四川省巴中市普通高中2024-2025学年高三上学期9月零诊考试 数学 PDF版含答案(可编辑)

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数学·第1页(共4页)巴中市普通高中2022级“零诊”考试数学试题(满分150分120分钟完卷)注意事项:1.答题前,考生务必将自己的姓名、班级、考号填写在答题卡规定的位置.2.答选择题时请使用2B铅笔将答题卡上对应题目的答案标号涂黑;非选择题答题时必须用0.5毫米黑色墨迹签字笔,将答案书

写在答题卡规定的位置,在规定的答题区域以外答题无效,在试题卷上答题无效.3.考试结束后,考生将答题卡交回.一、单选题:本大题共8小题,每小题5分,共40分.每小题给出的四个选项中,只有一项符合题目要求的.1.已知复数21iz,则||z()A.22B.1C.2D.2

2.设lmn、、为直线,且mn、在平面内,则“l”是“lmln且”的()A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件3.已知集合4{|,,}1PxyxyxN,{|14}Qxx≤≤,则PQ()A.{1,2,4}B.{0,1,3}C.{|0

3}xx≤≤D.{|14}xx≤≤4.已知nS是等差数列{}na的前n项和,若481240SS,,则12S()A.44B.56C.68D.845.设函数(4),0,()(4),0.xxxfxxxx≥若2(3)(1)fafa,

则实数a的取值范围是()A.(,1)(2,)B.(,2)(1,)C.(,1)(3,)D.(,3)(1,)6.有4名志愿者参加社区服务,服务星期六、星期日

两天.若每天从4人中任选两人参加服务,则恰有1人连续参加两天服务的概率为()A.34B.23C.13D.14{#{QQABaYSAgggAAoAAARgCUwVqCEMQkACCASgGBFAAIAAB

ARNABAA=}#}数学·第2页(共4页)7.已知函数1()31fxxx的图象与直线(1)4ykx有两个交点11(,)xy,22(,)xy,则1212xxyy()A.6B.8C.10D

.128.已知12FF,是椭圆2222:1(0)yxCabab的左右焦点,A,B是椭圆C上的两点.若122FAFB,且124AFF,则椭圆C的离心率为()A.13B.23C.33D.23二、多选题:本大题共3小

题,每小题6分,共18分.每小题给出的四个选项中,有多项符合题目要求.全部选对得6分,选对但不全得部分分,有选错的得0分.9.设离散型随机变量X的分布列如下表X01234P0.10.2m0.20.1若离散型随机变量Y满足21YX,则()A.0.4

mB.()2,()1.2EXDXC.()3,()3.4EYDYD.()5,()4.8EYDY10.已知函数()sincosfxaxx的图象关于3x对称,下列结论中正确的是()A.()6fx是奇函数B.62(

)44fC.若()fx在[,]mm上单调递增,则03m≤D.()fx的图象与直线23yx有三个交点11.已知A,B为双曲线22:12yCx的左,右顶点,12,FF分别为双曲线C的左,右焦点.下列命题中正确的是()A.若R为双曲线C上一点,且1|4RF|,则2|6RF

|B.2F到双曲线C的渐近线的距离为2C.若P为双曲线C上非顶点的任意一点,则直线PA、PB的斜率之积为2D.双曲线C上存在不同两点M,N关于点(1,1)Q对称{#{QQABaYSAgggAAoAAARgCUwVqCEMQkA

CCASgGBFAAIAABARNABAA=}#}数学·第3页(共4页)三、填空题:本大题共3小题,每小题5分,共15分.12.41(2)xx的展开式中2x的系数是.13.已知正四棱台的高为2,上下底面的边长分别

为22和42,其顶点都在同一球面上,则该球的表面积为.14.已知向量,ab满足||2a,|2|||6abb,则||ab的取值范围为____________.四、解答题:本大题共5小题,共77分.解答应写出文字说明、证明过程或演算步骤.15.(本小

题满分13分)已知数列{}na的首项112a,且满足132nnnaaa.(1)证明:数列1{1}na为等比数列;(2)若123111150naaaa,求满足条件的最大整数n.16.(本小题满分15分)

在直三棱柱111ABCABC中,122AAAB,90ABC,D在1BB上,且12BD.(1)证明:1ACAD;(2)当四棱锥1ABCCD的体积为54时,求平面1ACD与平面ABC所成二面角的正弦值.ABC1A1B1C

