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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11221nnannT(),....................................
.....................................................................................................................
.........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)nTnnn111()22123nn,.................
.....................................................8分数学参考答案第2页(共8页)所以1111111111()()23557212323233(23)nnSnnnn
.................10分18.(本小题满分12分)解:(1)依题意得sinsincoscossincosBCCBCC,................................................................
................................2分所以sinsinsincoscossincoscosBCBCBCBC,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2222
abcbc,所以2222128bcbcbc,即22221228bcbcbcbc≥,...............................................10分所以822bc≥,当且仅当bc时取“等号”,.
.........................................................................11分而1π2sin244ABCSbcbc△,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SBO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23CO,........................
......................2分在BCO△中,由4BC,23CO,2BO,可得222BCCOBO,从而COBO,且6BCO,.............................................
...........................................................................................3分在ACO△中,4AC,6ACO,所以ACOBCO△△,从而2A
OBO,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AB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1b........................................................................................................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)2yyxx,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)111112222122
1(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232()()()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)420uxu≥≥,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''()4e20xux,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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