吉林地区2023-2024高三一模数学试题参考答案(1)_2023年11月_0211月合集_2024届吉林省吉林市高三上学期第一次模拟考试_吉林省吉林市2024届高三上学期第一次模拟考试数学

吉林地区普通高中 2023—2024 学年度高三年级第一次模拟考试 如图2，当函数 y  ax与 y  log x有一个公共点(x ,y )在直线 y  x上，且在该公共点处的 a 0 0  ax 0  x ① 切线为 y  x，所以有 0 ，①代入②消ax 0得x lna  lnax 0 1，ax 0  e，代入 ax 0 lna 1② 0 数学试题参考答案  1 一、单项选择题：本大题共8小题，每小题5分，共40分．   a  ee ②中得 即 y  ax与 y  log x的公共点为(e，e)  x  e a 1 2 3 4 5 6 7 8 0 A C B D A C B C 结合图象得到以下结论： 二、多项选择题：本大题共4小题，共20分．全部选对的得5分，部分选对的得2分，有选错的得0 1 （1）当a  ee时，函数 y  ax与 y  log x的图象无公共点（如图1） 分． a 1 （2）当a  ee时，函数 y  ax与 y  log x的图象有一个公共点（如图2） 9 10 11 12 a 1 BCD ACD BC ABD （3）当1 a  ee时，函数 y  ax与 y  log x的图象有两个公共点（如图3） a （二）当0 a 1时函数图象呈现以下三种情况 三、填空题：本大题共4小题，每小题5分，共20分．其中第15题的第一个空填对得2分，第二 个空填对得3分． 2 13． 5 14． 5 15． 68； 108 16． (1e4,e4 1) （注：或写成1e4  m  e4 1） 图4 图5 图6 7．教学建议：请教师们注意结合向量运算掌握三角形的“四心”问题． 如图5，当函数 y  ax与 y  log x有一个公共点(x ,y )在直线 y  x上，且在该点处有公切 a 0 0 9．教学提示：建议教学中指导学生甄别a范围不同时，指、对函数的位置关系． 线（斜率为1）， （一）当a 1时函数图象呈现以下三种情况  ax 0  x ① 1 所以有 0 ，①代入②消ax 0得x lna  lnax 0  1，ax 0  ，代入②中得 ax 0 lna  1② 0 e a  ee  1 1  1 ，即 y  ax与 y  log x的公共点为( ，) x  a e e  0 e 结合图象得到以下结论： （4）当0 a  ee时，函数 y  ax与 y  log x的图象有三个公共点（如图4） a 图1 图2 图3 高三年级第一次模拟考试数学试题参考答案 第 1 页 （共 8 页）（5）当a  ee时，函数 y  ax与 y  log x的图象有一个公共点（如图5） h(e2) 0 h(e2) 0 a   令h(t) t2 mt e4，则h(1) 0 或h(1) 0 （6）当ee  a 1时，函数 y  ax与 y  log a x的图象有一个公共点（如图6）  h(e2) 0  h(e2) 0 故0 m  e4 1或1e4  m  0 15．教学建议：（1）关注《数学课程标准》中分层随机抽样的教学要求． 综上，实数m的取值范围是(1e4,e4 1)． （2）学生掌握推导过程． 方法二： 摘自《数学课程标准》： ②分层随机抽样 ()式化为m  t e4 ，令h(t) t e4 (t  0)， t t 通过实例，了解分层随机抽样的特点和适用范围，了解分层随机抽样的必要性，掌握各层样本量比例 易知 y  h(t)在(，0)，(0,)上单调递增， 分配的方法．结合具体实例，掌握分层随机抽样的样本均值和样本方差． (x2)ex 且h(1)1e4，h(1) e4 1，h(e2) h(e2) 0 16．教学提示：当x  0且x 1时， f(x) ， f(2) 0 (x1)2 其图象大致如图： 当0 x  2且x 1时， f(x) 0；当x  2时， f(x) 0． 当0 m  e4 1或1e4  m  0时，满足 故 f(x)在(0,1)，(1,2)上单调递减，在(2,)上单调递增， e2  t  1 t  e2  1 或 1  t  e2  1 t  e2 2 2 当x  2时， f(x)取得极小值 f(2) e2， 综上，实数m的取值范围是(1e4,e4 1)． 当x 1时， f(x) ；当x 1时， f(x) ；当x  时， f(x) ． 由 f(x)解析式可知， f(x)为奇函数．画出 f(x)图象大致如下： 四 、解答题 17．【解析】 令g(x) 0得 f 2(x)mf (x)e4  0，设t  f(x)， （Ⅰ） a // b  3sinxcosxcos2 x  0····························································1分 得关于t的方程t2 mt e4  0() 即 cosx( 3sinxcosx) 0   m2 4e4  0恒成立，设()式有两个不等实根t ,t ， 1 2 3 当t  e2,t  e2时，即m  0，满足题意， cosx  0或tanx  ························································································3分 1 2 3 e2  t  1 t  e2 当 1 或 1 ,满足题意， x(0,)  t  e2  1 t  e2 2 2 方法一： 高三年级第一次模拟考试数学试题参考答案 第 2 页 （共 8 页）    x  或x  ···································································································5分  f(x)的单调递增区间是[  k,  k](kZ)·················································10分 2 6 3 6 （注：少一个解扣1分，没有角的范围表述扣1分.） （注：单调区间没有写成区间形式扣1分，没有注明k 的范围扣1分．） 方法二： a // b  3sinxcosxcos2 x  0·························································1分 18．