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文科数学参考答案第1页（共6页）绵阳市高中2020级第二次诊断性考试文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．DDCAABCDBACA二、填空题：本大题共4小题，每小题5分，共20分．13．214．115．316．[1，3)三、解答题

：本大题共6小题，共70分．17．解：（1）由23(cos)sinbaCaC，及正弦定理可得，3sin3sincossinsinBACaAC，················································

2分∵3sin3sin()3sincos3cossinBACACAC·······································4分∴3cossinsinsinACaAC，··········

······················································6分即sin3cosaAA，且3A，可得3a；············································

·8分（2）由2121)cos(bcAbcACBA，可得1cb，·······················10分由余弦定理2222cos4bcabcA.·······················

····························12分18．解：（1）由题意知，2nS=2na+na，①···············································1分当n=1时，21a=21a+1a，则11a；·········

··················································2分当2≥n时，21nS=21na+1na，②··········································

···········3分①②相减可得，2an=2na−21na+na−1na，················································4分∴an+1na=2na−21na，则an-1na=1，∴数列na是以11a

为首项，1为公差的等差数列，··································5分所以，an=n(n∈N∗).·······································································

·6分（2）2()3nnnabn，············································································7分设nnncab，

则1112232()(1)()()3333nnnnnnccnn，·························8分∴当3n时，10nncc，所以1nncc，·········

···································9分当3n时，10nncc，所以1nncc，·············································10分文科数学参考答案第2页（共6页）当3n时，

10nncc，所以1nncc，··············································11分则12345ccccc，∴存在23或m，使得对任意的，≤nnm

mnNabab恒成立．·····················12分19．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∵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分由ˆ1.90.12111n，得ˆ0.5690.

57n，即ˆ0.120.57yu，···········································································11分∴所求非线性回归方程为：2ˆ0.120.57yx．······

··································12分20．解：（1）设11()Bxy，，22()Cxy，，直线BC的方程为：4xmy，其中1=mk，·············

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

所以：1222434myym，1223634yym，·············································3分从而121212121222(6)(6)yyyyk

kxxmymy12212126()36yymyymyy文科数学参考答案第3页（共6页）2222236134361444363434mmmmm所以：12kk为定值14．·

·······································································5分（2）直线AB的方程为：)2(211xxyy，···············

·······························6分令4x，得到66261111myyxyyM，··················································7分同理：2266Nyymy．·····················

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

·····························9分又221212122124||()434myyyyyym，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）解：(1)a=2时，2()ln32fxxxx

，2231(21)(1)()xxxxfxxx，·······················································2分由()0fx解得：x>1或102x；由()0fx解得：112x

．·················3分故f(x)在区间(1)，，1(0)2，上单调递增，在区间1(1)2，上单调递减．···········4分所以f(x)的极大值是13()ln224f，极小值是f(1)=0；·····

··························5分文科数学参考答案第4页（共6页）（2）2(1)1(1)(1)()axaxaxxfxxx，且10≥x，································6分

①当≥1a时，10≥ax，(1)(1)()0≥axxfxx，故f(x)在区间[1，2]上单调递增，所以min()()0fxha，·····························7分②当102≤a时，10

ax，(1)(1)()0≤axxfxx，故f(x)在区间[1，2]上单调递减，所以min()()(2)ln2102≥afxhaf，显然()ha在区间1(0]2，上单调递增，故13()()ln224≤hah<0．··········

····························································9分③当112a时，由()0fx解得：12≤xa；由()0fx解得：

11≤xa．故f(x)在区间1(2]，a上单调递增，在区间1[1)，a上单调递减．此时min11()()()ln22afxfhaaaa，则222111(1)()0222≥ahaaaa，故()ha在区间1(1)2，上单调

递增，故h(a)<h(1)=0.·······································11分综上：011()ln2102211ln1222，≥，≤，aahaaaaaa

，且h(a)的最大值是0.··························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∵3≤≤2，即20≤≤.综上所述，曲线C的极坐标方程为：文科数学参考答案第5

页（共6页）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，∴1(12)≤≤Mttt，且1Mtt在12，上是增函数，······················9分∴22221≤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分文科数学参考答案第6页（共6页）（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22222244

164(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