,0),DB(3,1,0)，······································9分z设平面PCD的法向量为n(x1,y1,z1)nPD03xy4z0P则，即111，nCD02y10令z13，得n(4,0,3)·····························10分AB设平面PBD的法向量为m(x2,y2,z2)ymPD03x2y24z20C则，即，DExmDB03x2y20令z23，得m(2,23,3)···································································11分|mn|11设平面PCD与平面PBD的夹角为，则cos|m||n|1911所以平面PCD与平面PBD夹角的余弦值为·············································12分1920．（本小题满分12分）解：（1）因为前2个矩形面积之和为(0.010.03)100.40.5，前3个矩形面积之和为(0.010.030.04)100.80.5，则中位数在(80,90)内，设为m，则(m80)0.040.50.40.1，解得m82.5，即中位数为82.5．·································································3分11（2）因为成绩在[90，100]的频率为，所以概率为，55114则X~B(10,)，所以P(Xk)Ck()k()10k，········································5分5105514Ck()k()10kP(Xk)10115k所以551，······································6分k11k1411kP(Xk1)C()()4k1055当1k2时，P(Xk)P(Xk1)，P(X0)P(X1)P(X2)；当k3时，P(Xk)P(Xk1)，P(X2)P(X3)，所以k2时，P(Xk)取到最大值．····························································7分数学评分标准第3页（共8页）C3C1C43C3C1142433（3）甲进入复赛的概率P14，乙进入复赛的概率P24，C65C6531故甲、乙两人进入复赛的概率分别为，．····················································8分55248由题意可得：的可能取值为0，1，2，则有：P(0)(1P)(1P)，125525342114313P(1)P(1P)(1P)P，P(2)PP，1212555525125525所以的分布列为：0128143P252525················································································11分81434所以E()012························································12分252525521．（本小题满分12分）解:（1）记|F1F2|2c，由题意知：|AF1||AF2|a，2c2a······················1分12所以Sa1，解得a2·······························································2分AF1F22所以b1,c1····························································································3分x2所以椭圆C的标准方程为：y21····························································4分2（2）（ⅰ）选②③为条件：设P(x1,y1)，Q(x2,y2)，当直线l的斜率不存在时，根据椭圆的对称性不妨设点P在第一象限，12则由kk，可得k，122122x22此时直线WP的方程为yx，联立y21，解得P(1,)2222所以S·······························································································6分2当直线l的斜率存在时，设直线l的方程为：ykxt，y1y21则k1k2，即x1x22y1y20x1x22x2将ykxt代入y21得：(12k2)x24ktx2t22024kt2t22所以xx,xx,····························································7分1212k21212k2数学评分标准第4页（共8页）t22k2所以yy(kxt)(kxt)k2xxkt(xx)t2