【解析】 3 1cos2x  sin2x  0 2 2 1 （Ⅰ）解： f(x) 2 (x  0)·············································································1分 x  1 sin(2x ) ··································································································3分 6 2 所求切线斜率为 f(1) 1，切点为(1,2)··································································3分   11 故所求切线方程为 y(2) (x1)，即x y1 0·················································5分 x(0,) 2x ( , ) 6 6 6 （注：将切线方程表示成 y  x1也给分）    5 2x  或2x  （Ⅱ）方法一：分离变量 6 6 6 6 lnx   由 f(x) ax2 2x得a  在(0,)恒成立··························································· 6分 x  或x  ···································································································5分 x2 6 2 （注：少一个解扣1分，没有角的范围表述扣1分.） lnx 令g(x) (x  0)，则a  g(x) x2 max 1 1 （Ⅱ） f(x) ab  3sinxcosxcos2 x ················································6分 12lnx 2 2 g(x) ，g( e) 0·················································································8分 x3 3 1  sin2x cos2x 当0 x  e 时，g(x) 0；当x  e 时，g(x) 0 2 2 故g(x)在(0, e)上单调递增，在( e,)上单调递减   sin(2x )······················································································7分 6 1 故当x  e 时，g(x)取最大值 ···········································································11分 2e    令  2k 2x   2k (kZ) 2 6 2 1 1 故a  ，即a的取值范围是[ ,)······································································12分 2e 2e     k x   k························································································9分 3 6 1 (注：表示成a  不扣分） 2e 高三年级第一次模拟考试数学试题参考答案 第 3 页 （共 8 页）方法二：分类讨论 1 g(x) 2ax,h(x) x 由 f(x) ax2 2x得ax2 lnx  0在(0,)恒成立····················································6分 若a  0，当x 1时，ax2  0,lnx  0,g(x) h(x)，不合题意；·······························7分 1 2ax2 1 令g(x) ax2 lnx(x  0)，则g(x) 2ax  x x 若a  0，g(x) h(x) ①当a  0时，g(x) 0恒成立，g(x)在(0,)上单调递减， 曲线 y  g(x)与曲线 y  h(x)有且只有一个公共点，且在该公共点处的切线相同. 又g(1) a  0，故当x 1时，g(x) 0，不合题意； （或直接由g(2) 4aln2 0，不合题意 设切点坐标为(x , y ) 0 0 或当x 1时，ax2  0,lnx  0,g(x) 0，不合题意）···············································8分 y  ax2  lnx x  e  0 0 0  0 则 1 解得 1  2ax  a  ②当a  0时，令g(x) 0得x  1 ，  0 x 0  2e 2a 1 故当a  时，g(x) h(x) 1 1 2e 令g(x) 0得x  ，令g(x) 0得0 x  , 2a 2a 1 即a的取值范围是[ ,).····················································································· 12分 1 1 2e 故g(x)在(0, )上单调递减，g(x)在( ，)上单调递增 2a 2a （教学建议：1.教师应强调第二问的法三比较适合选填题； 1 1 1 1 故当x  时，g(x)取最小值g( ) ln  0·········································11分 2.在教学中，教师应注意强调 f(x) ln(kx)(k  0)的导函数的相关内容； 2a 2a 2 2a 变式：若x[1,0],xln(12x)(a1)x恒成立，求a的取值范围. 1 1 故a  ，即a的取值范围是[ ,) 2e 2e 19．【解析】 1 综上所述，a的取值范围是[ ,)··········································································12分 2e （Ⅰ）选择① 方法三：数形结合 由已知可得： 由 f(x) ax2 2x得ax2  lnx在(0,)恒成立·························································6分 当n1时，S  a  2························································································1分 1 2 令g(x) ax2,h(x) lnx，则当x  0时，g(x) h(x)恒成立 当n 2且nN时，a  S  S  a a n n n1 n1 n 高三年级第一次模拟考试数学试题参考答案 第 4 页 （共 8 页）a 1 a  2a 即 n1  2····························································································3分 