2022年巴中市零诊考试理科数学参考答案（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理数试题

巴中市普通高中 级“零诊”考试 2020 数学阅卷参考答案（理科） 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】D．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D． 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 |b b210| F( 1b2, 0)，双曲线的渐近线为ybx，即bxy0，则 b2，即b24a2，所以离心 b21 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】A．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A． 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】A．设公差为 d ，则 S na  n(n1) d  d n2(a d)n ，故 S n d n(a d ) ；又 n 1 2 2 1 n 2 1 2 S S S 2023  2022 1 ， a 2022 ，故 { n} 是以 2022 为首项，1 为公差的等差数列，于是得 2023 2022 1 n S 2023 2022(20231)10，所以S 0．选A．本题也可用基本量法求解，借助等差数列前n项 2023 2023 和的性质运算更为简洁． 9．【解析】D．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 ， 从 而 得 ： y          ADBCAD(ACAB)[(1x)ABxAC](ACAB) C 3 x  A  C 2 (x1)  A  B 2 13x4，由0≤x≤1得4≤13x4≤9，故  A  D    B  C  [4, 9]． D 方法二：以A为原点，AB，AC所在直线分别为x, y轴建立直角坐标系（如图），则       AB(2, 0), AC(0, 3), BC(2, 3)，设D(x, y)，则AD(x, y)，故ADBC2x3y B A 2 x 理科数学答案第1页 共11页（*），由点D在BC上得：3x2y60, 0≤x≤2（可借助初中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（*）式得：ADBC2x3y913 x, 0≤x≤2，故 2   ADBC[4, 9]．       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．选D．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】B．由 f(x1)2f(x)得： f(x)2f(x1)．又当x(0, 1]时， f(x)1 sinx[1 , 0] ，故当x(1, 2]时， f(x)[1 , 0]；类推得：当 4 4 2 x(2, 3]时，f(x)4f(x2)sinx[1, 0]，且x(2, 3]．如图． 由sinx 3 得sinx 3 ，解得x21或x22，解得 2 2 3 3 x7或x8．故若对任意x(, m]，都有 f(x)≥ 3，则m≤ 7．选B． 3 3 2 3 12．【解析】C．要比较a, b, c的大小，可先比较lna, lnb, lnc的大小．又lna22ln20，lnb21ln21， lnc20ln22． 方法一：由22202121202242 ，令函数 f(x)(42x)lnx, x≥20，则 f(x)lnx421在 x [20, )上单调递减，所以 f(x)≤ f(20)ln2011 ；因为e232920，所以ln202，ln202， 10 所 以 f(x)≤ f(20)ln2011211 9 0 ， 由 此 知 f(x) 在 [20, ) 上 单 调 递 减 ， 故 10 10 10 f(20) f(21) f(22)，即lnalnblnc，故cba．故选C． 方法二：先比较lna与lnc的大小，易证：函数g(x)lnx在(e, )上单调递减，故ln20ln22，所以 x 20 22 lna22ln2020ln22lnc ，从而ac ；再比较比较lna与lnb 的大小，令h(x) lnx , x≥20 ，则 x1 x1lnx h(x) x ，记y x1lnx，则y 1 10，故y x1lnx在(0, )上是减函数，所以 (x1)2 x x2 x x 当x≥20时，y≤ 21ln200，从而h(x)0，由此知h(x)在[20, )上单调递减，故h(20)h(21)， 20 即 ln20ln21 ，变形得22ln2021ln21，所以lnalnb ，由此得ab ；同理可比较得到bc；故 21 22 abc．故选C． 二、填空题：本大题共4个小题，每小题5分，共20分． 13．【解析】1．利用二项展开式的通项公式及已知建立m的方程求得m的值．因为展开式中含x3的 项为C322(mx)340m3x3，所以a 40m340，解得m1．注：本题原型为人教A版选修2-3 5 3 例2（1）题，主要考查二项式定理及其通项公式，及数学运算核心素养和运算求解能力． 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． 理科数学答案第2页 共11页15．【解析】8 6．由题意有：BD平面 ADC ， AD, DC平面 ADC ，故 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】，[ 2 3 , 3)．以三角形边角关系的射影定理为背景，综合考查正弦、余弦定理、三角 3 3 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 1  1 的范围问题，可 tanB tanC 切化弦后转化为求sinBsinC的范围，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 此时可直接代入消元化简也可用对称设元简化；也可用三角形中三内角的正切关系 tanAtanBtanC tanAtanBtanC 转化；还可以构造几何图形作几何法或坐标法求解． （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 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 1  1 范围的过程与方法． tanB tanC 策略一：切化弦后转化借助正弦型函数的图象与性质． 1  1 cosBcosC cosBsinC cosCsinB sin(BC)  sinA  3 ，由A 得 tanB tanC sinB sinC sinBsinC sinBsinC sinBsinC 2sinBsinC 3 BC 2，故C2 B，又B为锐角，所以 B ． 3 3 6 2 ①sinBsinCsinBsin( 2 B)sinB( 3 cosB1 sinB) 3 sin2B1 cos2B1 1 sin(2B  )1 ．因 3 2 2 4 4 4 2 6 4 为 2B  5 ，故 1sin(2B  )≤1 ，当且仅当 B  取等号，所以 sinBsinC( 1 , 3 ] ，故 6 6 6 2 6 3 2 4 1  1 [ 2 3 , 3)． tanB tanC 3 ② 令 B  x, C  x ， 由 B 、 C 均 为 锐 角 得   x  ， 故 3 3 6 6 sinBsinCsin(  x)sin(  x) ( 3 cosx1 sinx)( 3 cosx1 sinx)3 cos2x1 sin2x3sin2x( 1 , 3 ]， 3 3 2 2 2 2 4 4 4 2 4 下同． ③由和、差角的余弦公式可得：2sinBsinCcos(BC)cos(BC)cos(BC) 1 ，由已知得 2  B, C ，故  BC ，所以cos(BC)( 1 , 1]，故2sinBsinC(1, 3 ]，下同． 