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数学参考答案第1页(共8页)南平市2023届高中毕业班第三次质量检测数学参考答案及评分标准说明:1.本解答给出了一种或几种解法供参考,如果考生的解法与本解答不同,可根据试题的主要考查内容比照评分标准制定相应的评分细则.2
.对计算题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内容和难度,可视影响的程度决定后继部分的给分,但不得超过该部分正确解答应给分数的一半;如果后继部分的解答有较严重的错误,就不再给分.3.只给整数分数.选择题
和填空题不给中间分.一、选择题:本题考查基础知识和基本运算,每小题5分,满分40分。1.C2.B3.C4.C5.D6.A7.B8.A二、选择题:本大题共4小题,每小题5分,满分20分.在每小题给出的四个选项中,有
多项符合题目要求。全部选对的得5分,部分选对的得2分,有选错的得0分。9.BD10.ACD11.AC12.BCD三、填空题:本题考查基础知识和基本运算,每小题5分,满分20分。13.2014.(1,1)(答案不
唯一)15.2ee0xy16.82π四、解答题:本大题共6小题,共70分,解答应写出文字说明、证明过程或演算步骤。17.(10分)解:(1)依题意,{}nnaT是以1为首项,2为公差的等差数列,所以11221nna
nnT(),.....................................................................................................................
.............................................2分即(21)nnnTa,从而当2n≥时,有11(23)nnnTa,..................................
.....................3分两式相除得,12123nnnnaana,由已知得0nT,从而0na,所以当2n≥时,121123nnna,即12321nnan,.....................................
.................................5分所以2121nnan..............................................................................
......................................................6分(2)设{}23nTn的前n项和为nS,由(1)得,21nnaTn,所以121nTn,..................7分故123(21)(23)nTnnn1
11()22123nn,......................................................................8分数学参考答案第2页(共8页)所以1111111111()()2355
7212323233(23)nnSnnnn.................10分18.(本小题满分12分)解:(1)依题意得sinsincoscossincosBCCBCC,......................
..........................................................................2分所以sinsinsincoscossincoscosBC
BCBCBC,所以sincos0BCBC,.......................................................................................................4分所以
tan1BC即tan1A,........................................................................................
............5分又因为0πA,所以π4A............................................................................
..................................6分(2)由11π2sin224ABCSabc△,...........................................................................
........................7分所以24abc.由余弦定理得222π2cos4abcbc,.......................................................
........................................8分即2222abcbc,所以2222128bcbcbc,即22221228bcbcbcbc≥,.............
..................................10分所以822bc≥,当且仅当bc时取“等号”,..........................................................................11分而1π2s
in244ABCSbcbc△,故min28224244ABCS△....................................12分19.(12分)数学参考答案第3页(共8页)(1)证明:作SO平面ABC,垂直为O,则SB
O为SB与平面ABC所成角,所以3SBO,在RtSBO△中,由4SB可得2BO,23SO.........................................1分因为SO平面ABC,所以SOCO,所以在RtSCO△中,由26SC,23SO,可得23C
O,..............................................2分在BCO△中,由4BC,23CO,2BO,可得222BCCOBO,从而COBO,且6BCO,...............
.........................................................................................................................3分在ACO△中,4
AC,6ACO,所以ACOBCO△△,从而2AOBO,所以4AOBO,又4AB,所以点O必在AB上,且O为AB的中点...............................................................................
............................................................................4分因为SO平面ABC,所以SOAB,又ABCO,SOCOO,,SOCO平面SCO,所以A
B平面SCO,从而SCAB.................................................................................................6分(2)解:以点
O为坐标原点,,,OCOBOS所在直线分别为zyx,,轴建立如图所示的空间直角坐标系xyzO,.......................................................................................
....................................7分(0,2,0)B,(23,0,0)C,(0,0,23)S,(3,3,0)D,则133()(,,3)222OMOSOD,即33(,,3)22M................
...................................8分37(,,3)22BM,(23,2,0)BC,设平面BCM的法向量为),,(zyxn,由,,00BCnBMn得
.0232032723yxzyx,取3y,可得)4,3,1(n,....................................................................
..................................10分取平面SBC的一个法向量)001(,,m,........................................................
............................11分设所求的角为,则105|201||||||||,cos|cosnmnmnm,因此所求平面SBC数学参考答案第4页(共8页)与
平面SAD夹角的余弦值为105.....................................................................................................12分20.(12分)解:(1)因为散点,1,
2,,6iivi集中在一条直线附近,设回归直线方程为ˆˆˆbva,由4.1,3.05v,则1222175.364.13.051ˆ101.464.12niiiniivnvbvnv,....
