广西南宁市2023届高中毕业班第一次适应性测试数学（文科）试题

南宁市2023届高中毕业班第一次适应性测试数学（文科）答案一、选择题：1.【答案】Ｃ【解析】∵A{xN|1„x„3}{0,1,2,3},∴AB{0,1,2,3,4},故选Ｃ.2.【答案】A3i3i1i24i【解析】由题意z1i3i,可变形为z12i,1i1i1i2则复数z12i,故选A.3.【答案】B【解析】5件产品中的2件次品记为a,b,3件合格品记为A,B,C,“从这5件产品中任取2件”,则该试验的样本空间Ω={(a,b),(a,A),(a,B),(a,C),(b,A),(b,B),(b,C),(A,B),(A,C),(B,C)},n(A)6即n()10.设事件A=“恰有一件次品”,则n(A)6,故P(A)0.6.n()104.【答案】B22【解析】sin1cos1cos2cos1,cos2cos20,(cos1)(cos2)0,cos1或cos2(舍)3又sincos1.25.【答案】D【解析】对于A,f(x)tanx为奇函数,在定义域内不单调,不符合题意,A错误.1对于B,f(x),定义域为(,0)(0,),f(x)f(x),所以f(x)为奇函数,在定义域x内不单调,B错误.对于C,f(x)xcosx,f(x)xcos(x)xcosxf(x),故函数f(x)xcosx不是奇函数,不符合题意C,错误.故选D.6.【答案】B【解析】符合题目要求的分类方法共：“甲3张乙1张”,“甲2张乙2张”,“甲1张乙3张”,三类①“甲3张乙1张”的基本事件为：甲123乙4；甲124乙3,甲134乙2,甲234乙1,共4类；②“甲2张乙2张”的基本事件为：甲12乙34；甲13乙24,甲14乙23,甲23乙14,甲24乙13,甲33乙12,共6类；③“甲1张乙3张”的基本事件为：乙123甲4；乙124甲3,乙134甲2,乙234甲1,共4类；故选B.7.【答案】B【解析】已知圆锥的侧面展开图为半径是3的扇形，如图，一只蚂蚁从A点出发绕着圆锥的侧面爬行一圈回到点A的最短距离为AA，设ASA，2圆锥底面周长为2，所以AA32所以，在SAA中，由3SASA3,得AASA2SA22SASAcos13232233()332故选：B.文数答案第1页共10页8.【答案】Ｃ【解析】设报告厅的座位从第排到第排各排的座位数依次排成一列构成数列其120,,an,前项和为根据题意数列是一个公差为的等差数列且nSn.,and2,a1041,故a1a109d411823.20(201)由S20a2840,因此,则该报告厅总有座位数为840个座位.故选Ｃ.20129.【答案】Bπ4【解析】已知sin，65πππ2π167cos2cos2cos212sin12,33662525故选B.10.【答案】Ｄ【解析】由f(x)x2,得f(x)2x,则f(1)2,又f(1)1,所以函数f(x)x2的图象在x1处的切线方程为y12(x1),即y2x1.ex设y2x1与函数g(x)的图象相切于点x,y,a00ex0xgx02,3ea31ee由g(x),可得解得x,ae2,故选Ｄ.x0ae0222gx2x1,0a011.【答案】A【解析】设A(x1,y1),B(x2,y2)则由于AB的斜率存在,设AB的斜率为k.ppA,B,都在x轴上方,由题意知k0,由抛物线定义AFx1,BFx222px1222则x1x24,由弦长公式AB1kxx,p12x42253所以AB1k2xx51k2k,故选A.124412.【答案】Ｃlnx1lnx【解析】设f(x),则f(x),xx2当0xe时,f(x)0,函数单调递增,当xe时,f(x)0,函数单调递减,1故当xe时,函数取得最大值f(e),e3(2ln3)e2ln31因为af(),cf(3),bf(e),e233elnx故ba,bc,设函数ym的零点为x,x,且0xx,x1212则mx1lnx1,mx2lnx2,所以lnx2lnx1m(x2x1),lnx2lnx1lnx1x2m(x2x1)①,文数答案第2页共10页2(x1)(x1)2令g(x)lnx,x1,则g(x)0,x1x(x1)22(x1)故g(x)在(1,)单调递增,g(x)g(1)0,所以,当x1时,lnx,x1x2(21)x2x112从而ln,即(lnx2lnx1)②,xx2xxxx112121x122e①代入②得,x1x2e,令x,则x23,13故f(x1)f(x2)f(3),故ac,综上acb.故选：Ｃ.二、填空题：13.【答案】2【解析】由约束条件作出可行域如图所示,由目标函数z3xy可知当目标函数过点C(2,4)时,z取得最大值,最大值3242.π14.【答案】214【解析】函数f(x)cos(3x)的图象关于点,0对称,234ππ7π则有3kπ,kZ,于是得kπ,kZ,3227π显然kπ对于kZ是单调递增的,2ππ而k3或4时,||,所以||的最小值为.2215.【答案】e3PF1PF2PF1sinPFF【解析】由正弦定理得,所以22sinPF2F1sinPF1F2PF2sinPF1F2即PF12PF2,由双曲线的定义可得PF1PF2PF22a,所以PF22a,PF14a；222因为F1PF260,由余弦定理可得4c16a4a24a2acos60,c2整理可得4c212a2,所以e23,即e3.a216.