【文档说明】中学生标准学术能力诊断性测试2024-2025学年高三上学期10月月考试题 数学 PDF版含答案(可编辑).pdf,共(8)页,705.096 KB,由管理员店铺上传
转载请保留链接:https://www.doc5u.com/view-55c4218dfab3bc5de44977e642e4b05d.html
以下为本文档部分文字说明:
第1页共4页第2页共4页标准学术能力诊断性测试2024年10月测试数学试卷本试卷共150分一、单项选择题:本题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.已知集合==−−AxBx4|24,2,1,0,1,21
,则AB=A.−1,0,1C.0,1B.−−2,1,0,1,2D.−1,12.若−=+zz1i1,则=zA.2B.22C.1D.213.已知单位向量a和b,若()2aab⊥+,则ab+=A.2B.1C.2D.34.已知圆柱的底面半径和球的半径相等
,圆柱的高与球的半径相等,则圆柱与球的表面积之比为A.1:2B.1:1C.3:4D.2:35.已知+==3tansin,21tan)(,则−=sin)(A.−31B.−91C.31D.916.已知函数−
=fxxfxxx21,112,01)()(,则函数=−xgxfx2)()(的零点个数为A.2B.0C.3D.无穷7.将=yxsin的图象变换为=−yx6sin3的图象,下列变换正确的是A.将图象上点的横坐标变为原来的31倍,再将图象向
右平移6个单位B.将图象上点的横坐标变为原来的3倍,再将图象向右平移18个单位C.将图象向右平移6个单位,再将图象上点的横坐标变为原来的31倍D.将图象向右平移6个单位,再将图象上点的横坐标变为原来的3倍8.定义在R上的函
数fx)(满足:−+−−−=fxfx110)()(,且++−=fxfx110)()(,当−x1,1时,=−fxax2)(,则fx)(的最小值为A.−6B.−4C.−3D.−2二、多项选择题:本题共3小题,每小题6分,共18分.在每小题给
出的四个选项中,有多项符合题目要求.全部选对得6分,部分选对但不全得3分,有错选的得0分.9.从1,2,3中随机取一个数记为a,从4,5,6中随机取一个数记为b,则下列说法正确的是A.事件“+ab为偶数”的概率为94B.事件“ab为偶数”的概率为97C
.设=+Xab,则X的数学期望为=EX6)(D.设=Yab,则在Y的所有可能的取值中最有可能取到的值是1210.在直棱柱−ABCDABCD1111中,底面ABCD为正方形,==CDCC331,P为线段BC1上动点,EF,分别为AD11和BC的中点,则下列说法正确的是A.若0CPCB
=311,则经过PEF,,三点的直棱柱的截面为四边形B.直线BC1与AC11所成角的余弦值为46C.三棱锥−PADC11的体积为定值D.+APBP1的最小值为711.一条动直线l1与圆+=xy122相切
,并与圆+=xy2522相交于点AB,,点P为定直线+−=lxy:1002上动点,则下列说法正确的是A.存在直线l1,使得以AB为直径的圆与l2相切{#{QQABaQyQogCAAhAAAQhCQQniCgMQkhCCCYgOREAIoA
AASBFABAA=}#}第3页共4页第4页共4页B.+PAPB22的最小值为−150202C.APPB的最大值为−+27102D.+PAPB的最小值为83三、填空题:本题共3小题,每小题5分,共15分.12.若−xxxm1的展开式中存
在x2项,则由满足条件的所有正整数m从小到大排列构成的数列an的通项公式为.13.设双曲线−=abCabxy:10,02222)(的右顶点为F,且F是抛物线=yx:42的焦点.过点F的直线l与抛物线交于AB,两点,满足2AFFB=,若点A也在双曲线C上,则双曲线C的离心率
为.14.已知=−−+−xfxaxalnln21)(,则fx)(的最小值为.四、解答题:本题共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.(13分)记ABC的内角ABC,,的对边分别是abc,,,满足++=abc2321222)
(.(1)若==bcA4,cos3,求ABC的面积;(2)记BC边的中点为D,=ADx,若A为钝角,求x的取值范围.16.(15分)如图所示,在四棱锥−PABCD中,===PAACBC2,1,=AB3.(1)若⊥AD平面PA
B,证明:AD平面PBC;(2)若⊥PA底面ABCD,⊥ADCD,二面角−−ACPD的正弦值为36,求AD的长.17.(15分)已知椭圆+=abCabxy:102222)(,C的下顶点为B,左、右焦点分别为F1和F2,离心率为21,过F2的直线l与椭圆C相交于DE,两点.若直线l垂直于
BF1,则BDE的周长为8.(1)求椭圆C的方程;(2)若直线l与坐标轴不垂直,点E关于x轴的对称点为G,试判断直线DG是否过定点,并说明理由.18.(17分)已知函数=+fxaxxxsin,0,)(.(1)若=
