2022年巴中市零诊考试文科数学参考答案（20220904定稿）_2023年8月_01每日更新_2号_2023届四川省巴中市高三9月零诊_2023届四川省巴中市高三零诊考试文数试题

巴中市普通高中 级“零诊”考试 2020 数学阅卷参考答案（文科） 一、选择题：本大题共12小题，每小题5分，共60分．在每小题给出的四个选项中，只有 一个是符合题目要求的． 1．【解析】B．先写出集合M ，然后逐项验证即可．由U {1, 2, 3, 4, 5}且ð M {1, 2}得M {3, 4, 5}， U 故选B．备注：2022年全国乙卷理数第1题改编. 2．【解析】C．利用复数四则运算，先求出z，再依照复数的概念求出复数z的虚部．选C． 方法一：由题意有z34i (34i)(i) 43i，故复数z的虚部为3． i i(i) 方法二：由iz34ii(3i4)，得z43i，故复数z的虚部为3． 3．【解析】A．l ∥l m1，故“m1”是“l ∥l ”的充分不必要条件．选A． 1 2 1 2 4．【解析】D．不妨取双曲线的右焦点(c, 0)，渐近线ybx，由点到直线距离公式得b24，然后利 用离心率的变通公式c 1b2  5，进而求得离心率e的值．由题意得，不妨取双曲线的右焦点 F( 1b2, 0)，双曲线的渐近线为 ybx，即bxy0，则 |b b210| b2，即b24a2，所以离 b21 心率e b21 5．选D． 5．【解析】C．充分利用长方体中的棱、面之间的关系直观感知，同时结合空间中线面间平行及垂直 的判定与性质推理论证，需注意相应定理的条件的完备性．对于A选项，n也可能；对于B选项， 由条件得不到m，故不能推断出；对于C选项，则法线与法向量垂直则两个平面垂直知正确； 对于D选项，条件中缺少m，故得不到m． 6．【解析】D．由任意角的三角函数定义，得tanab，故B(2, 2a)，|OB|2 1tan22|OA|．由 1 2 cos23得：cos2cos2sin2 cos2sin2 3，变形得：1tan2 3，解得tan24，所 5 cos2sin2 5 1tan2 5 以|OB|2 5 ．或者，设|OA|r ，则 r21a2 ，sina , cos1 , |OB|2r ；由 cos23 得 r r 5 cos2cos2sin21a2 1a2 3 ，解得：a24，故|OB|2r2 5．选D． r2 1a2 5 7．【解析】D．借助判断函数的奇偶性、对称性和有界性，正弦型函数的符号变化规律，均值不等式 2sin(x) 等知识进行推断．由 f(x) , x[2, 2] 知 f(x)为奇函数，且在(0, 1)内恒正，故A、B选项不 exe x 正确；又2sin(x)≤2，exe x≥2且等号不同时成立，由不等式的性质知| f(x)|1，排除C选项．选 D． 8．【解析】B．设公差为 d ，则由 a a a 6 得 3(a 8d)6 ，即 a 8d a 2 ，故 2 8 17 1 1 9 17(a a ) S  1 17 17(a 8d)34．选B．或者由a 8d a 2得S 17a 34．作为选择题由于满 17 2 1 1 9 17 9 足条件的数列不唯一，可举常数列取a 2验证作出选择． n 9．【解析】A．本题考查平面向量的线性运算、数量积及其几何意义，数量积的坐标表示，数形结合 思想、化归与转化思想、函数与方程思想，运算求解能力．   方 法 一 ： 由 点 D 在 BC 上 ， 设 BDxBC , 0≤x≤1 ， 则           ADABBDABxBCABx(ACAB) (1x)ABxAC ， 故  A  D    B  C    A  D  (  A  C    A  B  )[(1x)  A  B  x  A  C  ](  A  C    A  B  ) x  A  C 2 (x1)  A  B 2 13x4，由   y 0≤x≤1得4≤13x4≤9，所以ADBC[4, 9]．选A． C 3 方法二：以A为原点，AB，AC所在直线分别为x, y轴建立平面直角坐标系（如图），     则 AB(2, 0), AC(0, 3), BC(2, 3) ， 设 D(x, y) ， 则 AD(x, y) ， 故 D   ADBC2x3y（*），由点D在BC上得：3x2y60, 0≤x≤2（可借助初中 B 的一次函数知识或必修2第三章直线的方程获得x, y满足的方程），用x表示 y代入 A 2 x 文科数学答案第1页 共8页    （*）式得：ADBC2x3y913 x, 0≤x≤2，故ADBC[4, 9]．选A． 2       方法三：设AD 与与BC的夹角为，则由题意得ADBC 13|AD|cos，故|AD|cos取最大值时      ADBC最大，|AD|cos取最小值时ADBC最小，结合上图，用运动变化的观点分析易知：D在斜边     BC上移动时，当D与C重合时AD的模最大且与BC的夹角最小（ACB），故此时ADBC取得最大          值，且ADBCmax ACBC AC(ACAB)9；当 D与B重合时AD 的模最小且与BC的夹角最大          （ABC），故此时ADBC取得最小值，且ADBCmin=ABBCAB(ACAB)4．应注意，由向   量夹角的定义知ABC 不是向量AB与BC的夹角！！这是向量问题中的易错点！ 10．【解析】B．将函数ycos(x  )的图象向左平移个单位长度，得ycos[(x  )  ]的图象．而 3 3 3 3 ycos[(x  )  ]cos(x    )sin[  (x    )]sin(x  5 ) ， 故 由 题 意 知 3 3 3 3 2 3 3 3 6 sinx sin(x  5 )，所以x  5 2kx (kZ)，解得6k5 (kZ)，由0知： 3 6 3 6 2 当k1时取最小值，故 7．选B．或者，由ycos(x  )知x  2时y1，由ysinx min 2 3 3 知当x 时y1，故由题意得 5     ，解得 7 ． 2 3 3 2 2 11．【解析】D． f(x)3x23的变号零点为x1和x1，故A正确；由 f(1)301 f(1) 知B 正确；由yx33x是奇函数，其图象向上平移1个单位长度得到函数 f(x)的图象，故C正确；由于 函数 f(x)在x1处取极小值1，故直线xy0与曲线y f(x)不相切，故D错误，选D．也可借助 函数的图象直观感知作出判断． 12．【解析】A．由已知得：alog 61log 3, blog 121log 4, clog 201log 5 ，故a, b, c 2 2 3 3 4 4 的大小顺序与log 3, log 4, log 5的大小一致．由log 3log 9log 5知ac，排除B、D．由2332 2 3 4 2 4 4 得log 33 ；由4233得2log 43，即log 43 ，所以ab，排除C．故选A． 2 2 3 3 2 ln(x1) 或者利用函数 f(x) (x1)的单调性比较log 3, log 4, log 5的大小．