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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）吉林地区普通高中2023—2024学年度高三年级第一次模拟考试数学试题参考答案一、单项选择题：本大题共8小题，每小题5分，共40分．12345678ACBDACBC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分．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．13．514．5215．68；10816．)1,1(44ee（注：或写成1144

eme）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．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．（一）当1a时函数图象呈现以下三种情况图1图2图3如图2，当函数xay与xlogya有一个公共点)(00y,x在直线xy上，且在该公

共点处的切线为xy，所以有②①1000alnaxaxx，①代入②消0xa得eaalnalnxxx0010，，代入②中得exeae01即xay与xlogya的公共点为)(e

e，结合图象得到以下结论：（1）当eea1时，函数xay与xlogya的图象无公共点（如图1）（2）当eea1时，函数xay与xlogya的图象有一个公共点（如图2）（3）当eea11时，函数xay

与xlogya的图象有两个公共点（如图3）（二）当10a时函数图象呈现以下三种情况图4图5图6如图5，当函数xay与xlogya有一个公共点)(00y,x在直线xy上，且在该点处有公切线（斜率为1），所以有②①

1000alnaxaxx，①代入②消0xa得eaalnalnxxx11000，，代入②中得exeae10，即xay与xlogya的公共点为)11(ee，结合图象得到以下结论：（4）当eea0时，函数xay与xlogya

的图象有三个公共点（如图4）9101112BCDACDBCABD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）（5）当eea时，函数xay与xlogya的图象有一个公共点（如图5）（6）当1ae

e时，函数xay与xlogya的图象有一个公共点（如图6）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．（2）学生掌握推导过程．摘自《数学课程标准》：②分层随机抽样通过实例，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．16．教学提示：当10xx且时，2)1()2()(xexxfx

，0)2(f当120xx且时，0)(xf；当2x时，0)(xf．故)(xf在)1,0(，)2,1(上单调递减，在),2(上单调递增，当2x时，)(xf取得极小值2)2(ef，当

1x时，)(xf；当1x时，)(xf；当x时，)(xf．由)(xf解析式可知，)(xf为奇函数．画出)(xf图象大致如下：令0)(xg得0)()(42exmfxf，设)(xft，

得关于t的方程)(042emtt0442em恒成立，设)(式有两个不等实根21,tt，当2221,etet时，即0m，满足题意，当22121ette或,12221etet满足题意，方法一：令42)(emttth，则

0)(0)1(0)(22ehheh或0)(0)1(0)(22ehheh故104em或014me综上，实数m的取值范围是)1,1(44ee．方法二：)(式化为tetm4，令)0()(4ttetth，易知)(thy在)

0(，，)0(,上单调递增，且41)1(eh，1)1(4eh，0)()(22eheh其图象大致如图：当104em或014me时，满足22121ette或

22211etet综上，实数m的取值范围是)1,1(44ee．四、解答题17．【解析】（Ⅰ）b//a032xcosxcosxsin··································

··························1分即0)3(xcosxsinxcos0xcos或33xtan····························································

····························3分)(0,x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）2x或6x·································································

··································5分（注：少一个解扣1分，没有角的范围表述扣1分.）方法二：b//a032xcosxcosxsin·························································1分02

21223xcosxsin21)6(2xsin··················································································

················3分)(0,x)611,6(62x662x或6562x6x或2x··························································

·········································5分（注：少一个解扣1分，没有角的范围表述扣1分.）（Ⅱ）21)(baxf2132xcosxcosxsin·······································

·········6分xcosxsin221223)6(2xsin·································································

·····················7分令kxk226222)(Zkkxk63··················································································

······9分)(xf的单调递增区间是)](63[Zkk,k·················································10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．）18．【解析】（Ⅰ）解：)0

(12)(xxxf·············································································1分所求切线斜率为1)1(f，切点为)2,1(··················

················································3分故所求切线方程为)1()2(xy，即01yx·················································5分（注

：将切线方程表示成1xy也给分）（Ⅱ）方法一：分离变量由xaxxf2)(2得2lnxxa在),0(恒成立···················································