−a1,证明:fx0)(;(2)若fx0)(,求a的取值范围;(3)若a0,记=−+agxfxxln11)()()(,讨论函数gx)(的零点个数.19.(17分)乒乓球比赛有两种赛制,其中就有“5局3胜制”和“7局4胜制”
,“5局3胜制”指5局中胜3局的一方取得胜利,“7局4胜制”指7局中胜4局的一方取得胜利.(1)甲、乙两人进行乒乓球比赛,若采用5局3胜制,比赛结束算一场比赛,甲获胜的概率为0.8;若采用7局4胜制,比赛结束算一场比赛,甲获胜的概率为0.9.
已知甲、乙两人共进行了Nmm)(场比赛,请根据小概率值=0.010的K2独立性检验,来推断赛制是否对甲获胜的场数有影响.(2)若甲、乙两人采用5局3胜制比赛,设甲每局比赛的胜率均为p,没有平局.记事件“甲只要取得3局比赛的胜利比赛结束且甲获胜”为A,事件“两人赛满
5局,甲至少取得3局比赛胜利且甲获胜”为B,试证明:=PAPB)()(.(3)甲、乙两人进行乒乓球比赛,每局比赛甲的胜率都是pp0.5)(,没有平局.若采用“赛满−n21局,胜方至少取得n局胜利”的赛制,甲获胜的概率记为Pn)(.若采用“赛满+n21局,胜方至
少取得+n1局胜利”的赛制,甲获胜的概率记为+Pn1)(,试比较Pn)(与+Pn1)(的大小.附:()()()()++++=−abcdacbdKnadbc22)(,其中=+++nabcd.PKk02)(0.050.0250.010k03.8415.0
246.635DCBAP(第16题图){#{QQABaQyQogCAAhAAAQhCQQniCgMQkhCCCYgOREAIoAAASBFABAA=}#}第1页共6页标准学术能力诊断性测试2024年10月测试数学参考答案一、单项选择题:本题共8
小题,每小题5分,共40分.在每小题给出的四个选项中,只有一项是符合题目要求的.12345678ACBBDACB二、多项选择题:本题共3小题,每小题6分,共18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6分,部分选对但不全的得3
分,有错选的得0分.91011ABDBCDBCD三、填空题:本题共3小题,每小题5分,共15分.12.=ann413.33314.2四、解答题:本题共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.(13分)解:(1)由余弦定理知:+=+bcbcA5214cos22)(,又==bc
A4,cos3,代入等式中可得:=+bcbc10213,即得=bc3,所以==bc3··························································
·············4分所以ABC的面积为==bcA2248sin13737·············································5分(2)因为D为线段BC的中点,所以()1ADABAC=+2,两边平方得
:=++xbcbcA42cos1222)(,由余弦定理可得:=+−bcAbca2cos222,代入上式得:=+−xbca42212222)(,再由++=abc2321222)(,可得=−ax761222,+=+bcx738222·············
·····10分因为A为钝角,所以+abc222,可得−+xx776312822,解得x100105.{#{QQABaQyQogCAAhAAAQhCQQniCgMQkhCCCYgOREAIoAAASBFABAA=}#}第2页共6页所以,x的取值范围为
xx100105·····················································13分16.(15分)解:(1)因为⊥AD平面PAB,AB平面PAB,所以⊥AD
AB,由===ACBCAB2,1,3,可得=+ACABBC222,所以⊥BCAB,所以在平面四边形ABCD中,由⊥⊥ADABBCAB,,可得ADBC,因为AD平面PBC,BC平面PBC,所以AD平面PBC·····················
·····························································6分(2)【方法一】因为⊥PA底面ABCD,CD底面ABCD,所以⊥PACD,因为ADCDPAADA⊥=,,所以⊥CD平面PAD,可得⊥
CDPD,即=PDC90.以直线DA为x轴,直线DC为y轴,过点D且垂直于平面ABCD的直线为z轴,建立空间直角坐标系,如图所示:························································8分设==ADaDCb
,,则DAaCbPa0,0,0,,0,0,0,,0,,0,2)()()()(,在坐标平面xDz中,直线DP的法向量就是平面PDC的法向量,可得其中一个法向量为(2,0,na=−1).设平面PAC的一个法向
