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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3

A，可得3a；·······································································6分（2）由11cos()22BAAC

cbAcb，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②2ˆˆˆymxn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∵14

916250.81.11.52.43.7111.955=，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ˆ0.120.5

7yu，∴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231bbq，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，≤nnmmnNabab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()4nna，···················································································8分所以1122112()()44()4433663

≤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)24360m

ymy，·····················2分∴1222434myym，1223634yym·················································

·····3分从而：121212121222(6)(6)yyyykkxxmymy12212126()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11116626Myyyxmy，······················································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22ma

xmin()()(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)()0xfxxea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.∴2m

in1()(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≤ae，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2

1()(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∴2sin2cos，······································································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