········6分令)0(ln)(2xxxxg，则maxxga)(3ln21)(xxxg，0)(eg····························································

·····················8分当ex0时，0)(xg；当ex时，0)(xg故)(xg在),0(e上单调递增，在),(e上单调递减故当ex时，)(xg取最大值e21··············

·····························································11分故ea21，即a的取值范围是),21[e·················

·····················································12分(注：表示成ea21不扣分）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）方法二：分类讨论由xaxxf2)(2得0ln2xax在),0(恒成立·

···················································6分令)0(ln)(2xxaxxg，则xaxxaxxg1212)(2①当0a时，0)(xg恒成立，)(xg在),0(上单调递减，

0)1(ag又，故当1x时，0)(xg，不合题意；（或直接由02ln4)2(ag，不合题意或0)(,0ln,012xgxaxx时，当，不合题意）······················

·························8分②当0a时，axxg210)(得令，axxg210)(得令，axxg2100)(得令,故)(xg在)21,0(a上单调递减，)(xg在)

21(，a上单调递增故当ax21时，)(xg取最小值021ln21)21(aag·········································11分故ea21，即a的取值范围是),21[e综上所述，a的取值范围是),21[e···········

·······························································12分方法三：数形结合由xaxxf2)(2得xaxln2在),0(恒成立··································

·······················6分令xxhaxxgln)(,)(2，则当0x时，)()(xhxg恒成立xxhaxxg1)(,2)(若0a，)()(,0ln,012xhxgx

axx时，当，不合题意；·······························7分若0a，)()(xhxg曲线)(xgy与曲线)(xhy有且只有一个公共点，且在该公共点处的切线相同.设切点坐标

为),(00yx则00020012lnxaxxaxy解得eaex210故当ea21时，)()(xhxg即a的取值范围是),21[e.···························

··························································12分（教学建议：1.教师应强调第二问的法三比较适合选填题；2.在教学中，教师应注意强调)

0)(()(kkxlnxf的导函数的相关内容；变式：若xaxlnx,,x)1()21(]01[恒成立，求a的取值范围.19．【解析】（Ⅰ）选择①由已知可得：当1n时，221aS···················

·····································································1分当2n且Nn时，nnnnnaaSSa11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）nnaa21即21nnaa·················································································

···········3分又112aa不符合上式Nnn,,n,ann且22121················································································

····5分选择②1nnaSnnnSSS1···········································································

··························1分nnSS21即21nnSS·············································································

·············3分又211aS}{nS是以2为首项，2为公比的等比数列．nnnS2221)(Nn····················································································5

分（Ⅱ）由（Ⅰ）可知nnnaS21，122nna121121)12)(12(2)1)(1(1121nnnnnnnnnaaSb···································9分nnbbbb

T321)121121()15171()7131()311(1nn12111n·······················································

··········································11分101211nnTNn························································

··············12分（注：若（Ⅰ）选择①，且未讨论1a的情况，扣2分；但若（Ⅱ）运算正确，（Ⅱ）问正常赋分．）关于劣构问题：劣构问题的常见形式：（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；（3）多个类似条件的结构不良问题；（4）问题条件或数据冗余的结构不良问题；（5）不同求解目标的结构不良问题．条件的选择原则：（1）应先分析所有条件，优先选择自己会做的、有把握的；（2）在有把握的前提下，优先选择难度最小的；（3）在选择条件时，要通盘考虑条件对整个

问题的影响．教学建议：（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；（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；（3）注重变式训练，提高学生的辨析能力和应变能力．20．【解析】（

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55.0，该指标在90以下的概率为8.0，因此该项指标的第70百分位数为a一定在)90,80[内7.080025.010025.010023.010005.01

0002.0）（a（也可以用7.0190025.01002.0）（a）得86a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）所以Q型芯片I级品该项指标的第70百分位数为86························

······························3分（Ⅱ）当临界值65c时，Q型芯片Ⅱ级品应用于A型手机的概率为05.0657001.0）（····································6分（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），]6003010010[300]500050100020[700）（）（c..c..yc5.5409]60,50[c·············

