【文档说明】2023届四川省绵阳市二诊理数试题及答案.pdf，共(10)页，458.142 KB，由小喜鸽上传

转载请保留链接：https://www.ichengzhen.cn/view-170627.html

以下为本文档部分文字说明：

理科数学参考答案第1页共6页绵阳市高中2020级第二次诊断性考试理科数学参考答案及评分意见一、选择题：本大题共12小题，每小题5分，共60分．DACBDABCADCA二、填空题：本大题共4小题，每小题5分，共20分．13．914．1315．516．[1，3)三、解答题：本大题共

6小题，共70分．17．解：（1）由23cossin3aCaCb，可得3sincossinsin3sinACaACB，················································

··2分又3sin3sin()3(sincoscossin)BACACAC，····································4分∴sin3cosaAA，5分且3A，可得3a；·

······································································6分（2）由11cos()22BAACcbAcb

，可得1cb，······················8分由余弦定理2222cos4cbabcA.····················································9分∵1()2ATABAC

，平方可得2221()[()()2]4ATABACABAC，··································10分即22215()(2cos)44ATcbbcA

，所以52AT．·······························12分18．解：（1）因为0.92<0.99，根据统计学相关知识，2R越大，意味着残差平方和521ˆ()iiyy越小，那么拟合效果越好，因此选择非线性回归方程②2ˆˆˆ

ymxn进行拟合更加符合问题实际．·························································································

4分（2）令2iiux，则先求出线性回归方程：ˆˆˆymun，·································5分∵14916250.81.11.52.43.7111.955=，uy

，······················7分2222221()(111)(411)(911)(1611)(2511)niiuu=374，············9分∴121()()45.1ˆ0.121374()niiiniiuuyymuu

，···················································10分理科数学参考答案第2页共6页由ˆ1.90.12111n，得ˆ0.5690.57n，········

·····································11分即ˆ0.120.57yu，∴所求非线性回归方程为：2ˆ0.120.57yx．········································12分19．解：设{}nb的公比为q，则231

bbq，所以282273q，所以23q，····················································································

·2分因为{}nb的各项都为正，所以取23q，···················································3分所以2()3nnb．··············

·····································································4分若选①：由21(1)nnaSn，得1121(2)≥nnaSn，·······················

······5分两式相减得：1220nnnaaa，整理得12(2)≥nnaan，························6分因为110a，所以{}na是公比为2，首项为1的等比数列，·····················7分∴12nna，··

···················································································8分∴14()23nnnab，·········

·····································································9分∵14()23xy在R上为增函数，··························

·································10分∴数列{}nnab单调递增，没有最大值，··················································11分∴不存在mN，使得对任意的，≤nn

mmnNabab恒成立．···················12分若选择②：因为211(2)nnnaaan，且1200，aa，∴{}na为等比数列，·····································

······································6分公比2114aqa，················································································7分∴11()4nn

a，···················································································8分所以1122112()()44()44336

63≤nnnnnnnab．·································10分当且仅当1n时取得最大值23，··························································11分∴存在1m，使得对

任意的，≤nnmmnNabab恒成立．··························12分若选择③：因为11nnaa，所以11nnaa(2)≥n，·····························5分∴{}na是以1为公差

的等差数列，又11a，············································6分∴nan，所以2()3nnnabn，·······················································

·····8分理科数学参考答案第3页共6页设nnncab，则1112232()(1)()()3333nnnnnnccnn，····················9分∴当3n时，10nncc，所以1nncc，当3n

时，10nncc，所以1nncc，当3n时，10nncc，所以1nncc，则12345ccccc，·····························································

···11分∴存在23m或，使得对任意的，≤nnmmnNabab恒成立．······················12分20．解：（1）设11()，Bxy，22()，Cxy，直线BC的方程为：4xmy(1=mk)，·····························

······················1分联立224143xmyxy，消x整理得：22(34)24360mymy，·····················2分∴1222434myym，1223

634yym······················································3分从而：121212121222(6)(6)yyyykkxxmymy122

12126()36yymyymyy2222236134361444363434mmmmm∴12kk为定值14.···································································

·············5分（2）直线AB的方程为：11(2)2yyxx，················································6分令4x，得到11116626Myyyx

my，······················································7分同理：2266Nyymy．······································

····································8分从而121266||||||66MNyyMNyymymy122121236|||6()36|yymyymyy···············

·······································9分又221212122124||()434myyyyyym，理科数学参考答案第4页共6页212122144|6()36|34myymyym，················

·····································10分所以2||34MNm，···························································

············11分因为：1[34]，mk，所以||[3563]，MN，即线段MN长度的取值范围为[3563]，．········································