.......................................2分所以1ˆˆ3.054.112abv,.........................................................................
............................4分所以变量关于v的回归方程为1ˆ12v,令ln,ln,vxy所以1lnln1,2yx所以12ˆeyx,.............................................
...................6分综上,y关于x的回归方程为12eyx.(2)由1212eeee,97yxxxx,解得4981x≤≤,所以50,60,65,72x,所以B,C,D,E为“热门套票”
,................................................................................................8分则三人中购买“
热门套票”的人数X服从超几何分布,X的可能取值为1,2,3.1221342424333666CCCCC131(1),(2)(3)C5C55,CPXPXPX.....................................................10分所以X的分
布列为:131()1232555EX...........................................................................
....................12分X123P153515数学参考答案第5页(共8页)21.(12分)解:(1)由题意得2,2()()1,caacac................................
..................................................................................................2分解得2a,1c,所以1
b...................................................................................................
.....3分椭圆C的方程为2212xy......................................................................................
.......................4分(2)依题意得直线2l的方程为1x.设直线1l的方程为1xmy,1122(,y),(,)PxQxy.由2212xmyxy,=1,得22(2)210mymy.........................
.........................................................................................5分12222myym,12212yym,.........
.................................................................................6分所以12122yymyy.1AP的方程为:11(2)2
yyxx,由111(2)2xyyxx,,解得11(12)2Myyx.......................................................
...........7分2AQ的方程为:22(2)2yyxx,由221,(2),2xyyxx解得22(12)2Nyyx...............................................................
...8分数学参考答案第6页(共8页)1111122221221(12)(12)||||2(2)21(2)||||(12)(21)22yFNAFSxyxySyxFMAFx..................
......................10分12112121211221221()(12)(m12)(12)21(m12)(12)()(12)2yyyyymyyyyymyyyyyy..
..............................11分121212123+22322322322322yyyyyyyy322...........................................................
...............................................................................12分22.(12分)解:(1)2()6exfxa
,...................................................................................................
.............1分当0a≥时,()0fx,fx单调递增,无极值,........................................................................2分当0a时,由()0fx,得1ln()26ax,...
................................................................................3分当1(0,ln())26ax时,'()0fx,函数()fx单
调递减;当1(ln(),)26ax时,'()0fx,函数()fx单调递增;所以当1ln()26ax时,函数()fx的极小值为ln()226aaa,无极大值..........................4分(2)由(1)得ln()3226aaa
,令()ln()226aaaa0a,则1'()ln()26aa,由'()0a,得6a,数学参考答案第7页(共8页)当(,6)a时,'()0a,函数()a单调递增;当(6,0)a时,'()0a,函数()a单调递减;所以当且
仅当6a时,函数()a的最大值为3,故由ln()3226aaa,得6a.............................................................
..................................5分不妨设1x2x,当x(0,)时,2'()6(e1)0xfx,fx单调递增;当x(0,)时,2'()660gxxx
,gx单调递增;所以1212|()()||()()|fxfxgxgx≥,可化成2121()()()()fxfxgxgx≥,..............6分即2211()()()()fxgxfxgx≥,对10x2x恒成立,令2
32()()()3e236xhxfxgxxxx,x(0,),则()hx在(0,)上单调递增,所以22'()6e6660xhxxx在(0,)上恒成立,..............................
..........................7分设22()e1xuxxx,则(0)0u,2'()2e2xuxx,'(0)2u,当2≤时,因为2''()4e2xux在(0,)上单调递增;所以''()''(0)420ux
u≥≥,所以2'()2e2xuxx在(0,)上单调递增;所以'()'(0)20uxu≥≥,所以22()e1xuxxx在(0,)上单调递增;所以()(0)0uxu≥,数学参考答案第8页(共8页)所以'()6()0hxux≥在(0,)上恒
成立..................................................................................10分当2时,令2''()4e20xu
x,得1ln22x,当1(0,ln)22x时,''()0ux,所以当1(0,ln)22x时函数'()ux单调递减,'()'(0)0uxu,所以当1(0,ln)22x时函数()ux单调递减,()(0
)=0uxu,所以当1(0,ln)22x时'()6()0hxux,矛盾.综上所述,2≤...................................................................................
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