6 2 3 3 2 2 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，由已知得0≤|sinBsinC| 1 ，故 4 4 2 1sinBsinC≤3，仅当sinBsinC 时取等号；下同． 2 4 策略三：用公式tanAtanBtanC tanAtanBtanC 转化． ⑤由A 得：tan(BC)tanA 3 tanBtanC ，变形得： 3(tanBtanC1)tanBtanC ， 3 1tanBtanC 理科数学答案第3页 共11页故 1  1 tanBtanC  3tanBtanC 3  3 3 ； 由  B, C  知 ： tanB tanC tanBtanC tanBtanC tanBtanC 6 2 3(tanBtanC1) tanBtanC≥2 tanBtanC ，仅当 tanBtanC 取等号，解得 tanBtanC≥ 3 ，故 tanBtanC≥3，所以 3 ≤ 3 0，从而 1  1 [ 2 3 , 3)． 3 tanBtanC tanB tanC 3 策略四：几何法或坐标法 ⑥不妨设a 3，过点A作ADBC于D，如右图．设BDx, 0x 3， y 则CD 3x，tanB AD tanC AD ，所以 1  1  x 3x  3 ． A x 3x tanB tanC AD AD A1 A2  B, C 得1AD≤3，故 1  1 [ 2 3 , 3)． 6 2 2 tanB tanC 3 ⑦不防设△ABC的边a 3，则其外接圆半径为1．如右图建立直角坐标系， B D O C x 则△ABC的外接圆的方程为x 2(y1 )21，设A(x, y)，由已知得 3 x 3 ， 2 2 2 3 3 x x 1 y≤3， 1  1  2  2  3 [ 2 3 , 3)． 2 tanB tanC y y y 3 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 ∴ 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 方法三：（归纳猜想，然后用数学归纳法证明） 由a 2S 且S 2S 2得：S 6232 1 1 2 1 2 由S 2S 2得：S 14242 3 2 3 同理可得：S 30252····················································································1分 4 猜想：S 2n12·····························································································2分 n 证明：当n1时，S a 2222 ，结论正确 1 1 假设当nk时，有S 2k12成立 k 那么，当nk1时，S 2S 22(2k12)22(k1)12 ··················3分 k1 k 故当nk1时，结论正确 综上可知，S 2n12··············································································4分 n 理科数学答案第4页 共11页下同方法二 （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人····················································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 ∴ X的所有可能取值为0，1，2，且X服从超几何分布··········································7分 由组合数公式及等可能事件的概率公式得： P(X 0) C0 6 C2 2  1 ···························································································8分 C2 28 8 P(X 1) C1 6 C1 2 12 ···························································································9分 C2 28 8 P(X 2) C 6 2C0 2 15 ·························································································10分 C2 28 8 ∴ X的分布列为 ····························································································11分 X 0 1 2 1 12 15 P 28 28 28 EX 0 1 112215 3 ．··························································12分 28 28 28 2 理科数学答案第5页 共11页19．（12 分） F 解：（1）证明 方法一： E 由正方形ABCD的性质得：AB∥CD······································································1分 又 AB平面DCF，CD平面DCF ∴ AB∥平面DCF ·························································································· 2分 ∵ BE∥CF ，BE平面DCF，CF 平面DCF C B ∴ BE∥平面DCF······················································································································3分 ∵ ABBEB, AB, BE平面ABE D A ∴ 平面ABE∥平面DCF············································································································4分 ∵ AE平面ABE ∴ AE∥平面DCF····························································································6分 方法二： 在CF取点G使得CG2BE，连结EG、DG ∵ BE∥CF F ∴ 四边形BEGC是平行四边形···········································································1分 故 EG∥BC，且EGBC ····················································G······························ 2分 E 又 AD∥BC, ADBC ∴ AD∥EG, ADEG······················································································· 3分 ∴ 四边形ADGE是平行四边形···········································································4分 ∴ AE∥DG·······································································C················· B ·············5分 又 AE平面DCF，DG平面DCF ∴ AE∥平面DCF····································································D··························A························6分 方法三： ∵ 平面BCFE平面ABCD，平面BCFE平面ABCD=BC BEBC ，BF 平面BCFE ∴ BE平面ABCD··························································································1分 又 BE∥CF ∴ CF 平面ABCD ·························································································2分 又 CDBC ，故CB，CF，CD两两垂直    以C为原点，CD, CB, CF 的方向分别为x, y, z轴的正方向建立空间直角坐标系Cxyz 由ABBCCDDABE2，CF 3得： C(0, 0, 0), D(2, 0, 0), B(0, 2, 0), A(2, 2, 0), E(0, 2, 2), F(0, 0, 3)·································3分  ∴ AE(2, 0, 2)·····························································································4分   ∵ CBAE(0, 2, 0)(2, 0, 2)0   ∴ CBAE·····································································································5分  又 CB(0, 2, 0)是平面CDF的一个法向量，且AE平面CDF ∴ AE∥平面DCF····························································································6分       或者，由AE(2, 0, 2)DC 2 CF 知：向量AE, DC, CF 共面 3 又 AE平面CDF，DC, CF 平面CDF ∴ AE∥平面DCF （2）在（1）中方法三的坐标系下，有：    AE(2, 0, 2), EF (0, 2, 1)，平面CEF的一个法向量为CD(2, 0, 0)····················8分  设平面AEF的一个法向量为m(x, y, z)，则由   mAE0, 2x2z0,    得： 取y1得：m(2, 1, 2) ·········································10分 mEF 0. 2yz0.     ∴ cosm, CD  m  C  D   221020 2 ·····················································11分 |m||CD| 2 221222 3 故由几何体的空间结构知：二面角AEFC的余弦值为2．··································12分 3 20．（12 分） 理科数学答案第6页 共11页解：（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 21．（12 分） 解：（1）证明 方法一: 由题意得：g(x) f(x) x2ax1 , x0······························································1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··················································2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 ·············································3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·······························································4分 由amm21及amm21得：am 1 ≥0，解得m≥1 m ∴ h(m)≥h(1)0，故g(x) lnm1 m21 ≥0···················································5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····························································6分 方法二： 理科数学答案第7页 共11页由题意得：g(x) f(x) x2ax1 , x0 ····························································1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························································2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 ·············································3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 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·····································································4分 2 2 由amm21及a≥0得：m 1 ≥0，解得m≥1······················································5分 m ∴ h(m)≥h(1)0，即当a≥0时，g(x) ≥0成立 max ∴ g(x)有零点·································································································6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················································1分 2 x x 设(x)1 (x1 )lnx , (x0)，则(x)1 (1 1 )1lnx x212lnx ·····················2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···························3分 x 又 H(1)0，故(x)0有唯一零点x1······························································4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 ∴ (x) (1)1 (11 )ln10，即(x)≥0······················································5分 min 2 1 1 ∴ 当a≥0时，函数g(x) f(x)1 有零点·····························································6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 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， ∴ 