·····12分21．解：（1）由a=1时，()(1)(1)xfxxe，·············································1分由()0fx解得：x>0或x<−1；由()0fx解得：−1<x<0.···

······················3分故f(x)在区间(1)(0)，，，上单调递增，在区间(−1，0)上单调递减.··········4分∴f(x)的极大值是f(−1)=22ee，极小值是f(0)=0；·····································

··5分（2）()(1)()[02]，，xfxxeax时，2[1]xee，，且2(2)24(0)0，feaf，12[02]，，xx，恒有212()()2≤fxfxae等价于2maxmin()()2≤fxfxae．i)若1≤a

时，0≥xea，故()0≥fx，所以f(x)在区间[0，2]上单调递增，故22maxmin()()(2)(0)242≤fxfxffeaae，解得：01≤≤a，·············6分ii)若2≥ae时，0≤xea，故()0≤fx，所以f(

x)在区间[0，2]上单调递减，故22maxmin()()(0)(2)422≤fxfxffaeae，解得：2243≤≤eae，·······7分iii)若21ae时，由()(1)()0

xfxxea解得：ln2≤ax，故f(x)在区间(ln2]，a上单调递增；·····································································

················8分由()(1)()0xfxxea解得：0ln≤xa，故f(x)在区间[0ln)，a上单调递减.∴2min1()(ln)(ln)2fxfaaa，max()(2)fxf或f(0)．又f(2)−f(0)=224ea，··············

···························································9分①当212≤ea时，f(2)−f(0)0≥，故222maxmin1()()(2)(ln)24(ln)22≤fxfxffaeaaaae，解得：100≤a

e，又212≤ea，故此时212≤ea．································10分②当222eae时，f(2)−f(0)<0，故22maxmin1()()(0)(

ln)(ln)22≤fxfxffaaaae，令221()(ln)22haaaae，则21()(ln)ln12haaa，又ln(2ln22)，a，理科数学参考答案第5页共6页故21()(ln)ln12haaa>0，即h(a)在

区间22()2，ee上单调递增，又22()0hee，则221(ln)22≤aaae恒成立．·····································11分综上：2403≤≤ae．···········

································································12分22．解：（1）①当B在线段AO上时，由|OA|‧|OB|=4，则B（2，）

或（2，23）；②当B不在线段AO上时，设B（ρ，θ），且满足|OA|‧|OB|=4，∴A4（，），···············································

······························1分又∵A在曲线l上，则44cos()sin()2，···························3分∴2si

n2cos，······································································4分又∵32≤≤，即20≤≤.综上所述，曲线C的极坐标方程为：2sin2cos2（0≤≤），

或32()2=或=．·························5分（2）①若曲线C为：32()2=或=，此时P，Q重合，不符合题意；②若曲线C为：2sin2cos2（0≤≤），设l1：2（0≤≤），又l1与曲

线C交于点P，联立2sin2cos，，得：2sin2cosP，·······················································

·············6分又l1与曲线l交于点Q，联立sincos2，，得：2sincosQ，··························································

·············7分又∵M是P，Q的中点，1sincos(0)2sincos2≤≤PQM，·······························8分令sincost，则2sin()4t，又∵20≤≤

，则3444≤≤，且12≤≤t，理科数学参考答案第6页共6页∴1(12)≤≤Mttt，且1Mtt在12，上是增函数，······················9分∴222210≤≤M，且当42时，即4时等号成立．∴OM的最大值

为22．····································································10分23．解：（1）由()fx≤3的解集为[n，1]，可知，1是方程()fx=3的根，∴(1)f=3+|m+1|=3，则

m=−1，······························································1分∴()fx=|2x+1|+|x−1|，①当x≤12时，()fx=−3x≤3，即x≥−1，解得：−1≤x≤12，·········

·········2分②当112x时，()fx=x+2≤3，解得：112x，·································3分③当x≥1时，()fx=3x≤3，解得：x=1．········································

········4分综上所述：()fx的解集为[−1，1]，所以m=−1，n=−1．······························5分（2）由（1）可知m=−1，则1222ab．·················

·····································6分令12xa，2yb，则12ax，2by，又a，b均为正数，则2xy（00，xy），由基本不等式得，22≥xyxy，······

·················································7分∴1≤xy，当且仅当x=y=1时，等号成立．所以有11≥xy，当且仅当x=y=1时，等号成立．··

······································8分又22222244164(2)ababxy224482≥xyxy(当且仅当x=y时，等号成立)．·······9分∴

22168≥ab成立，(当且仅当，122，ab时等号成立)．·····················10分