【文档说明】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）.pdf，共(11)页，822.795 KB，由小赞的店铺上传

转载请保留链接：https://www.doc5u.com/view-8bd2d76992cb88b41c3b87e48ff3327d.html

以下为本文档部分文字说明：

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．22．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{|,,}1Pxyxy

xN，{|14}Qxx≤≤，则PQ（）A．{1,2,4}B．{0,1,3}C．{|03}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{#{QQABaYSAgggAAoAAARgCUwVqCEMQkACCASgGBFAAIAABARNA

BAA=}#}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．0.4m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{#{QQABaYSAgggAAoAAARgCUwVqCEMQkACCASgGBFAAIAABARNABAA=}#}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．ABC1A1B1CD{#{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．{#{QQABaYSAgggAAoAAARgCUwVqCEMQkACCASgGBFAAIAABARNABAA=}#}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．ABD；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．解：（1）由112a=且132nnnaaa+=+知0na且1

121133nnaa+=+········································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∴

12321()31111233()2313nnnnnaaaa−++++=+=+−−······································8分∴123111150naaaa++++等价于233()503nn+−

即23()473nn−······················································································9分

令2()3()3nfnn=−，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1

BB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===∴1tan2tanADBABA==，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122AAAB==，1

2BD=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)2ADAC=−=−············

················································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1ACD与平面ABC所成二面角的正弦值为33．······························1

5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）ABC1A1B1CDABC1A1B1CDxyABC1A1B1CDz{#{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sinsin2sincosACC

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

C=−=−···············································4分由,02CB得22BC−−·······························································5分∴CB

C=−，故2BC=········································································7分方法二由2cosaccB−=及余弦定理得22222acbaccac+−−=，化简得22bc

ac−=··········1分由上式及正弦定理得22sinsinsinsinBCAC−=··············································2分∵221cos21cos2cos2cos2sinsin222BCCBBC−−−−=−=···········

···················3分又cos2cos[()()]cos()cos()sin()sin()CBCBCBCBCBCBC=+−−=+−++−cos2cos[()()]cos()cos()sin()sin()BBCBCBCBCBCBC=

++−=+−−+−∴22sinsinsin()sin()BCBCBC−=+−∴sinsinsin()sin()ACBCBC=+−···························································4分由BCA+=−，,,02ACB

知sin()sin0BCA+=，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22sinsin()sincoscossin2sincos(12sin)sinABCBC

BCCCCC=+=+=+−····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12cos1

2cos12BBca++==①·································8分由正弦定理得sinsinBCbc=·····································

···································9分代入2BC=得sin2sinCCbc=，化简得cos12Cbc=②·····································11分由①②得cos331cos42CBbc+=+··

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

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

···········14分∴cos1Cbc+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∴cossin14sin42CAabC

cc===，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cos

1Cbc+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,)M

AxymBFy=+−=−，2114yx=·································7分∴21121122122(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(22)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)2mm+=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111MMM

B⊥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()111AMstsstsstktssssst−+−+=====−+++·······························

·········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)fxx

xax=−−．（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，214a=−△当0△，即214

0a−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−−=，221

1412axa+−=···················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)0gxg=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()()l

n(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()1

2ln0Hxxx=−−，()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11lnkkkkk−−=，变形得ln11kkk−·········

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

kkkk==−−−················································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2nT

n=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(2kkkkak

kkk=+−−−−=≥····································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2ln22222−·····························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)1ln(1)12222211kkkkkkkk+++−+−+=−++····························14分由重要不等式知2(1)

(1)21kkkkk+++=+∴2(1)12(1)2111kkkkkk+++=+++··························································15分∴ln(1)ln3lnln4ln2

212231kkkkk+++++++−−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=}#}