【答案】225【解析】取的中点的中点连结正方体BB1G,A1B1H,GH,C1G,C1H,A1B,EG,HF.的棱长为为中点所以ABCDA1B1C1D12.E,F,G,H,EF//A1B,GH//A1B,所以EF//GH且EFGH2.因为为分别为的中点F,HAB,A1B1,所以且所以四边形为平行四边形FH//CC1,FHCC1,FHC1C,所以HC1//CF.因为面面HC1CD1EF,CFCD1EF,所以面HC1//CD1EF.文数答案第3页共10页同理可证：面HG//CD1EF.又面面GHHC1H,HC1C1GH,GHC1GH,所以面面C1HG//CD1EF.所以点在正方体表面上运动所形成的轨迹为三角形PC1HG.因为正方体ABCDABCD的棱长为2,所以22,1111HC1GC1215所以三角形的周长为.C1HGGHHC1GC1255225三、解答题：117.【答案】（1）a0.025，360人（2）2【解析】（1）依题意,成绩在[40,50]的概率为p10.1,成绩在[50,60)的概率为p20.15,成绩在的概率为成绩在的概率为60,70p30.15,70,80p40.3,成绩在的概率为成绩在的概率为80,90p5a10,[90,100]p60.05,故p5a1010.10.150.150.30.050.25,∴a0.025；······························································································3分该年级生涯规划大赛初赛成绩“优秀”等级的概率为pp5p60.3.4分该年级生涯规划大赛初赛成绩优秀”等级的学生人数为0.31200360人；············6分（2）在评为“优秀”等级的学生中采用分层抽样抽取6人，其中成绩在80,90应抽5人,成绩在[90,100]应抽1人,分别设为A1,A2,A3,A4,A5,和B,·················································································································7分从6人中随机抽取3人的基本事件为：A1,A2,B,A1,A3,B,A1,A4,B,A1,A5,B,A2,A3,B,A2,A4,B,A2,A5,B,A3,A4,B,A3,A5,B,A4,A5,B,A1,A2,A3,A1,A2,A4,A1,A2,A5,A1,A3,A4,A1,A3,A5,，A1,A4,A5,A2,A3,A4,A2,A3,A5A2,A4,A5,A3,A4,A5,共20种,·····································································································9分其中恰有1人成绩在[90,100]的基本事件为：A1,A2,B,A1,A3,B,A1,A4,B,A1,A5,B,A2,A3,B,A2,A4,B,A2,A5,B,A3,A4,B,A3,A5,B,A4,A5,B,共10种,···································································································10分文数答案第4页共10页101故所求概率为P.·············································································12分202π【答案】（）B（）(5,6]18.132【解析】（1）由(bc)(sinBsinC)(sinAsinC)a根据正弦定理可得(bc)(bc)(ac)a,·························································1分所以,a2c2b2ac,··················································································2分a2c2b21由余弦定理可得cosB,··························································3分2ac2∵B(0,π),································································································4分π因此B．·······························································································5分3(2)由余弦定理,得b2a2c22accosB,3a2c2ac,····································6分即a2c23acacb3由正弦定理得2,sinAsinCsinB3,·····························