D{#{QQABaYSAgggAAoAAARgCUwVqCEMQkACCASgGBFAAIAABARNABAA=}#}数学·第4页(共4页)17.(本小题满分15分)已知锐角ABC△中,角A,B,C的对边分别为a,b,c,若2cosaccB.(1)证明

:2BC;(2)若2a,求cos1Cbc的取值范围.18.(本小题满分17分)已知动圆Q经过点(1,0)F且与直线1x相切,记圆心Q的轨迹为曲线C.(1)求曲线C的方程;(2)设过点F且斜率为正的直线l交曲线C于A,B两点(点A

在点B的上方),AB的中点为M,①过M,B作直线1x的垂线,垂足分别为11,MB,试证明:11AMFB∥;②设线段AB的垂直平分线交x轴于点P,若FPM△的面积为4,求直线l的方程.19.(本小题满分17分)设函数2()ln(1)fxxxax.(1)

若曲线()yfx在点(1,0)处的切线方程为10xy,求a的值;(2)当1x时()0fx恒成立,求实数a的取值范围;(3)证明:222(*)ln1nkkknnN.{#{QQABaYSAgggAAo

AAARgCUwVqCEMQkACCASgGBFAAIAABARNABAA=}#}数学答案与评分标准第1页(共7页)巴中市高2022级零诊考试数学参考答案与评分标准一.单选题:本大题共8小题,每小题5分,共40分.答案

:1.C;2.A;3.B;4.D;5.A;6.B;7.C;8.B.二.多选题:本大题共3小题,每小题6分,共18分.每小题给出的四个选项中,有多项符合题目要求.全部选对得6分,选对但不全得部分分,有选错的得0分.答案:9.AB

D;10.AC;11.BC.三.填空题:本大题共3小题,每小题5分,共15分.答案:12.32−;13.80;14.,[53].四.解答题:本大题共5小题,共77分.15.(本小题满分13分,第(1)问6分,第(2)问7分)已知数列{}na的首项112a=,且满足132

nnnaaa+=+.(1)证明:数列1{1}na−为等比数列;(2)若123111150naaaa++++,求满足条件的最大整数n.解:(1)由112a=且132nnnaaa+=+知0na且1121133nnaa+=+····················

····················2分变形得11211(1)3nnaa+−=−········································································4分由

112a=得1111a−=·················································································5分∴数列1{1}na−是以1为首项,23为公比的等比数

列····································6分(2)由(1)得1121()3nna−−=,故1121()3nna−=+···················································7分∴1

2321()31111233()2313nnnnnaaaa−++++=+=+−−······································8分∴123111150naaaa++++等

价于233()503nn+−即23()473nn−······················································································9分令2()3()3nf

nn=−,则2(1)()1()03nfnfn+−=−∴()fn单调递增·····································································

···········10分∵472221()()3392=,故47202()13∴484722(48)483()4712()4733f=−=+−·························

······················11分又472(47)473()473f=−······································································

·12分∴使()47fn成立的最大整数为47{#{QQABaYSAgggAAoAAARgCUwVqCEMQkACCASgGBFAAIAABARNABAA=}#}数学答案与评分标准第2页(共7页)∴使123111150naaaa++++的最大整数

为47.······································13分16.(本小题满分15分,第(1)问7分,第(2)问8分)在直三棱柱111ABCABC−中,1AB=,12AA=,90ABC=,D在1B

B上,且12BD=.(1)证明:1ACAD⊥;(2)当四棱锥1ABCCD−的体积为54时,求平面1ACD与平面ABC所成二面角的正弦值.解:(1)证明连结1AB,在矩形11ABBA中,1,1,122ABBDAA===∴1tan2tanADBAB

A==,故1ADBABA=·········································2分又90ADBBAD+=,故190ABABAD+=∴1ADAB⊥···········

·············································································3分∵1,BBCBABBC⊥⊥1,BBAB平面11ABBA,1ABBBB=∴BC⊥平面11ABBA·····················

························································4分又AD平面11ABBA,故BCAD⊥····························

·································5分∵1,ADABADBC⊥⊥,1,ABBC平面1ADC,且1BCABB=∴AD⊥平面1ABC········································

·6分又1AC平面1ABC,故1ACAD⊥························7分(2)由题意知,1,,ABBCBB两两垂直以B为原点,以向量1,,BCBABB分别为,,xyz轴的正方向建立空间直角坐标系设(0)BCa=,由122AAA

B==,12BD=得:1,,,,,,,,,,,,1(00)(010)(002)(00)2CaABD11,,,,,(012)(02)ACa······································