nN  0T 1······································································ 12分 n1 n a 2n1 1 n n （注：若（Ⅰ）选择①，且未讨论a 1的情况，扣2分； a 又 2 1不符合上式 但若（Ⅱ）运算正确，（Ⅱ）问正常赋分．） a 1 关于劣构问题： ···················································································· 5分 2, n1, 劣构问题的常见形式： a   n 2n1,n 2且nN （1）目标界定不明确的结构不良问题； （2）具有多种解法、途径的结构不良问题； 选择② （3）多个类似条件的结构不良问题； S n  a n1 （4）问题条件或数据冗余的结构不良问题； S  S  S ·····································································································1分 （5）不同求解目标的结构不良问题． n n1 n 条件的选择原则： S S n1  2S n 即 S n1  2··························································································3分 （1）应先分析所有条件，优先选择自己会做的、有把握的； n （2）在有把握的前提下，优先选择难度最小的； 又S  a  2 1 1 （3）在选择条件时，要通盘考虑条件对整个问题的影响． {S n }是以2为首项，2为公比的等比数列． 教学建议： S  22n1  2n (nN)····················································································5分 （1）引导学生从知识的习得记忆转向问题的解决、策略的选择，使数学应用在思维层面真正发生； n （2）注重渗透不良结构问题，采用开放式、互动式的教学方式，引导学生关注数学问题情境的变化； （Ⅱ）由（Ⅰ）可知S a  2n，a  2n1 n n1 n2 （3）注重变式训练，提高学生的辨析能力和应变能力． S 2n 1 1 b  n    ···································9分 n (a 1)(a 1) (2n 1)(2n1 1) 2n 1 2n1 1 n1 n2 20．【解析】 T  b b b b n 1 2 3 n （Ⅰ）设Q 型芯片I级品该项指标的第70百分位数为a， 则该指标在80以下的概率为0.55，该指标在90以下的概率为0.8，因此该项指标的第70百分位数 1 1 1 1 1 1 1 (1 )(  )(  )(  ) 3 3 7 7 15 2n 1 2n1 1 为a一定在[80,90)内 1 1 ·································································································11分 0.002100.005100.023100.025100.025（a80） 0.7 2n1 1 （也可以用0.02100.025（90a）10.7） 得a  86 高三年级第一次模拟考试数学试题参考答案 第 5 页 （共 8 页）所以Q 型芯片I级品该项指标的第70百分位数为86······················································3分   2sin(C  ) 2 即 sin(C  ) 1··································································3分 6 6 （Ⅱ）当临界值c 65时，   7   Q 型芯片Ⅱ级品应用于A型手机的概率为0.01（7065） 0.05····································6分 C(0,) C  ( , ) C   6 6 6 6 2 （Ⅲ）设直接将Q 型芯片Ⅰ级品、Ⅱ级品应用于A型、B型手机时，该芯片生产商支出为 y（万元），  C  ················································································································5分 3 y  700[0.002100.005（c50）]300[0.01100.03（60c）] (注：没有角的范围表述扣1分.)  4095.5c c[50,60]·····················································································8分  （Ⅱ）方法一：AB  BC ，C  ABC 为等边三角形 3 所以当50 c  56时， y 101， AB  BC  AC  2 当c  56时， y 101， CE  PE  x,BE  2 x,BP  m,m(0,2) 当56 c  60时， y 101·············································································10分 在BPE 中，由余弦定理得 PE2  BP2  BE2 2BPBEcosB······································································· 7分 综上：为降低芯片生产商的成本，当临界值c[50,56)时，选择方案二； 当临界值c  56时，选择方案一和方案二均可； 1 即x2  m2 (2 x)2 2m(2 x) 2 当临界值c(56,60]时，选择方案一．······························································12分 （注：以上三种情况，少一种扣1分，没有文字表述扣2分） 整理可得x  m2 2m4 ,m(0,2)·········································································8分 4m 21．