115225中學(xué)生標(biāo)準(zhǔn)學(xué)術(shù)能力診斷性測(cè)試2023年3月測(cè)試數(shù)學(xué)試卷A．B．C．D．10157．在矩形中，已知24，==E是的中點(diǎn)，將1C，連接．當(dāng)二面角沿直線翻折成本試卷共150分，考試時(shí)間120分鐘。1??C的平面角的大小為1?外接85分．在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是球的表面積為符合題目要求的．A．B．?x21123（第7題圖）?4x+3，A=xx2B=yy=1．已知集合A，則C．D．3)()+)D．+)2,3C．A．B．12，a1A=x2x2logaxB=xy=x+?x，AB8．已知a0且3+z1+izz是實(shí)數(shù)，則的虛部為2．設(shè)是純虛數(shù)，若a則實(shí)數(shù)的取值范圍是A．3?B．1?C．1D．3111114()=3sinx+)?cosx+))fx()fx3．已知函數(shù)A．B．4“=?”的3111,1,1C．D．A．充分不必要條件B．必要不充分條件D．既不充分也不必要條件44C．充要條件45分．在每小題給出的選項(xiàng)中，有多項(xiàng)符合題目要求，全部選對(duì)的得5分，部分選對(duì)的得2分，有選錯(cuò)的得0分．(?)4．若圓xa2y3=+(?)20上有四個(gè)點(diǎn)到直線2x?y+1=0的距離為5a的取值范2ab0，滿足a+b=1，下列說(shuō)法正確的是9．設(shè)圍是121A．a(chǎn)b的最大值為B．+的最小值為83131722?,??,??,A．B．24ab22113nC．a(chǎn)2+b2的最小值為D．a(chǎn)2+b的最小值為123737?,C．D．2222aSa+a+a=S=2510．已知等差數(shù)列的前項(xiàng)和為，滿足，5，下列說(shuō)法正確的是nn123?++nn1n5．若7n+C1n17n1C是9的倍數(shù)，則自然數(shù)為a=2n+3S=?B．nn2+10nA．C．nA4的倍數(shù)B．3的倍數(shù)C．奇數(shù)D．偶數(shù)110SS?的前10項(xiàng)和為的最大值為D．n5aa99nn16．現(xiàn)將0-9十個(gè)數(shù)字填入右方的金字塔中，要求每個(gè)數(shù)字都使用一的內(nèi)角,,C所對(duì)邊的長(zhǎng)分別為a,,c，已知b=c=6，的面積S滿足11．已知a(b+c2=(438Sa+)+2，點(diǎn)O為的外心，滿足AO=AB+AC，則下列結(jié)論正為bc字為d，則滿足abcd的填法的概率為確的是（第6題圖）21233（2）若二面角A??C的平面角的大小為，求直線與面所成角的正弦值．A．S=6B．=C．AO=D．=2?33x29y220．（12分）為提高學(xué)生的數(shù)學(xué)應(yīng)用能力和創(chuàng)造力，學(xué)校打算開(kāi)設(shè)“數(shù)學(xué)建?！边x修課，為了解學(xué)生對(duì)“數(shù)學(xué)建模”的興趣度是否與性別有關(guān)，學(xué)校隨機(jī)抽取該校名高中學(xué)生進(jìn)行問(wèn)卷調(diào)查，其中認(rèn)為感興趣的人數(shù)占70%．()()+12=?2，則下Px,y,Qx,y+=1上兩個(gè)不同點(diǎn)，且滿足xx91y212．已知是橢圓112244列說(shuō)法正確的是2x+3y?3+2x+3y?3A．B．的最大值為6+251122（12285%的把握認(rèn)為學(xué)生對(duì)“數(shù)學(xué)建模”選修課的興趣度與性別有關(guān)？2x+3y?3+2x+3y?3的最小值為3?511222105x?3y+5+x?3y+5C．D．的最大值為25+1122感興趣不感興趣合計(jì)x?3y+5+x?3y+51男生女生合計(jì)12的最小值為22?1225三、填空題：本題共4小題，每小題5分，共20分．30=8xN為圓x2+(y?4)=5M到y(tǒng)軸的213．已知點(diǎn)M為拋物線y2（243名進(jìn)行二次訪談，記選出高三女生的人數(shù)為X，求X的分布列與數(shù)學(xué)期望．距離與點(diǎn)M到點(diǎn)的距離之和最小值為N．()fx()()+)fx在上單調(diào)遞增，則不等式hxx214知為R上的偶函數(shù)，函數(shù)(?)nadbc2(?)2(?)?(+)2(+)1xf1x3xf3x0的解集為．=n=a+b+c+d．K2附：，其中(+)(+)(+)(+)abcdacbd154,5這六個(gè)數(shù)字組成無(wú)重復(fù)數(shù)字的六位數(shù)，要求任意兩個(gè)偶數(shù)數(shù)字之間至少有一個(gè)奇()PK2k00.150.100.050.0255.0240.0106.6350.0057.8790.001數(shù)數(shù)字，則符合要求的六位數(shù)的個(gè)數(shù)有(?)+kxx3x(+)kxex16．若關(guān)于的不等式對(duì)任意的的最大值為．k02.0722.7063.84110.828四、解答題：本題共6小題，共70分．解答應(yīng)寫出文字說(shuō)明、證明過(guò)程或演算步驟．21．（12分）已知雙曲線C以2x（1）求雙曲線C5y0為漸近線，其上焦點(diǎn)F坐標(biāo)為0,3=()4中，1=，(n+9)(n+)2an1=(n+2)an．3．a(chǎn)17．（10分）在數(shù)列n9的方程；（1）求的通項(xiàng)公式；an,Q兩點(diǎn)，y軸于點(diǎn)T（2l過(guò)F與雙曲線C交于的中垂線交52n+5?（2a的前n項(xiàng)和為Sn，證明：S．nn4nTF4是否為定值，若是，請(qǐng)求出定值，若不是，請(qǐng)說(shuō)明理由．,,CbA?a=c．PQ18．（12分）已知的內(nèi)角a,,c的對(duì)邊分別為，且x()=()22．（12分）設(shè)fxxR．（1）求角B；ex（212的角平分線交于點(diǎn)D，若=2的面積的最小值．()()fx的單調(diào)性，并求在fxx=（1（2處的切線方程；19．（12分）如圖所示，在三棱錐A中，滿足CD33，?==()()(+)x+)xfxkx1在上恒成立，求k的取值范圍．點(diǎn)M在CD上，且=，為邊長(zhǎng)為6的等邊三角形，E為的中點(diǎn)，F(xiàn)為的三等分點(diǎn)，且2=．；（1）求證：面（第19題圖）中學(xué)生標(biāo)準(zhǔn)學(xué)術(shù)能力診斷性測(cè)試2023年3月測(cè)試數(shù)學(xué)參考答案85項(xiàng)是符合題目要求的．12345678ADBDCCAB45符合題目要求．全部選對(duì)的得5分，部分選對(duì)但不全的得2分，有錯(cuò)選的得0分．91012ACBCDABDAD三、填空題：本題共4小題，每小題5分，共20分．13．5?215．14．(??),116．1四、解答題：本題共6小題，共70分．解答應(yīng)寫出文字說(shuō)明、證明過(guò)程或演算步驟．172(+)3（1）n1a=n2a(+)nn(+)(+)n2an3n3an1=(+)n22(+)n12(+)1(+)n1=3(+)n3an2an即·····································································2分(+)n222n1(+)13()n+2a12111=n2是首項(xiàng)為，公比為的等比數(shù)列又，所以數(shù)列(+)112(+)33n1(+)n2n1n2(+)n1(+)==，則a··················································5分從而(+)n123nnn23(+)2n+1n+1nn1(+)（2）=···························································6分n23nn23+nn2312313nSnT=++++323n31323nn=++設(shè)，則n322333兩式相減得：n1111?9323211n2=+n+1n=++?31323n+1313n11?n1211n+152n+5?