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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x

y，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)(ee，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②①1000alnaxaxx，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aee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xta

n························································································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0221223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xaxxf

2)(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,012x

hxgxaxx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nnbbbbT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7.080025.010025.010023.010005.010002.0

）（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5

6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AsinCAsinAsinAcosCsin2)()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2

222·······································································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32

4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)(xcosme

xfx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)(，xcosxsinexcosxsinxcosexgxx22)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，m043

2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)(000

000xsinxsinxcosxsinexfx又)243(0，x，)43(40，x，)01()4(20,xsin即0)(10xf·····································

·····························································12分获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com