8分由题意可知AB⊥平面11BCCB,平面ABC的一个法向量为,,(001)m=···············9分∴111155()32124ABCCDVCCBDBCBC−=+==,解得3BC=·······················10分1,,,,,1(01)(312)2AD

AC=−=−····························································11分设平面1ACD的一个法向量为,,()uxyz=由10,0uADuAC==

得10,2320,yzxyz−+=−+=取1y=得,,(112)u=−·······························13分∴,|62cos3|||114mumumu=

==++···················································14分设平面1ACD与平面ABC所成二面角的大小为,则2,3sin1cos3mu=−=∴平面1A

CD与平面ABC所成二面角的正弦值为33.······························15分注:第(1)问也可用坐标法证明.17.(本小题满分15分,第(1)问7分,第(2)问8分)ABC1A1B1CDAB

C1A1B1CDxyABC1A1B1CDz{#{QQABaYSAgggAAoAAARgCUwVqCEMQkACCASgGBFAAIAABARNABAA=}#}数学答案与评分标准第3页(共7页)在锐角ABC△中,角A,B,C的对边分别为

a,b,c,若2cosaccB−=.(1)证明:2BC=;(2)若2a=,求cos1Cbc+的取值范围.解:(1)方法一由2cosaccB−=与正弦定理得sinsin2sincosACCB−=······

···························1分又sinsin[()]sin()sincoscossinABCBCBCBC=−+=+=+····························3分∴sinsincossincossin()CBCCBBC=−=−··········

·····································4分由,02CB得22BC−−·········································

······················5分∴CBC=−,故2BC=········································································7分方法二由2cosacc

B−=及余弦定理得22222acbaccac+−−=,化简得22bcac−=··········1分由上式及正弦定理得22sinsinsinsinBCAC−=············································

··2分∵221cos21cos2cos2cos2sinsin222BCCBBC−−−−=−=······························3分又cos2cos[()()]cos()cos()s

in()sin()CBCBCBCBCBCBC=+−−=+−++−cos2cos[()()]cos()cos()sin()sin()BBCBCBCBCBCBC=++−=+−−+−∴22sinsinsin()sin()BCBCB

C−=+−∴sinsinsin()sin()ACBCBC=+−···························································4分由BCA+=−,,,02ACB知sin()sin0B

CA+=,22BC−−·········5分∴sinsin()CBC=−··············································································6

分∴CBC=−,故2BC=········································································7分(2)方法一由正弦定理及2a=得,sin2sinsin2sinsinsinsinsincaBBaCCbAAAA=

===∴cossincossin12sin2sinCACAbcBC+=+·····························································8分由(1)知2BC=,故sin2sincosBCC=·········

············································9分∴cossinsin3sin14sin2sin4sinCAAAbcCCC+=+=··················································10分由22sin

sin()sincoscossin2sincos(12sin)sinABCBCBCCCCC=+=+=+−····11分∴22sin2cos12sin12cos2sinACCCC=+−=+·························

·····················12分由,,02ACB,2BC=,ABC++=得64C,故232C·········13分∴10cos22C,故112cos22C+,即sin12sinAC·

····························14分∴,cos3sin331()4sin42CAbcC+=,即cos1Cbc+的取值范围为,33()42.············15分方法二由2cosaccB−=及2a=得12cos12cos12BBca++==①·

································8分由正弦定理得sinsinBCbc=········································································9分代入2BC=得sin2sinCCbc=,化简得c

os12Cbc=②·····································11分由①②得cos331cos42CBbc+=+·········································

·······················12分由,,02ACB,2BC=,ABC++=得32B·······························13分{#{QQABaYSA

gggAAoAAARgCUwVqCEMQkACCASgGBFAAIAABARNABAA=}#}数学答案与评分标准第4页(共7页)∴10cos2B,故,3333cos()4242B+··················································14分∴co

s1Cbc+的取值范围为,33()42.························································15分方法三由(1)知sin2sincosBCC=······································

·······························8分由2a=及正弦定理得sin2sin4sincossinsinsinaBBCCbAAA===·································10分∴cossin14sin42

CAabCcc===,故cos312Cbcc+=·········································11分如右图,1290AABC==由2BC=,且2a=得12,12ABAB==·

······················································12分由于ABC△为锐角三角形且2BC=故12ABcAB,即12c····························

·······································13分∴333422c,即cos1Cbc+的取值范围为,33()42.···································15分18.(本小题满分17分,第(1)问5分,第(2①)5分,第(2②

