南充市高2023屆高考適應性考試（零診）文科數(shù)學參考答案及評分細則一．選擇題：本題共12小題，每小題5分，共60分．1234567891012CAADDBBADCCA.填空題：本題共4小題，每小題5分，共20分．13.314.515.416.3.解答題17.1）x123453，51201051009580y100...................................2分55i1xyii23455.......................................................................................................3分2x149ii15xy5xyii，∴b9i1522x5xii1由ybxa過10027127，∴y9x.........................................5分6月份獲得“金牌銷售稱號的有9612773（人）........................................................6分2）m60n........................................................................................................................8分22...................................................................................10分

K806010095%的把握認為獲得“金牌銷售”稱號與性別有關................................................12分18.1）解：由題可得a23a428，2(a2)aa，324則2(aa，3338a2a420............................................................................................................2分aaqaq311于是有aq21132,aa1或1.........................................................................4分q2q.2高三文科數(shù)學（零診）參考答案第頁（共5頁）1q22na..................................................................................................5分當n11aqa()6........................................................................................6分n當1n222）因為a是遞增的數(shù)列，所以a2n.nnbaan2n，...................................................................................................8分nn2nT12...b22232...n2①123nnn2122223...(2n2n1Tnn.......................................................10分則n②①，得T122...2nn2n1n即數(shù)列bnTn.......................................................................12分2(2n1nn19.1的中點F，連接EF，.E是的中點，所以1=，21CDABAB..................................................................................................3分2且，所以四邊形CDEF是平行四邊形，............................................................................................4分，又CF面，面.....................................................................5分.....................................................................................................................6分2）解：已知,AG4AG3，BG1.G到平面PBC距離為d.CD，2，CD222......................................................8分，2，2526.................................................9分CB22CBCP，，則CF.VV，所以GP11SdS.3311113CFdd.......................................11分32323所以點G到平面距離為33..........................................................................................12分高三文科數(shù)學（零診）參考答案第頁（共5頁）c3ea2abc22220.ab1得a2，b1.故橢圓E的方程為：x24y21............................................................................................4分2）設直線l：yk(xk0）P(4,0),設(x,y)，11B(x,y)，Q(x,y)...........................................................5分22QQx2y214yk(x(4kx32kxk0.2222330k(,0))6632k2xx1224k1k2xx1224k1.....................................................................................8分由PAQAPBQB4xxx11Q4xxx2Q2.....................................................................................10分則xQ32k4k22424(xx)2xx4k14k112212128(xx)32k21284k12點Q橫坐標為定值1.............................................................................................................12分21.1）f(x)2a(xxx（axRf(x)2x(x.....................1分20a1a0.f(x)在(,0)單調(diào)遞增，a)單調(diào)遞減，(,)單調(diào)遞增；.................2分a1a0.高三文科數(shù)學（零診）參考答案第頁（共5頁）f(x)在(,)單調(diào)遞增；.............................................................................................3分a1a0.f(x)在(,a)單調(diào)遞增，(,0)單調(diào)遞減，)單調(diào)遞增；.................4分2）g(x)f(x2axex(x，12x)，不等式g(x)xx2x0恒成立.得2xx1h(x)xxx1，(x..................................................6分x(xxx)h(x)p(x)xx.xe2x11

p(x)xx在)單調(diào)遞減，p()10，p10.

ee1x(，使得p(x)(xx)0.................................................8分則存在唯一0000eh(x)xx1在x)單調(diào)遞增，(x,)單調(diào)遞減.00xh(x)0得由0000...........................................................................................10分xx1xx12h(x)1.00000xxlnxxee0000a1(,)..............................................................................................................12分2e1xtt22.（）解：因為曲線C的參數(shù)方程為（t為參數(shù)）1yt所以曲線C的普通方程為：x2y24.........................................................................2分xysinl的極坐標方程為sincos30所以直線l的直角坐標方程為3yx30x3y30............................5分3xm22）點Pl上，直線l的參數(shù)方程為my12（m為參數(shù)..............6分高三文科數(shù)學（零診）參考答案第頁（共5頁）A、B對應參數(shù)分別為m、m12m，1m.2由x2y24和直線l的參數(shù)方程為3xm2my12（mmm020mm212mm12.....................................................................................................................8分||mmmm..............................................................................10分121223.1）|x2||x44x42x4x2或或............................................................................3分2x642462x4....