事实上，当x1时 2 3 4 lnx lnx  ln(x1) f(x) x1 x 0 ，故 f(x) 在(1, ) 上是减函数，所以 f(2) f(3) f(4) ，由换底公式得 ln2x log 3log 4log 5，故abc．选A． 2 3 4 二、填空题：本大题共4个小题，每小题5分，共20分． 13．【解析】2．由抛物线y2 2px (p0)的几何性质知，其焦点到准线的距离为 p，本题中 p2． 14．【解析】57．计算得x1 (2356)4，y1 (28314148)37，则样本中心点是(4, 37)， 4 4 代入回归方程得ay5x375417，所以回归方程是y5x17，将x8代入得y57． 15．【解析】8 6．由BD平面ADC，AD, DC平面ADC，得BDAD, BDCD；由BD2， AB2 2 ， BC2 5 及勾股定理得： AD2, CD4 ，又 AC2 5 ，故 AD2CD2AC2 ，所以 ADDC ，即 BD，AD，CD 两两垂直，所以三棱锥 ABCD的外接球与以BD，AD，CD分别为长、宽、高的长方体的外接相同（如 右图，O为球心），所以球半径R 222242  6，从而V 4R38 6． 2 3 16．【解析】，(0, 3 ]．以三角形边角关系的射影定理为背景，综合考查正弦、 3 4 余弦定理、三角变换的基本公式与方法，三角函数的图象与性质等知识，求角A时，既可用正弦定理 边化角，也可用余弦定理角化边，还可直接用教材中习题的结论——射影定理简化；对于sinBsinC的 范围问题，可利用BC2且0B, C 转化只含一个角变量的函数的值域． 3 2 （1）求角A的过程与方法． ①由已知及正弦定理得：2sinAcosAsinCcosBsinBcosC sin(BC)sinA ，又0A ， 2 文科数学答案第2页 共8页故cosA1，所以A ． 2 3 ②由已知及射影定理得：2acosAccosBbcosCa，故cosA1，又0A ，所以A ． 2 2 3 ③由已知及余弦定理得：a2c2b2 a2b2c2 2acosA，化简得cosA1，又0A ，所以 2a 2a 2 2 A ． 3 （2）求sinBsinC范围的过程与方法． 策略一：利用正弦型函数的图象与性质． 由A 得BC2，故C2 B，且0B2． 3 3 3 3 ①sinBsinCsinBsin( 2 B)sinB( 3 cosB1 sinB) 3 sin2B1 cos2B11 sin(2B  )1． 3 2 2 4 4 4 2 6 4 因为  2B  7，故1sin(2B  )≤1，当且仅当B 时取等号，故sinBsinC(0, 3 ]． 6 6 6 2 6 3 4 ②令B  x, C  x，由题意得  x ， 3 sinx 3 ． 3 3 3 3 2 2 故sinBsinCsin(  x)sin(  x) ( 3 cosx1 sinx)( 3 cosx1 sinx) 3 3 2 2 2 2 3 cos2x1 sin2x 3sin2x( 1 , 3 ]． 4 4 4 2 4 因为  x ，所以 3 sinx 3 ，0sin2x3，当且仅当x0，即BC 时取等号， 3 3 2 2 4 3 故sinBsinC 3sin2x(0, 3 ]． 4 4 ③由和、差角的余弦公式可得：2sinBsinCcos(BC)cos(BC)cos(BC)1， 2 由已知得0B, C2，故2 BC2，所以cos(BC)(1 , 1]，当且仅当BC 时取 3 3 3 2 3 等号，故sinBsinC(0, 3 ]． 4 策略二：用余弦定理转化． ④在△ABC中，由正弦、余弦定理得：sin2Bsin2C2sinBsinCcosAsin2A ，代入A 得： 3 3sin2Bsin2CsinBsinC，变形得sinBsinC3(sinBsinC)2， 4 4 又sinBsinCsin( BCBC )sin( BCBC ) 2 2 2 2 2cos BC sin BC 2cos  sin BC sin BC ， 2 2 3 2 2 由已知得0B, C2，故   BC  ，所以sin BC ( 3 , 3 ) 3 3 2 3 2 2 2 所以0≤|sinBsinC| 3 ，当且仅当BC 时取等号， 2 3 故sinBsinC(0, 3 ]． 4 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．第17－21题为必考题， 每个试题考生都必须作答．第22、23题为选考题，考生根据要求作答． （一）必考题：60分． 17．（12 分） 解：（1）方法一： 由S 2S 2得：S 2S 2·······································································1分 n1 n n2 n1 ∴ S 2S S 2S ，变形得S S 2(S S ) ···································2分 n1 n n2 n1 n2 n1 n1 n ∴ a 2a ①···························································································3分 n2 n1 又 a 2且S S S 2 1 2 1 1 ∴ a 2a ②··································································································4分 2 1 由①②知：对任意nN*，恒有a 2a ，且a 2 n1 n 1 文科数学答案第3页 共8页∴ 数列{a }是首项与公比均为2的等比数列·························································5分 n ∴ a 2n·······································································································6分 n 方法二： 由S 2S 2变形得：S 22(S 2) ·····························································2分 n1 n n1 n 又 a 2，故S 2a 24·············································································3分 1 1 1 ∴ 数列{S 2}是以4首项，2为公比的等比数列 n ∴ S 242n12n1，故S 2n12······························································4分 n n ∴ 当n≥2时，a S S 