全国100所名枚最所惠考棋拟示花格全国100所名枚最新高考停18(2分)记△ABC的内角A,B,C的对边分别为a,bc,且2bsin(C+看)=a+c如图，在三棱锥P-ABC中，△PAC是边长为4的正三角形，AB=2,∠ACB一20.(12分)(1)求B;(2)若=2，求△ABC的面积的最大值.点P在底面ABC上的投影为点M,0府捍2snBn(ct)=(n+snC(1)求证：面PAB⊥面PBC,2sinB (inc+os c)=sin(Btctsin(2)求二面角A-PC-B的余弦值，43B)门J3sinBsincesinct cosBsinc1正明AC=4A9-2,LACB=2又：SinC0)由赶浮几AB⊥BCi-万imQ-coB=J.2"=a'+C'-accaB=a'rc-ae又：卢P在所面A?C上白的书/为点MSin(B-=±≥20C-0C=0S7-0C≤4冫-PM1干面/ABC且B∈(0，工)当日仅30=c=2取“三之ARLPM19.(12分)7B=Sa=sacsinB-Fac叉：PMnBe=M已知数列(a,)的前n项和为S且S。一2a.十2=0，数列(6，}为等差数列，6=a,-AB⊥干面gPBQb5=b2+63.~面P/ABL干面PBC(1)求数列{a},{b,}的通项公式；M(0.01/A(-2y.0,0(2)数列{c.}是由数列{b.}的项删去数列(a}的项后按从小到天的顺序排列构成c(,0.0)P(0.0,+5新数列，求数列{cn}的前50项和T5o.(2)在Rt△ABC中.BC=2J乃月n-20nt2o-PB=25AC=(2J5-2.0)0.01-20mt1十2：0®)i0664.01=1287b6s=112SABC=4J2J)=4元可(-5045)@0号0时1-20nt1十20m=0广刊fGn1白0江玩力列:SAPBO=土IBc|IM法罪万(×yZ)bn伯前6T册1去51f0m六0nt1=20n=25*PM∫25×-2y=0y当n1日T,01=2的前6M.|PM=正广cn是12方项4乙=b1tb寸寸br6-(0t0对PM:45T(1J5,马)十061见/0n:2”=56x(2寸112)2×(1-2)-CMI=JIPC-IP=(8

4已阅向量0的来角为号b与00垂直，b1=2,则a1乐20.(12分15.某产品的质量检验过程依次为进货检验(1QC)、生产过程检验(PQC)、出货检验(OQC)某剧场三个环节.已知某产品1QC的单独通过率为；，I0C的单独通过率为子，规定上一类检验位：元数据得不通过则不进入下一类检验，未通过可修复后再检验一次（修复后无需从头检验，通过率不变且每类检验最多两次)，且各类检验间相互独立，则一件该产品能进入OQC环节的概率为16.已知函数f(x)与g(x)的定义域均为Rf(x+1)+g(x-2)=3,f(x-1)-g(-x)=1,且(I)(-1)=2,8(x-)为偶函数，则名)+8(k)]=(Ⅱ四、解答题：共70分.解答应写出文字说明，证明过程或演算步骤参考17.(10分)51在①bn=√an+√an+i;②6.=；③6，=24，这三个条件中任选一个补充在下面横线anQn+1参方上，并解答问题知数列1a,的前n项和S,三na。二)n2+了22距(I)证明：数列{an}是等差数列；（Ⅱ)若a1=2,设2,求数列{bn}的前n项和Tn18.(12分)已知△ABC的内角A,B,C的对边分别为a,b,c,且3b=c(3cosA+sinA).(I)求C;(I)若AB⊥AC,AC=3,角C的分线交AB于点D,点E满足D2=C⑦，求sin∠AEB.19.(12分)如图，在三棱柱ABC-A,B,C,中，四边形AAB,B为菱形，E为棱CC,的中点，△AB,C为等边三角形，（I)求证：AB1⊥B,C1;(Ⅱ)若AC⊥BC,AC=4,BC=3,求面AAB,B和面AB,E夹角的余弦值.4B公正，允2黑小京，小