)7分)已知动圆Q经过点(1,0)F且与直线1x=−相切,记圆心Q的轨迹为曲线C.(1)求曲线C的方程;(2)设过点F且斜率为正的直线l交曲线C于A,B两点(点A在点B的上方),AB的中点为M,①过M,B作直线1x=−的垂线,垂足分别

为11,MB,证明:11AMFB∥;②设线段AB的垂直平分线交x轴于点P,若FPM△的面积为4,求直线l的方程.解:(1)设,()Qxy,由Q到直线1x=−的距离为|1|x+···················

···························1分由动圆Q经过点(1,0)F且与直线1x=−相切得|||1|QFx=+·····························3分即22(1)|1|xyx−+=+

············································································4分化简得24yx=∴曲线C的方程为24yx=·········································

····························5分(2)设直线l的方程为1xmy=+,其中0m,112200,,,,,()()()AxyBxyMxy由21,4xmyyx=+=消去x得2440ymy−−=,显然216(1)0m=

+△恒成立∴12022yyym+==,124yy=−································································································

6分①方法一由题意得1(1,2)Mm−,12(1,)By−∵11112(1,2),(2,)MAxymBFy=+−=−,2114yx=·································7分∴2112112

2122(2)(1)24404yymyxymyyyym−++=−++=+−=················9分∴11AMFB∥,故11AMFB∥······························································10分②由02ym

=得200121xmym=+=+设(,0)Pt,由MPAB⊥得1FMPMkk=−,故2221PMmkmmt=−=+−·············11分解得223tm=+,故2||22FPm=+·················

········································12分∴FPM△的面积为201||||(22)2FPymm=+··········································13分由FPM△的面积为4得2(2

2)4mm+=,即2(1)2mm+=···························14分设3()2fmmm=+−,则2()310fmm=+,()fm在,(0)+上单调递增·····15分又(1)0f=

,故()fm在,(0)+内有唯一零点1m=1ABC2A{#{QQABaYSAgggAAoAAARgCUwVqCEMQkACCASgGBFAAIAABARNABAA=}#}数学答案与评分标准第5页(共7页)即方程2(1)2

mm+=有唯一解1m=·························································16分∴直线l的方程为10xy−−=.··············································

···········17分由(1)知,曲线C是以F为焦点,1x=−为准线的抛物线①方法二设点A在准线上的射影为1A,则11||||,||||AAAFBBBF==···························6分由M为AB的中点且111MMMB⊥得1||||||2ABMMAM==·····

························7分又111BBMB⊥,故11MMBB∥(如右图)···················································8

分∴等腰三角形11BBFMMA∽△△∴11BFBMAM=···········································································9分∴11AMFB∥··························

························································10分方法三设22(,2),(,2)(0)AssBttts,则22,2MMstystx+=+=,11(1,2),(1,)BtMst−−+·········

································6分由A,F,B共线得FBFBkk=∴222211tsts=−−,化简得(1)()0stts+−=,故1st=−································7分∴12222222()111AMstsstsst

ktssssst−+−+=====−+++········································8分又1021(1)FBtkt−==−−−,故11FBAMkk=······················································

···9分∴11AMFB∥··················································································10分19.(本小题满分17分,第(1)问4分,第(2)6分,第(3)7分)已知函数2()ln

(1)fxxxax=−−.(1)若曲线()yfx=在点(1,0)处的切线方程为10xy+−=,求实数a的值;(2)当1x时()0fx恒成立,求实数a的取值范围;(3)证明:222(*)ln1nkkknn=−−N.

解:(1)()1ln2fxxax=+−,故(1)12fa=−····························································1分∵曲线()yfx=在点(1,0)处的切线方程为10xy+−=又直线10xy+−=的斜率为1−·····

·······························································2分∴(1)1f=−,故112a−=−····························

··········································3分∴1a=·····························································································

···4分(2)由1x时()0fx即2ln(1)0(1)xxaxx−−等价于1()ln0(1)axxxx−−···································································5分令

1()()ln(1)gxaxxxx=−−,则2221()(1)(1)1axxagxaxxxx−+=+−=方法一令2()(1)hxaxxax=−+若0a≤,则当1x时()0hx恒成立,此时()0gx,(

)gx是减函数·················6分∴当1x时,()(1)0gxg=,不合题意·····················································7分若0a,214

a=−△当0△,即2140a−时,021aAMFB1B1MxyO1A{#{QQABaYSAgggAAoAAARgCUwVqCEMQkACCASgGBFAAIAABARNABAA=}#}数学答案与评分标准第6页(共7页)由()0hx=且121xx=知