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 （2）方法一： 由已知得 f(x)1xa，故 f(x ) 1 x a x 0 x 0 0 理科数学答案第8页 共11页f(x ) f(x ) 又 f(x ) 2 1 ，且 0 x x 2 1 (lnx 1 x 2ax )(lnx 1 x2ax) f(x 2 ) f(x 1 )  2 2 2 2 1 2 1 1  lnx 2 lnx 1 x 1 x 2 a x x x x x x 2 2 1 2 1 2 1 ∴ 1 x  lnx 2 lnx 1 x 1 x 2 ①·····································································7分 x 0 x x 2 0 2 1 令F(x)(x1)lnx2(x1), (x1)，则F(x)lnx11, (x1) x 设yF(x)lnx11, x1，则x1时，y x10，故F(x)是增函数 x x2 ∴ F(x)F(1)0，故F(x)在(1, )上是增函数 ∴ F(x)F(1)0······························································································8分 x 由x x 0得： 2 1, x x 0·········································································9分 2 1 x 2 1 1 x x x x 取x 2 得：( 2 1)ln 2 2( 2 1)0 ，变形得：(x x)(lnx lnx)2(x x) x x x x 2 1 2 1 2 1 1 1 1 1 ∴ lnx 2 lnx 1  2 ②···············································································10分 x x x x 2 1 2 1 由①②得： 1 x  2  x 1 x 2 ······································································11分 x 0 x x 2 0 2 1 设k(x)1x (x0)，则上式即为k(x )k( x 1 x 2) x 0 2 ∵ 函数k(x)1x在(0, )上是减函数 x x x ∴ x  1 2 ．·····························································································12分 0 2 方法二： ∵ f(x)1xa，故 f(x ) 1 x a x 0 x 0 0 f(x ) f(x ) lnx lnx x x 由 f(x ) 2 1 得： f(x ) 2 1 1 2 a 0 x x 0 x x 2 2 1 2 1 ∴ 1 x  lnx 2 lnx 1 x 1 x 2 ①·······································································7分 x 0 x x 2 0 2 1 x x x x 下面先证明：当0x x 时， 2 1  1 2 ② 1 2 lnx lnx 2 2 1 由0x x 得：lnx lnx ，故lnx lnx 0 1 2 2 1 2 1 x 2( 2 1) 2x x x x 要证②成立，需证：lnx lnx  2 1 ，即证：ln 2  1 ③···························8分 2 1 x x x x 1 2 1 2 1 x 1 x 2(t1) 令t 2 ，则③式即为lnt (t1) x 1t 1 设G(t) 2(t1) lnt, t1，则G(t) 2 1 1t2 0 t1 (t1)2 t t(t1)2 ∴ G(t)在(1, )上是减函数·············································································9分 2(t1) ∴ G(t)G(1)0，即：lnt (t1)成立 1t x x x x ∴ 2 1  1 2 成立···············································································10分 lnx lnx 2 2 1 ∴ 当0x x 时， lnx 2 lnx 1  2 成立 1 2 x x x x 2 1 1 2 结合①得： 1 x  2  x 1 x 2 ······································································11分 x 0 x x 2 0 1 2 理科数学答案第9页 共11页设k(x)1x (x0)，则上式即为k(x )k( x 1 x 2) x 0 2 ∵ 函数k(x)1x在(0, )上是减函数 x x x ∴ x  1 2 ．·····························································································12分 0 2 方法三： ∵ f(x)1xa，故 f(x ) 1 x a x 0 x 0 0 f(x ) f(x ) lnx lnx x x 由 f(x ) 2 1 得： f(x ) 2 1 1 2 a 0 x x 0 x x 2 2 1 2 1 ∴ 1 x  lnx 2 lnx 1 x 1 x 2 ···········································································7分 x 0 x x 2 0 2 1 由 f(x)1xa在(0, )上是减函数知： x x x x x 当0x x 时， 1 2 与x 的大小关系同 f( 1 2)与 f(x )的大小关系一致·············8分 1 2 2 0 2 0 故只需比较 2 与 lnx 2 lnx 1 的大小 x x x x 1 2 2 1 x 2( 2 1) 作差变形得： 2  lnx 2 lnx 1 1 [ x 1 ln x 2]··········································9分 x x x x x x x x 1 2 2 1 2 1 2 1 1 x 1 x 由0x x 知：t 2 1, x x 0 1 2 x 2 1 1 设G(t) 2(t1) lnt, t1，则G(t) 2 1 1t2 0 t1 (t1)2 t t(t1)2 ∴ G(t)在(1, )上是减函数············································································10分 x 2( 2 1) x x 故G(t)G(1)0，即： 1 ln 2 0 x x 2 1 1 x 1 x 2( 2 1) ∴ 2  lnx 2 lnx 1 1 [ x 1 ln x 2]0，即 2  lnx 2 lnx 1 ··············11分 x x x x x x x x x x x x 1 2 2 1 2 1 2 1 1 1 2 2 1 x 1 x x x x ∴ f( 1 2) f(x )，故x  1 2 ．······························································12分 2 0 0 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 ∴ 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 理科数学答案第10页 共11页整理，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： 理科数学答案第11页 共11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 理科数学答案第12页 共11页