江西省八所重点中学2022届高三文科数学4月联考试卷（PDF版附答案）

\n\n江西省八所重点中学2022届高三联考文科数学答案故2022年2月20日该电商平台的预售人数146.8万人························12一、选择题18.解：（1）当n2时，由ann1a1a2a题号123456789101112得aaaa··················································1nn12-1答案BADCACDABCCD二、填空题an1两式作差得aaa，即2(n2)··························2n1nnan13.114.7215.316.52三、解答题a2又1a117.解：（1）由表中数据可得xy3,61·····································2所以数列{}a是从第二项开始公比为2的等比数列····························45n2所以(xxi)10·······················································3i11,n1故a····················································5nn22,n2552又(yyi)460，(xiixy)(y)66i1i1（2）当n1时，S1··················································6nn(xxy)(y)013nn21ii当n2时，Sn122222···························766i1所以r0.970.75······················5nn1046n1时，亦满足上式···············································822(xiix)(yy)n1ii11所以S2··························································9nn1所以该电商平台的第x天与到该电商平台参与预售的人数y（单位：万人）具有较高的线性相关即bSlogS2n1········································10nn2n程度即可用线性回归模型拟合人数y与天数x之间的关系．···························6Tnnb1b2b3bn012n1(xiixy)(y)(2222)(012n1)ˆ66（2）由表中数据可得bi16.6························8nn12(nn1)210·························12()xxi122i12nnn22则aˆybxˆ616.6341.2············································92所以yxˆ6.641.2·······················································10令x16，可得yˆ6.61641.2146.8（万人）····························111\n19.解：（1）取EF的中点G，连接DGCG，所以fx()在[1,2]上为增函数，四边形ADEF和BCEF都是菱形，DEFCEF6011221122E42mm即fx()f(2)，fx()f(1)·······························8max2minDGEF·································1Gee又BC2DC(4em)2fx()fx()的最大值为gm()fx()fx()·················912maxmin2eDGCG3································2F4[()fxfx()]恒成立················································10212又CDAB6ABe4(4em)2，22222ee即有DGCGCD，DGCG··········································32即m····························································11又CGEFG，CGEF，平面BCEF4eDG平面BCEF·······················································52又DG平面AFED又m0m0,4e平面AFED平面BCEF···············································6（2）连接DFDB，2由（1）知DG平面BCEF，同理CG平面AFED·······················7故m的取值范围0,···················································124e1且S2323，S233····························9菱形BCEFADF211c22212221221.解：（1）e，ca，baca.·············2VV=V=VV=SDGSCGDBCEFDABFDBCEFBADF菱形BCEFADFa2223311椭圆的中心O到直线xy2b0的距离为52，2333333·················································12|2|b22252，b5.b25,a2b50.322xy22mx2(m1)x4(mx2)(x2)20.解：（1）fx()(m0)··············1椭圆C的方程为1.································4xx5025ee（2）由（1）可知F(5,0)，由题可知直线AB的方程为yx2(5)，与椭圆C22令fx()0，解得x或x2，且2································2mmyx2(5)222的方程联立xy22，消去y得xx8100当x(,]时，fx()0，当x(,2)时，fx()0，1mm5025当x[2,)时，fx()0··················································4设Axy(,11)，Bxy(,22)，则有x1x28,xx1210.··························6设Qxy(,)，由OQOAOB得(x,y)(x,y)(x,y)(xx,yy)，2211221212即fx()的单调增区间为(,2)，单调减区间为(,]，[2,)··············5mmxx1x2····························································7yyy（2）由（1）知，当mx0,[1,2]时，fx()0恒成立·························6122\n22又点Q在椭圆上，∴xy2502即(xx)222(yy)50①xt212122整理得故直线l的参数方程为（t为参数），22(x22y2)2(x22y2)2(xx2yy)50②·········8yt2112212122点A,B在椭圆上，222222∴xy250，xy250yx1122把直线l的参数方程代入1，xx2yyxx4(x5)(x5)5xx20(xx)10010③48121212121212得到：2······································································10tt+12280············································7将②③代入①可得所以tt=122,tt8········································822121250502050222又211tt(tt)4tt28832121212故+5······10|MA||MB|tttt8221212即2050505010053xx3,1，当且仅当时取“”823.解：（1）fx()x5,1x2·······························25∴的最大值为.···················································123xx3,28x112xx221x22不等式fx()6等价于或或x2(tt)tt23x36x563x3622.解：（1）曲线C1的参数方程为（t为参数），整理得2，1ytt222解得1x3yt2t∴不等式的解集为[1,3]············································522yx两式相减得曲线C的直角坐标方程为：1·························3148（2）由（1）知：当x1时，f(x)6；当1x2时，3f(x)6；当x2时，f(x)3.xcos曲线C的极坐标方程为sin()220，根据ysin，可得曲线C直角坐标方故函数f(x)的值域为[3,)，即f(x)的最小值是3····················72242222xy∵不等式fx()2a5a对一切实数x恒成立，程为：xy40····················································521∴2aa53，解得：3a2（2）由于点M(2,2)满足直线l的方程，1故实数a的取值范围是3,·······································1023