=?=+1?················································9分363n1+62n1+52n+552n+5T=?S?·····················································10分從而，故nnn4n443418（1）由已知及正弦定理得：2sinBA?A=2sinC中，C=(A+B)=AB+AB又在·························2分2sinBA?A=2sinAB+2AB2sinAB=?A即1又A0，B=?········································································4分分20B，B=又，即角的大小為B··············································533123（2）acsinB=ac·····································································6分4BD是的角平分線，而S=S+S311ac=ABBDsin+BDsin422aca+c33=+=BD(ac)，·················································8分即ac44=，=(+)2ac2ac，ac4ac，即···············································10分13a=c=4=acsinB16=43當(dāng)且僅當(dāng)時(shí)取等號(hào)，則S24即的面積的最小值為43·······························································12分1912（1上取一點(diǎn)，使得NBN=NE，連接，NM1BN=BD=1=，=23，，61BNAF12，==2NEFE則···························································································2分，，面面面CMND15=，·······························································4分，面，面面，面面，面面··························································5分D，⊥A??C（2）AEC=············································6分所以二面角的平面角為3⊥面又面面⊥面ABDAEC，過(guò)點(diǎn)C作⊥，則⊥面面面3則CHCEsin==CE323362(233?32=32，CH=32=·······························8分236即C到面的距離為2553566CD，M的距離為=······························9到面分662431cosCDB==計(jì)算EM：33353==3，在中，DM22532+32?EM321512=EM=···············································分3532256534451=EM與面所成角的正弦值為·········································12分342（其他方法酌情給分）20（1）列聯(lián)表如下：感興趣不感興趣合計(jì)16男生129459女生合計(jì)142130··········································2分3012549(?)2K2=0.40822.0721614219所以沒(méi)有85%的把握認(rèn)為學(xué)生對(duì)“數(shù)學(xué)建?！边x修課的興趣度與性別有關(guān)············5分（2）由題意可知X的取值可能為，，23C35935(=0)====PX則···········································································6分············································································7分············································································8分C14521021(=)=PX193C2493C155(=2)==3)=PX14C34931(=PX················································································9分21故X的分布列為X01235102151P42142151021514()=+++=EX0123···············································12分421421321（1）因?yàn)殡p曲線C以2x5y=0為漸近線()2x+5y2x?5y=?5y=·························1分4x22設(shè)雙曲線方程為，即y2x2?=1，0，即：??54??=9，?=9，即=·····················································3分分54y2x2所以雙曲線C的方程為：?=1···························································445（2）設(shè)直線l:y=+3Px,y，()，()Qx,y21122?2=205y4xy=+3化簡(jiǎn)得：5k(+)532?4x2=20)2?4x2+30+25=0···························································6分30kx+x=?122?4kx,x，則此方程的兩根為12xx=12k2?4(?4)9k2?5k2900k2100PQ=1+k2?=101+k25k2?4)25k2?45k2?4)2(+)20k24k2+4=101+k2=··············································8分5k2?4)25k?42kPQ中點(diǎn)M坐標(biāo)為?,·······················································9分k2?4k2?4121k15k中垂線方程為：y+=?x+5k2?45k?42?2727x=0，y=T令，2?45k24?5k2715k5k22+15?4TF=3+==則····························································分5k2?415k5k22+15?4TF3=········································································12分(+)PQ20k5k242?422ex?ex1?x()fx==0（12···································································1分()e