量为(,,nxyz=2),则0nAPnCP==22,而()(0,0,2,,,2APCPab==−),可得=−=zaxby0,0.令=xb,则=ya,得(,,0nba=2)·····································
······················12分所以cos,nn=++−aabb4222212,依题可知,cos,nn=3312,可得()()++=abab43412222,因为+==abAC4222,所以−=bb83122,解得=b22,则=a22,得=AD2············
································································15分zyxDPCBA{#{QQABaQyQogCAAhAAAQhCQQniCgMQkhCCCYgOREAIoAAASBFABAA=}#}第3页共
6页【方法二】设点A到平面PCD的距离为d1,点A到直线PC的距离为d2,二面角−−ACPD的平面角为,则由二面角的平面角定义知=ddsin21.由题意计算可得=d22,所以=d2361,可得=d3231.由等体积公式可得=SPASdACDPCD33111,即=AD
CDPDCD33,得=PDAD3.因为=+=−PCPDCDCDACAD,222222,所以=+−ADAD83422,得=AD2.17.(15分)解:(1)由离心率为21,==BFaOFc,11,可得=BFOF2111,则=BFO601,可得BFF
12是正三角形,如图所示:若直线l垂直BF1,则直线l垂直平分线段BF1,可知BDE与FDE1全等,那么FDE1的周长为8.由椭圆定义可知:+=+=EFEFaDFDFa2,21212,所以FDE1的周长为a4,可得=a48,即=a2
.所以=c1,可得=b3,则椭圆C的方程为+=xy43122······································································6分(2)设l的方程为=+xmy1,DxyExy,
,,1122)()(,则−Gxy,22)(,如图所示:可得直线DG的方程为−−=−+xxyyxxyy121112)(,因为=+=+xmyxmy1,11122,将它们代入直线方程中,可得直线DG的方程为:()−=−−++myyyxmyyyy1121112)(,可整理得:()
−=+−−+myyyyyxmyyyy212121212)()((*)····································10分联立方程=++=xmyxy143122,得:++−=mymy3469022)
(,yxO2F1FBDEyxO2F1FDEG{#{QQABaQyQogCAAhAAAQhCQQniCgMQkhCCCYgOREAIoAAASBFABAA=}#}第4页共6页则+++=−=−mmyyyym3434,69221212,可得=+yymyy3
21212,=+myyyy231212)(,将其代入(*)式中,可得直线DG的方程为:()−=+−+myyyyyxyy4121212)()(()()+−=−−myyx3446122)(,可见直线DG过定点4,0)(,所以直线DG过定点,定
点坐标为4,0)(·······················································15分18.(17分)解:(1)若=−a1,则=−+fxxxsin)(,得=−+fxx1cos0)(,可知fx)(在0,单调递减
,可得fxf0)()(,而=f00)(,所以fx0)(········································································3分(2)依题意,必须f0)(,即a0,可得a0,求导得=+fx
axcos)(.若−a1,则fx0)(,得fx)(在0,单调递减,则fxf0)()(,而=f00)(,则fx0)(成立············································5分若−a10,
由于fx)(在0,单调递减,而=+fa010)(,=−fa10)(,可知fx)(在0,内有唯一零点,记为x1,当xx01时,fx0)(,可知fx)(在x0,1)单调递增,可得=fxf001)()(,这与fx0)(对任意
x0,恒成立矛盾,所以−a10不能成立,综上,实数a的取值范围为−−,1(······························································8分(3)有=+−+
agxxxxxsinln1,0,1)()(,观察知:=g00)(,可见=x0是gx)(的一个零点.下面我们考虑gx)(在0,(内的零点情况·······················································9分当x0,(时,若a0,则
