吉林省吉林市2023-2024学年高三上学期第一次模拟考试 数学答案

如图，当函数x与有一个公共点在直线上，且在该公共点处的吉林地区普通高中2023—2024学年度高三年级第一次模拟考试2yaylogax(x0,y0)yxax0x①切线为，所以有0，①代入②消x0得x0x0，代入yxax0lnalna1，aeax0lna1②数学试题参考答案1e一、单项选择题：本大题共8小题，每小题5分，共40分．ae②中得即x与的公共点为，xeyaylogax(ee)123456780ACBDACBC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01（1）当aee时，函数yax与ylogx的图象无公共点（如图1）分．a1（2）当aee时，函数yax与ylogx的图象有一个公共点（如图2）9101112a1（3）当e时，函数x与的图象有两个公共点（如图）BCDACDBCABD1aeyaylogax3（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．213．514．544；16．4415．68108(1e,e1)（注：或写成1eme1）图4图5图67．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，当函数x与有一个公共点在直线上，且在该点处有公切 5yaylogax(x0,y0)yx9．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）当时函数图象呈现以下三种情况xa1a0x①1所以有0，①代入②消x0得x0，x0，代入②中得xax0lnalna1aa0lna1②eaee11即x与的公共点为，1，yaylogax()x0eee结合图象得到以下结论：（4）当e时，函数x与的图象有三个公共点（如图）0aeyaylogax4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）22（5）当aee时，函数yax与ylogx的图象有一个公共点（如图5）h(e)0h(e)0a令h(t)t2mte4，则h(1)0或h(1)0（6）当e时，函数x与的图象有一个公共点（如图6）22ea1yaylogaxh(e)0h(e)0故0me41或1e4m015．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》：44②分层随机抽样ee()式化为mt，令h(t)t(t0)，tt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．易知yh(t)在(，0)，(0,)上单调递增，(x2)ex且h(1)1e4，h(1)e41，h(e2)h(e2)016．教学提示：当x0且x1时，f(x)，f(2)02(x1)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e21或122t2e1t2e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．【解析】令g(x)0得f2(x)mf(x)e40，设tf(x)，17（Ⅰ）2····························································1分24a//b3sinxcosxcosx0得关于t的方程tmte0()即24cosx(3sinxcosx)0m4e0恒成立，设()式有两个不等实根t1,t2，3当te2,te2时，即m0，满足题意，cosx0或tanx························································································3分123e2t1te2当1或1,满足题意，x(0,)22t2e1t2e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2636（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．）2方法二：a//b3sinxcosxcosx0·························································1分18．【解析】31cos2xsin2x0122（Ⅰ）解：f(x)2(x0)·············································································1分x1sin(2x)··································································································3分62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(,)666（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6666lnx由f(x)ax22x得a在(0,)恒成立···························································6分x或x···································································································5分x262（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)maxx211（Ⅱ）f(x)ab3sinxcosxcos2x················································6分12lnx22g(x)，g(e)0·················································································8分x331sin2xcos2x当0xe时，g(x)0；当xe时，g(x)022故g(x)在(0,e)上单调递增，在(e,)上单调递减sin(2x)······················································································7分61故当xe时，g(x)取最大值···········································································11分2e令2k2x2k(kZ)26211故a，即a的取值范围是[,)······································································12分2e2ekxk························································································9分361(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1g(x)2ax,h(x)2x由f(x)ax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12ax21令g(x)ax2lnx(x0)，则g(x)2axxx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；设切点坐标为（或直接由g(2)4aln20，不合题意(x0,y0)22或当x1时，ax0,lnx0,g(x)0，不合题意）···············································8分y0ax0lnx0xe0则1解得12ax0a1x②当a0时，令g(x)0得x，02e2a1故当a时，g(x)h(x)112e令g(x)0得x，令g(x)0得0x,2a2a1即a的取值范围是[,).·····················································································12分11故g(x)在(0,)上单调递减，g(x)在(，)上单调递增2e2a2a（教学建议：1.教师应强调第二问的法三比较适合选填题；1111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；2a2a22a变式：若x[1,0],xln(12x)(a1)x恒成立，求a的取值范围.11故a，即a的取值范围是[,)2e2e119．【解析】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：22由f(x)ax2x得axlnx在(0,)恒成立·························································6分当n1时，S