2n12(2n2)2n················································5分 n n n1 又 a 2也适合上式 1 ∴ a 2n·······································································································6分 n （2）方法一： 由（1）知，b na n2n··················································································7分 n n ∴ T 12222323n2n n 2T  122223(n1)2nn2n1···················································8分 n 两式相减得：T 22223 2nn2n1···························································9分 n  2(12n) n2n1(1n)2n12··············································11分 12 ∴ T (n1)2n12．···················································································12分 n 方法二： 由（1）知，b na n2n··················································································7分 n n 裂项变形得：b n2n(n1)2n1(n2)2n······················································9分 n ∴ T 12222323n2n n 223(22423)(325224)[(n1)2n1(n2)2n]···················10分 2(n2)2n1 即 T (n1)2n12．·················································································12分 n 18．（12 分） 解：（1）由题意得，总人数为200 45岁以上（含45岁）的人数为2003120，45岁以下的人数为80···························1分 5 一周内健步走少于5万步的人数为200 3 60·······················································2分 10 由此得如下列联表：···························································································3分 一周内健步走≥5万步 一周内健步走<5万步 总计 45岁以上(含45岁) 90 30 120 45岁以下 50 30 80 总计 140 60 200 故 K2 200(90305030)2  253 2.706·······················································5分 1406080120 7 ∴ 有90%的把握认为该市市民一周内健步走的步数与年龄有关································6分 （2）由题意，抽取的8人中一周内健步走≥5万步有6人，少于5万步的有2人··················7分 将一周内健步走≥5万步的6人编号为1，2，3，4，5，6，另外两人记为A, B， 则所有可能情况如下： 12，13，14，15，16，23，24，25，26，34，35，36，45，46，56， 1A，2A，3A，4A，5A，6A，1B，2B，3B，4B，5B，6B，AB．总共28种．···········10分 其中恰有一人一周内健步走步数不少于5万步所有可结果如下 ： 1A，2A，3A，4A，5A，6A，1B，2B，3B，4B，5B，6B．共12种·························11分 记“抽取的2人中恰有一人一周内健步走步数不少于5万步”这事件C 文科数学答案第4页 共8页由等可能事件的概率公式得：P(C)12 3．··············································· 12分 28 7 19．（12 分） 解：（1）证明 方法一： F 由正方形ABCD的性质得：AB∥CD······································································1分 又 AB平面DCF，CD平面DCF E ∴ AB∥平面DCF ··························································································2分 ∵ BE∥CF ，BE平面DCF，CF平面DCF ∴ BE∥平面DCF······················································································································3分 C ∵ ABBEB, AB, BE 平面ABE B ∴ 平面ABE∥平面DCF···········································································································4分 ∵ AE平面ABE D A ∴ AE∥平面DCF····························································································6分 方法二： 在CF取点G使得CG2BE，连结EG、DG，如右图 F ∵ BE∥CF ∴ 四边形BEGC是平行四边形·············································G······························1分 E 故 EG∥BC，且EGBC ··················································································2分 又 AD∥BC, ADBC ∴ AD∥EG, ADEG·······················································································3分 ∴ 四边形ADGE是平行四边形··············································C·····························4分 B ∴ AE∥DG·····································································································5分 又 AE平面DCF，DG平面DCF D A ∴ AE∥平面DCF······················································································································6分 1 （2）由体积的性质知：V FACE V ACEF  3 S △CEF h·················································8分 ∵ 平面BCFE平面ABCD，平面BCFE平面ABCD=BC ABBC ，AB平面ABCD ∴ AB平面BCFE··························································································9分 又 AB2 故 点A到平面CEF的距离为2，即三棱锥ACEF 底面CEF上的高h2···············10分 由题意，知BE BC, BE∥CF 且CF 3, BC 2 ∴ S△CEF 1 2 CFBC3··················································································11分 1 1 ∴ V FACE V ACEF  3 S △CEF h 3 322．············································12分 20．（12分 解：（1） f(x)1xa x2ax1 , x0········································································ 1分 x x f(x)1xa在(0, )是减函数·······································································2分 x 由在x2时取得极大值得： f(2)0，即12a0，解得：a3 ··························3分 2 2 ∴ f(x)lnx1 x23 x，故 f(1)3 , f(1)1 ·····················································4分 2 2 2 ∴ 曲线y f(x)在点(1, f(1))处的切线方程为y13 (x1)，即3x2y10···········5分 2 （2）证明 方法一: 由题意得：g(x) f(x) x2ax1 , x0······························································6分 x 由g(x)0得x2ax10，其判别式△a240 由一元二次方程根与系数的关系知，关于x的方程x2ax10有唯一正根 设x2ax10的唯一正根为m，则有amm21··················································7分 文科数学答案第5页 共8页当0xm时，g(x)0，故g(x)单调递增；当xm时，g(x)0，故g(x)单调递减 ∴ g(x) g(m)lnm1 m2am1lnm1 m21 ·············································8分 max 2 2 2 2 设h(x)lnx1 x21 , x0，则h(x)1x0 2 2 x ∴ h(x)在(0, )上是增函数且h(1)0·······························································9分 由amm21及amm21得：am 1 ≥0，解得m≥1 ······································10分 m ∴ h(m)≥h(1)0，故g(x) lnm1 m21 ≥0··················································11分 max 2 2 又 g(e a 2 2 3 2)a2 1 x2ax11 (xa)210且0e a 2 2 3 2 1 2 2 2 ∴ g(x)在(0, m]内有零点，即g(x)有零点···························································12分 方法二： 由题意得：g(x) f(x) x2ax1 , x0 ····························································6分 x 由g(x)0得x2ax10，其判别式△a240 由一元二次方程的根与系数的关系知，方程x2ax10有唯一正根 设x2ax10的正根为m，则有amm21························································7分 当0xm时，g(x)0，故g(x)单调递增；当xm时，g(x)0，故g(x)单调递减 ∴ g(x) g(m)lnm1 m2am1lnm1 m21 ·············································8分 max 2 2 2 2 ∵ g(e a 2 2 3 2)a2 1 x2ax11 (xa)210且0e a 2 2 3 2 1 2 2 2 ∴ g(x)有零点等价于g(x) ≥0，即lnm1 m21 ≥0·········································9分 max 2 2 由h(x)lnx1 x21 , x0在(0, )上是增函数且h(1)0知： 2 2 当且仅当m≥1时，lnm1 m21 ≥0···································································10分 2 2 由amm21及a≥0得：m 1 ≥0，解得m≥1····················································11分 m ∴ h(m)≥h(1)0，即当a≥0时，g(x) ≥0成立 max ∴ g(x)有零点································································································12分 