21114(0,1)2axa−−=,2211412axa+−=···················8分∴当21xx时,()0hx,故()0gx,()gx单调递减此时()(1)0gxg=不合题意··················

······················································9分当0≤△,即21a≥时,()0hx≥,()0gx≥,()gx单调递增此时()(1)0gx

g=综上可知,a的取值范围是1[,)2+···························································10分方法二∵(1)0g=,且()0(1)gxx∴(1)210ga=−≥,解

得12a≥································································7分下证当12a≥,且1x时,恒有()0fx由1x可知,10xx−∴当12a≥

时,111()()ln()ln(1)2gxaxxxxxxx=−−−−≥····························8分设11()()ln(1)2hxxxxx=−−,则22(1)()02xhxx−=

,故()hx在,(1)+上单调递增∴()(1)0hxh=,故1x时()0fx恒成立·················································9分综上可知,a的取值范围是1[,)2+·····

······················································10分方法三由1x知210x−,2ln()0(1)1xxfxaxx−≥···········································6分令

2ln()(1)1xxFxxx=−,则22221lnln()(1)(1)xxxxFxxx−−−=−令22()1lnlnGxxxxx=−−−,则1()2lnGxxxxx=−−令1()2lnHxxxxx=−−,则当1x时21()12ln0Hx

xx=−−,()Hx单调递减·····7分∴()(1())0HxxHG==,故()Gx在,(1)+上单调递减∴()(1)0GxG=,故()0Fx,()Fx在,(1)+上单调递减··························8分由导数的

定义知111lnlnln1limlim(ln)|111xxxxxxxx=→→−===−−∴2111lnln1limlimlim1121xxxxxxxxxx→→→==+−−·················································

··········9分∴12a≥,即a的取值范围是1[,)2+·····················································10分(3)方法一由(2)知,当1x时,恒有21ln(1)2xxx−··························

·····················11分令(1)xtt=得1ln(1)2ttt−,等价于1ln(1)tttt−······················12分∴当2k≥时,恒有11lnkkk

kk−−=,变形得ln11kkk−·······················13分又当2k≥时,1222(1)21kkkkkk==−−+−··································14分∴当2k≥时,22ln2(1

)1nnkkkkkk==−−−················································15分{#{QQABaYSAgggAAoAAARgCUwVqCEMQkACCASgGBFAAIAABARNABAA=}#}数

学答案与评分标准第7页(共7页)∵22(1)2[(21)(32)(1)]nkkknn=−−=−+−++−−2(1)22nn=−=−···························································

···················16分∴222(*)ln1nkkknn=−−N.······························································17分方法二设2nTn=是数列{}na的前n项和,则111,1,,1,

222221,221,nnnnnTnnTannTnn−=====−−−−−≥≥·······························11分12322nnnTaaaa−=−=+++···············································

···············12分故欲证222(*)ln1nkkknn=−−N只需证)ln222111(2kkkkakkkk=+−−−−=≥····································

13分由2k≥知,只需证2()ln12kkkk+−−················································14分由2k≥知,1()2kkk+−,故只需证1lnkkk−··················

·············15分由(2)知,当12a=,1x时,恒有21ln(1)2xxx−····································16分取,2kxk=≥得1ln(1)2kkk−,故1lnkkk−成立∴原不等式成立.··

··············································································17分方法三由21ln(1)2xxx−得12ln22,故2ln2

2222−·····························11分∴当2n=时,原不等式成立假设当nk=,且2n≥时原不等式成立即ln3lnln4ln222231kkk++++−−成立······································

············12分那么1nk=+时,ln(1)ln(1)ln3lnln4ln222231kkkkkkk+++++++−+−由21ln(1)2xxx−得1ln12kkk++,故ln(1)11kkk++··························13分∴2(1)1

ln(1)12222211kkkkkkkk+++−+−+=−++····························14分由重要不等式知2(1)(1)21kkkkk+++=+∴2(1)12(1)2111kkkk

kk+++=+++··························································15分∴ln(1)ln3lnln4ln2212231kkkkk++++++

+−−即1nk=+时,原不等式也成立·································································16分综上可知,对一切不小于2的正整数n,都有222ln1nkkkn=−

−成立.···············17分{#{QQABaYSAgggAAoAAARgCUwVqCEMQkACCASgGBFAAIAABARNABAA=}#}

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