文科数学答案 【四川省南充市高三零诊】

南充市高2023届高考适应性考试（零诊）文科数学参考答案及评分细则一．选择题：本题共12小题，每小题5分，共60分．1234567891012CAADDBBADCCA.填空题：本题共4小题，每小题5分，共20分．13.314.515.416.3.解答题17.1）x123453，51201051009580y100...................................2分55i1xyii23455.......................................................................................................3分2x149ii15xy5xyii，∴b9i1522x5xii1由ybxa过10027127，∴y9x.........................................5分6月份获得“金牌销售称号的有9612773（人）........................................................6分2）m60n........................................................................................................................8分22...................................................................................10分

K806010095%的把握认为获得“金牌销售”称号与性别有关................................................12分18.1）解：由题可得a23a428，2(a2)aa，324则2(aa，3338a2a420............................................................................................................2分aaqaq311于是有aq21132,aa1或1.........................................................................4分q2q.2高三文科数学（零诊）参考答案第页（共5页）1q22na..................................................................................................5分当n11aqa()6........................................................................................6分n当1n222）因为a是递增的数列，所以a2n.nnbaan2n，...................................................................................................8分nn2nT12...b22232...n2①123nnn2122223...(2n2n1Tnn.......................................................10分则n②①，得T122...2nn2n1n即数列bnTn.......................................................................12分2(2n1nn19.1的中点F，连接EF，.E是的中点，所以1=，21CDABAB..................................................................................................3分2且，所以四边形CDEF是平行四边形，............................................................................................4分，又CF面，面.....................................................................5分.....................................................................................................................6分2）解：已知,AG4AG3，BG1.G到平面PBC距离为d.CD，2，CD222......................................................8分，2，2526.................................................9分CB22CBCP，，则CF.VV，所以GP11SdS.3311113CFdd.......................................11分32323所以点G到平面距离为33..........................................................................................12分高三文科数学（零诊）参考答案第页（共5页）c3ea2abc22220.ab1得a2，b1.故椭圆E的方程为：x24y21............................................................................................4分2）设直线l：yk(xk0）P(4,0),设(x,y)，11B(x,y)，Q(x,y)...........................................................5分22QQx2y214yk(x(4kx32kxk0.2222330k(,0))6632k2xx1224k1k2xx1224k1.....................................................................................8分由PAQAPBQB4xxx11Q4xxx2Q2.....................................................................................10分则xQ32k4k22424(xx)2xx4k14k112212128(xx)32k21284k12点Q横坐标为定值1.............................................................................................................12分21.1）f(x)2a(xxx（axRf(x)2x(x.....................1分20a1a0.f(x)在(,0)单调递增，a)单调递减，(,)单调递增；.................2分a1a0.高三文科数学（零诊）参考答案第页（共5页）f(x)在(,)单调递增；.............................................................................................3分a1a0.f(x)在(,a)单调递增，(,0)单调递减，)单调递增；.................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)单调递减，p()10，p10.

ee1x(，使得p(x)(xx)0.................................................8分则存在唯一0000eh(x)xx1在x)单调递增，(x,)单调递减.00xh(x)0得由0000...........................................................................................10分xx1xx12h(x)1.00000xxlnxxee0000a1(,)..............................................................................................................12分2e1xtt22.（）解：因为曲线C的参数方程为（t为参数）1yt所以曲线C的普通方程为：x2y24.........................................................................2分xysinl的极坐标方程为sincos30所以直线l的直角坐标方程为3yx30x3y30............................5分3xm22）点Pl上，直线l的参数方程为my12（m为参数..............6分高三文科数学（零诊）参考答案第页（共5页）A、B对应参数分别为m、m12m，1m.2由x2y24和直线l的参数方程为3xm2my12（mmm020mm212mm12.....................................................................................................................8分||mmmm..............................................................................10分121223.1）|x2||x44x42x4x2或或............................................................................3分2x642462x4....