方法三： g(x)有零点等价于关于x的方程lnx1 x2ax10有正根 2 2 亦等价于关于x的方程a1 (x1 )lnx , (x0)有解················································6分 2 x x 设(x)1 (x1 )lnx , (x0)，则(x)1 (1 1 )1lnx x212lnx ·····················7分 2 x x 2 x2 x2 2x2 记H(x)x212lnx, x0，则H(x)2x20，故H(x)是增函数···························8分 x 又 H(1)0，故(x)0有唯一零点x1······························································9分 当0x1时，H(x)0，故(x)0，(x)是单调递减； 当x1时，H(x)0，故(x)0，(x)是单调递增···············································10分 ∴ (x) (1)1 (11 )ln10，即(x)≥0·····················································11分 min 2 1 1 ∴ 当a≥0时，函数g(x) f(x)1 有零点···························································12分 2 方法四： 要证：当a≥0时，函数g(x) f(x)1 有零点 2 只需证：当a≥0时，直线yax与函数h(x)1 x2lnx1 (x0)的图象有公共点·········7分 2 2 由h(x)x1 x21知： x x 文科数学答案第6页 共8页当0x1时，h(x)0，故h(x)单调递减；当x1时，h(x)0，故h(x)单调递··········8分 ∴ h(x) h(1)112ln110······································································9分 min 2 2 ∴ y0是曲线yh(x)在点(1, 0)处的切线··························································10分 即 当a0时，直线yax与函数h(x)的图象有唯一公共点 当a0时，直线yax与函数h(x)的图象在第一象限相交，有两个公共点．···············11分 综上，当a≥0时，直线yax与函数h(x)的图象有公共点． ∴ 当a≥0时，函数g(x) f(x)1 有零点．························································12分 2 21．（12 分） 解：（1）由点P(1, 3 )在C上得： 1  3 1 ①································································1分 2 a2 4b2 3 3 由椭圆的标准方程得A(a, 0)、B(a, 0)，故k  2 ，k  2 ····························2分 AP 1a A 2 E 1a 3 3 由k AP k BP 1 4 得： 2  2 1 ，解得：a24···················································3分 1a 1a 4 将a24代入①得：b21····················································································4分 ∴ 椭圆C的方程为 x2 y21············································································5分 4 （2）由题意知直线l不能平行于x轴 设直线l的方程为xtym，M(x, y ), N(x , y ) 1 1 2 2 由直线l与圆x2y21相切得： |m| 1，化简得m2t21···································6分 1t2 xtym,  由  x2 y21. 消去x整理得：(t24)y22tmym240  4 于是，△(2tm)24(t24)(m24)16(t2m24)16348 由求根公式得： y y  △  4 3 ·································································7分 2 1 t24 t24 ∴ MN  1t2 y y  4 3 1t2 ··································································· 8分 2 1 t24 令 1t2 n，则n≥1且 MN  4 3n  4 3 ≤ 4 3 2··············································9分 n23 n3 2 3 n 当且仅当n 3 ，即n 3时取等号······································································10分 n ∴ |MN| 2，此时由 1t2  3解得：t 2················································11分 max ∴ 直线l的斜率为 2 ．·····························································································12分 2 （二）选考题：共10分．请考生在第22、23题中任选一题作答．如果多做，则按所做的第 一题记分． 22．