【解析】 12 12 x  4m 6 2 (4m) 6  4 3 6 4m 4m （Ⅰ）由正弦定理得 sinC(cosA 3sinA) sinB2sinA·································································1分 当且仅当m  42 3 时取等号 sinC(cosA 3sinA) sin(AC)2sinA 即BP  42 3时，CE取最小值4 3 6 sinC(cosA 3sinA) sinAcosC cosAsinC 2sinA 此时BE  84 3 ································································································10分 即 3sinAsinC  sinAcosC  2sinA 1  1 3 S  BPBEsin  (42 3)(84 3) 14 3 24.·················12分 PBE 2 3 2 2 A(0,) sinA 0  方法二：AB  BC ，C  ABC 为等边三角形 3 高三年级第一次模拟考试数学试题参考答案 第 6 页 （共 8 页）AB  BC  AC  2  当0 x  时， f(x) 0恒成立，即 f(x)单调递增， CE  PE  x,BE  2 x,x(0,2) 2 m  0 2 设EPB ,(0, ) 3  ex 当  x 时，m   恒成立 2 cosx 在BPE 中，由正弦定理得  PE BE ex 2ex sin(x ) sinEBP  sinEPB ··························································································7分 设g(x)  cosx ， g(x)  ex(cosx sinx)   4 cos2 x cos2 x x 2 x 3  3  令g(x) 0，则  x ，令g(x) 0，则  x  即  sin 4 2 4 sin 3  3 3 g(x)在( ， )上单调递减，在( ，)上单调递增 3 2 2 4 4 x  ,(0, ) 整理可得 3 3 ········································································8分 sin 2 g(x)  g( 3 ) 2e 3 4  ，m  3 min 4 2e 4  当且仅当 时，x取最小值4 3 6 3 2 又m  0，0 m  2e 4 3 当CE取最小值4 3 6时，BE  84 3 综上，m的取值范围 (0, 2e 4 ] ········································································5分 （注：（1）不写成区间形式也给分；（2）没写m  0扣1分）  1 在RtBPE 中，BEP  ,BP  BE  42 3 ·················································10分 6 2 方法二：函数 f(x)在(0,)上单调递增 S  1 BPBEsin   1 (42 3)(84 3) 3 14 3 24.·················12分  f(x) ex mcosx  0在(0,)上恒成立（且不恒为0） PBE 2 3 2 2 1 cosx 又m  0，  在(0,)上恒成立································································2分 m ex 22．【解析】 cosx （Ⅰ）解： f(x) ex mcosx················································································1分 设h(x) ex 方法一：函数 f(x)在(0,)上单调递增  2sin(x ) sinxcosx h(x)   4  f(x) ex mcosx  0在(0，)上恒成立（且不恒为0）········································2分 ex ex 注：此处没取到等号，扣1分 3  3 令h(x) 0，则  x ，令h(x) 0，则  x  4 2 4 高三年级第一次模拟考试数学试题参考答案 第 7 页 （共 8 页） 3 3  h(x)在( ， )上单调递减，在( ，)上单调递增  f(x ) ex 0  sinx  cosx  sinx  2sin(x  ) 2 4 4 0 0 0 0 0 4 3 2 3   3  h(x)  h( )  又x ( ， )，x  (， )， 2sin(x  )(1,0) min 4 3 0 4 2 0 4 4 0 4 2e 4 1 2 即1 f(x ) 0··································································································12分    0 m 3 2e 4 3 又m  0，0 m  . 2e 4 3 综上，m的取值范围 (0, 2e 4 ] ·················································································5分 （Ⅱ）证明：m 1， f(x) ex  sinx， f(x) ex cosx 当x  0时，ex 1，cosx  1， f(x) ex cosx  0  f(x)在(0，)上单调递增，即 f(x)在(0，)上无极值点·······································7分 当 x  0时，设u(x) f(x)，u(x) ex sinx  0恒成立 u(x)在(，0)上单调递增 u(  2 ) e   2  0，u( 3 4  ) e  3 4   2 2  1 3  1 2  0 e 4 3  由零点存在性定理，存在唯一一个x ( ， )，使得u(x ) 0，即ex 0  cosx 0 4 2 0 0 当 x  x 时，u(x) 0， f(x) 0， f(x)在(，x )上单调递减 0 0 当x  x  0时，u(x) 0， f(x) 0， f(x)在(x ，0)上单调递增 0 0  f(x)在(，)上存在唯一极小值点x ·······························································10分 0 3  （注：此处x 所在区间必须是( ， )的子集，否则只给到10分位置） 0 4 2 高三年级第一次模拟考试数学试题参考答案 第 8 页 （共 8 页）