········································································8分所以当5650c时，101y，当56c时，101y，当6056c时，101y···················

··························································10分综上：为降低芯片生产商的成本，当临界值)56,50[c时，选择方案二；当临界值56c时，选

择方案一和方案二均可；当临界值]60,56(c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）21．【解析】（Ⅰ

）由正弦定理得AsinBsinAsinAcosCsin2)3(·································································1分AsinCAsinAsi

nAcosCsin2)()3(AsinCsinAcosCcossinAAsinAcosCsin2)3(即AsinCcossinACsinAsin230)(0,sinAA

2)6(2Csin即1)6(Csin··································································3分)67,6(6)(0,

CC26C3C··········································································································

······5分(注：没有角的范围表述扣1分.)（Ⅱ）方法一：BCAB，3CABC为等边三角形2ACBCAB(0,2),,2,mmBPxBExPECE在BPE中，由余弦定理得BcosBEBPBEBPPE2222···················

····················································7分即21)(22)(2222xmxmx整理可得(0,2),4422mmmmx

·········································································8分6346412)4(264124mmmmx当且仅当324

m时取等号即324BP时，CE取最小值634此时348BE································································································10

分.sinBEBPSPBE2431423)34(8)32(421321·················12分方法二：BCAB，3CABC为等边三角形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）2ACBCAB(0,2),2,

xxBExPECE设)32(0,,EPB在BPE中，由正弦定理得EPBsinBEEBPsinPE·················································································

·········7分即sinxsinx23整理可得)32(0,,233sinx························································

················8分当且仅当2时，x取最小值634当CE取最小值634时，348BE在BPEtR中，324216BEBP,BEP······························

···················10分.sinBEBPSPBE2431423)34(8)32(421321·················12分22．【解析】（Ⅰ）解：xcosmexfx)(···············

·································································1分方法一：函数)(xf在)0(,上单调递增0)(xcosmexfx在)0(，上恒成立（

且不恒为0）········································2分注：此处没取到等号，扣1分当20x时，0)(xf恒成立，即)(xf单调递增，0m当x2时，xcosemx恒成立设xcosexgx)(，xcosxsin

excosxsinxcosexgxx22)4(2)()(令0)(xg，则x43，令0)(xg，则432x)(xg在)432(，上单调递减，在)43(，上单调递增432)43()(egxgmin，

m432e又0m，m0432e综上，m的取值范围]20(43e,································································

········5分（注：（1）不写成区间形式也给分；（2）没写0m扣1分）方法二：函数)(xf在)0(,上单调递增0)(xcosmexfx在)0(,上恒成立（且不恒为0）又0m，xexcosm1在

)0(,上恒成立································································2分设xexcosxh)(xxexsinexcosxsinxh)4(2)(令0)(xh，

则x43，令0)(xh，则432x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）)(xh在)432(，上单调递减，在)43(，上单调递增4322)43()(ehxhmin43221em又0m

，m0432e.综上，m的取值范围]20(43e,·················································································5分（Ⅱ）证

明：1m，xsinexfx)(，xcosexfx)(当0x时，1xe，1xcos，0)(xcosexfx)(xf在)0(，上单调递增，即)(xf在)0(，上无极值点·················

······················7分当0x时，设)()(xfxu，0)(xsinexux恒成立)(xu在)0(，上单调递增0)2(2eu，021122)43(4343

eeu由零点存在性定理，存在唯一一个)243(0，x，使得0)(0xu，即00xcosex当0xx时，0)(xu，0)(xf，)(xf在)(0x，

上单调递减当00xx时，0)(xu，0)(xf，)(xf在)0(0，x上单调递增)(xf在)(，上存在唯一极小值点0x·······························································10分（注

：此处0x所在区间必须是)243(，的子集，否则只给到10分位置）)4(2)(000000xsinxsinxcosxsinexfx又)243(0，x，)43(40，x，)01()4(20,xsin即0

)(10xf······························································································

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