【选修4—4：坐标系与参数方程】（10分） 解：（1）∵ 直线l过点P(1, 0)，且倾斜角为 6 ∴ l的参数方程为    x1tcos  6 , (t为参数)，即    x1 2 3 t, (t为参数)····················2分 ytsin  . y1 t.  6  2 由2 2cos(  )，得2cos2sin······························································3分 4 ∴ 22cos2sin 将xcos, ysin代入上式得：x2y22x2y0 文科数学答案第7页 共8页∴ C的直角坐标方程为x2y22x2y0···························································5分 （2）设A，B两点对应的参数分别为t , t 1 2 将l的参数方程代入C的直角坐标方程，得( 3 t)2( 1 t1)22 2 2 整理，得t2t10··························································································· 6分 此时△(1)241(1)50 ，t t 1, tt 10·············································7分 1 2 12 ∴ |PA||PB||t ||t ||t t |···········································································8分 1 2 1 2  (t t )24tt  (1)241(1) 5·········································9分 1 2 12 即 |PA||PB| 5．·······················································································10分 23．【选修4—5 不等式选讲】（10分） 解：（1）方法一：  4x, x3 ,  2  f(x)6, 3 ≤x 3 ,·····················································································2分 2 2  4x, x≥ 3 .  2  x3，  3 ≤x3 ,  x≤ 3 , ∴ 不等式 f(x)≤8等价于 2 或 2 2 或 2 ···································4分 x≥2. 6≤8. x≤2. ∴ 不等式 f(x)≤8的解集为{x|2≤x≤2}······························································5分 方法二： 当x3 时， f(x)|2x3||2x3| (2x3)(2x3)4x ≤8， 2 解得2x3 ；······························································································1分 2 当3x3 时， f(x)|2x3||2x3| (2x3)(2x3)6 ≤8恒成立， 2 2 所以3x3 ；································································································2分 2 2 当x3时， f(x)|2x3||2x3| (2x3)(2x3)4x8， 2 解得3x2；··································································································3分 2 ∴ 不等式 f(x)≤8的解集为{x|2≤x≤2}······························································5分 方法三： 不等式 f(x)≤8等价于|x3||x3|4，····························································1分 2 2 由绝对值的几何意义知，|x3||x3|表示数轴上的x的对应点到3和3的对应点的距离 2 2 2 2 之和， 又数轴上的2和2的对应点到3和3的对应点的距离之和等于4， 2 2 而2和2之间的数均满足该不等式········································································3分 所以不等式|x3||x3|4的解集为{x|2≤x≤2}， 2 2 故不等式 f(x)≤8的解集为{x|2≤x≤2}·······························································5分 （2）求1 1  1 1的两种方法： a 2b 3c 方法一： ∵ f(x)|2x3||2x3|≥|(2x3)(2x3)|6 当且仅当(2x3)(2x3)≤0，即3 ≤x≤ 3时，取等号·······································6分 2 2 ∴ f(x)的最小值 M 6 从而1 1  1 1······························································································7分 a 2b 3c 方法二： 文科数学答案第8页 共8页 4x, x3 ,  2  由（1）知， f(x)6, 3 ≤x 3 , 2 2  4x, x≥ 3 .  2 作出 f(x)的图象（学生需画出图象）·····································································6分 ∴ f(x)的最小值 M 6，当且仅当3 ≤x≤ 3时取得， 2 2 从而1 1  1 1······························································································7分 a 2b 3c 证明a2b3c≥9的两种方法： 方法一： ∵ a，b，c均为正数 ∴ a2b3c(a2b3c)( 1 1  1 )3( a  2b )( a  3c )( 2b 3c ) ···················8分 a 2b 3c 2b a 3c a 3c 2b ≥32 a 2b 2 a 3c 2 2b3c 9················································9分 2b a 3c a 3c 2b 当且仅当a3，b 3，c1时等号成立 2 ∴ a2b3c≥9．·························································································10分 方法二： ∵ a，b，c均为正数 ∴ a2b3c(a2b3c)( 1 1  1 )≥( a 1  2b 1  3c 1 )29 ················9分 a 2b 3c a 2b 3c 当且仅当a2b3c，即a3，b 3，c1时等号成立 2 ∴ a2b3c≥9．·························································································10分 文科数学答案第9页 共8页