axsin01,可得+axxxsin1,令=−+Fxxxxln1,0,()()(,则+=xFxx10)(,得Fx)(在0,(单调递增,可得=FxF00)()(,即+xxln1)(,那么++axxxsinln11)(,
即gx0)(,{#{QQABaQyQogCAAhAAAQhCQQniCgMQkhCCCYgOREAIoAAASBFABAA=}#}第5页共6页故当a0时,函数gx)(在0,(内无零点·····································
··············12分若a0,则+=+−axgxx11cos11)(,①当x2,时,xcos0,则axcos01,而+−x1101,可得gx0)(;②当x20,时,()+=−+xagxx1si
n0112)(,可得gx)(在20,单调递增,因为+==−agg2200,1012)(,所以gx)(在20,内有唯一零点,记为x2,当xx02时,gx0)(
;当xx22时,gx0)(,综合①②,gx)(在x0,2)(单调递减,在x,2(单调递增.因为=g00)(,所以gx02)(,又由+xxln1)(可得=−+gln10)()(,所
以gx)(在0,(内恰有1个零点.综上所述,当a0时,gx)(有1个零点;当a0时,gx)(有2个零点··········17分19.(17分)解:(1)据题中条件,列出赛制和甲获胜情况列联表如下:甲获胜场数乙获
胜场数合计5局3胜m0.8m0.2m7局4胜m0.9m0.1m合计m1.7m0.3m2由计算公式得:==−mmmmKmmmm1.70.351220.080.182222)(,若m516.6352,即
m169.1925,故若m170时,根据小概率值=0.010的K2独立性检验,推断赛制对甲获胜的场数有影响,此推断犯错误的概率小于0.010.若m170,根据小概率值=0.010的K2独立性检验,没有证据认为赛制对甲获胜的场数有影响,此时赛制对甲获胜的场数没有影响·
·················································4分(2)依题意=+−+−PAppCpppCpp1134322222)()()(=+−+−+=−+ppppppppp31612615103332543)()(,{#{Q
QABaQyQogCAAhAAAQhCQQniCgMQkhCCCYgOREAIoAAASBFABAA=}#}第6页共6页又有=−+−+−PBCppCppCpp11155533445520)()()()(=−+−+ppppp101513452)()(=−++−+pppppp
10201055543455=−+ppp61510543所以=PAPB)()(··········································································7分(3)考虑赛满+n21局的情况,以赛完
−n21局为第一阶段,第二阶段为最后2局.设“赛满+n21局甲获胜”为事件C,结合第一阶段的结果,要使事件C发生,有两种情况:第一阶段甲获胜,记为A1;第一阶段乙获胜,且甲恰好胜了−n1局,记为A2,则=+CACAC12,得:=
+PCPACPAC12)()()(.若第一阶段甲获胜,即赛满−n21局甲至少胜n局,有两类情况:甲至少胜+n1局和甲恰好胜n局.第一类情况,无论第二阶段的2局结果如何,最终甲获胜;第二类情况,有可能甲
不能获胜,这种情况是第二阶段的2局比赛甲均失败,其概率值为:−−−−Cpppnnnn112112)()(,所以=−−−−−PACPnCpppnnnn1112112)()()()(.若第一阶段乙获胜,且甲恰好胜了−n1局,那么要使甲最终获胜,第二阶段的2局
比赛甲必须全部取胜,可得:==−−−−PACPAPCACpppnnnn122221112)()()()(,所以+==−−−+−−−−−−PnPCPnCpppCpppnnnnnnnn1111212111212)()()()()()(········
··············································14分可得+−=−−−−−−−−−PnPnCpppCpppnnnnnnnn1111212111212)()()()()(=−−−−−++CppCppnnnnnnnn11212111
)()(=−−−−Cppppnnnn1121)()()(=−−−Cpppnnnn221121)(因为p21,所以−−−Cpppnnnn2210121)(,可得+PnPn1)()(,综上:+
PnPn1)()(··································································17分{#{QQABaQyQogCAAhAAAQhCQQniCgMQk
hCCCYgOREAIoAAASBFABAA=}#}