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内蒙古赤峰市高三数学四月模拟考试试题文(PDF)答案内蒙古赤峰市届高三数学四月模拟考试试题文(PDF)内蒙古赤峰市届高三数学四月模拟考试试题文(PDF).pdf

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赤峰市高三4·20模拟考试试题文科数学参考答案2020.4说明：一、本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制订相应的评分细则．二、解答右端所注分数，表示考生正确做到这一步应得的累加分数．三、只给整数分数，选择题和填空题不给中间分．一、选择题：本大题共12小题，每小题5分，共60分1.C；2.D；3.A；4.C；5.B；6.A；7.C；8.B；9.D；10.C；11.A；12.B．二、填空题：本大题共4小题，每小题5分，共20分.1301350013．；14.（或写成60）；15.20，（或写成3）；16．3.8332三、解答题：共70分。解答应写出文字说明，证明过程或演算步骤。17.(12分)（1）证明：PAD为等边三角形，E为AD的中点，∴PEAD平面PAD底面ABCD，平面PAD底面ABCD=AD∴PE底面，BC平面，∴PEBC„„„„„3分由又题意可知ABCE为正方形，CEBC，又PEECE，BC平面PCE„„„„„„„„„„„„„„„„„„5分平面PBC，平面PBC平面PCE„„„„„„„„„„6分（2）解：过F作FGAB，垂足为G„„„„„„„„„„„„„„„„„„8分11111843VVSPEABFGPE13FABPPABF3ABF3232515„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„12分文科数学答案18.(12分)解：（1）由题设及正弦定理得sinBCABABsinsincos3sinsinABCsinCABsin()sinBABABABsin()sincos3sinsin„„„„„„„„„„„2分化简得sinBAA(3sincos1)0„„„„„„„„„„„„„4分1sinB0，3sinAAcos1，sinA620A，A„„„„„„„„„„„„„„„„„„6分3bc2212（2）由已知a23（1），根据余弦定理得cosA，即2bc1bc2212，bc=b22c12„„„„„„„„„„„„8分22bcb22c2bc,bc12（当且仅当bc=时取号）„„„„„„„„„10分11313SbcsinAbc1233„„„„„„„„12分ABC22222（当且仅当时取号）19.(12分)解：（1）m=30,n=20,x=50,y=50„„„„„„„„„„„„3分100(20203030)2K2=410.828（2）50505050，没有99.9%把握认为新冠肺炎的感染程度和性别有关.„„„„„„„„„„„„„„„„„„„„„7分文科数学答案（3）由于在“轻-中度感染”的患者中，按男女比例为2:3，设抽取的5人中3名女性患者用abc,,表示，2名男性患者用DE,表示，则所有组合为(,,)DEa，(,,)DEb，(,,)DEc，(,,)Dab，(,,)Dac，(,,)Dbc，(,,)Eab，(,,)Eac，(,,)Ebc，(,,)abc，可能的情况共有10种.其中至少抽到2名女性患者的情况有7种，设7至少抽到2名女性患者的事件为A，则PA()=.„„„„„„„„„„12分1020.(12分)解：（1）由已知得动点M到点F(0,1)的距离与到直线l:y1的距离相等„2分又由抛物线的定义可知，曲线C为抛物线，焦点为，准线为曲线C的方程为xy24„„„„„„„„„„„„„„„5分（2）设点A(,)x11y,B(,)x22y，P(t,1)„„„„„„„„„„„„„„6分11由xy24,即yx2,得yx．42x∴抛物线C在点A处的切线PA的方程为yy1(xx)121x1即y1xyx2．„„„„„„„„„„„„„„„„„8分21211x∵yx2，∴y1xy，∵点在切线PA上,14121xx∴11ty①，同理12ty②„„„„„„„„„„„10分2122x综合①、②得，点A(x,y),B(x,y)的坐标都满足方程1ty11222t即直线AB:yx1恒过抛物线焦点F(0,1)„„„„„„„„„12分2文科数学答案21.（12分）解：（1）函数fx()的定义域为(0，+).1(x1)(ax1)f(x)ax(1a)=„„„„„„„„„„„„„„„2分xx2ax1ax0,0，0，xf'x00x1，函数fx在(0,1)上为减函数；f'x0x1，函数在(1,)上为增函数a所以，fxf11，无极大