礼教学爱好者七年级同步训练下册（华师大版）。SHUXUE AIHAO ZHE4.如图，矩形ABCD中，AB=6,第1次移专题三图形的旋转将矩形ABCD沿AB的方向向右移5例⑤如图，在等边三角形ABC中，D个单位，得到矩形A1BC1D1;第2次移是边AC上一点，连结BD,将△BCD逆时针将矩形A1B1C1D1沿A1B1的方向向右旋转60°，得到△BAE,连结ED.若BC=10,移5个单位，得到矩形A2B2C2D2;…;第nBD=9,求△AED的周长，次移将矩形Am-1Bn-1Cm-1Dn-1沿Am-1B,n-1的方向移5个单位，得到矩形A,B,C D(n>2).DD,CD:CD CrABA,BA。Bn(1)求AB1和AB2的长；(2)若ABn的长为56，求n.·168·

在(-∞，b)上单调递增，在(b,+∞)上单调递减，所以p(x)的极大值点为x=b,所以2a=b,所以随机变量X服从正态分布，X~N(2a,a),故B正确：对于C,因为f(a)=P(X3a),又a+3a=2×2a,所以P(X3a),即f(a)=g(2a),故C正确：对于D,因为2a)=PX<20=2,8a=P(X>2@=2所以f2a)+g(20）=2+f@=8a)+fa),故D正确，11.BCD解：存在量词命题的否定方法是先改变量词，然后否定结论，命题：“3x0,x-x-10”的否定是“x0,x2-x-1<0”,故选A错故B正确：cos25°-sin25°_cos10cosl0°sin80=2sin40°sin50°sin40cos402sin801sin 802C9+2C+2C+2C+…+2C=(1+2=3",故C正确：如图所示，由PA=AB=PB=AC=2N5,CP=2√6，得PA⊥AC,由D是PB的中点，PA=AB=PB=2√3，解得AD=3,又CD=V21,所以CA2+AD2=CD2,得CA⊥AD,人AD∩AP=A,AP,ADC面PAB所以AC⊥面PAB.设球心为O,点O到底面PAB的距离为d=}AC=有，由正弦定理得2BPA2W5△PAB的外接圆半径=2sin60°2x6=2,在三角形0AE2中，球0的半径0A=R=VP+P=N5+22=V万，所以三棱锥P-ABC的外接球O的体积为y=号xR-号V列-287a故D正确：312.BCD:【解析】对于A当a=b时，函数f(x)定义域为R关于原点对称f(-x)=ae+be=f(x),故函数f(x)为偶函数，当函数f(x)为偶函数时，f(x)-f(-x)=0故(a-b)e+(b-a)e=0(a-b)e2=(a-b),又因为f()定义域为R,所以e2r不为1，故a=b所以a=b是函数f(x)为偶函数的充要条件，故①错误.对于B当a+b=0时，函数f(x)定义域为R关于原点对称f(x)+f(-x)=(a+b)e+(a+b)ex=0,故函数f(x)为奇函数当函数f(x)为奇函数时f(x)+f(-x)=(a+b)e+(a+b)e=0,因为e>0,e>0故a+b=0.所以a+b=0是函数f(x)为奇函数的充要条件，故②正确.对于c'(x)=ae-be因为ab<0若a>0,b<0则f'(x)=ae-be>0恒成立，则f(x)为单调递增函数，

由正弦定理，有二50sin'A 5sinA4ccosB sin2C4sinCcosBsim(-2B）5sim7-2B）cos22B 5cos2Bsin(受+B4sin受+Bcoscos2B 4cos2B-(2c0s2B-1)25(2c0s2B-1)cos2B4cos2B-4c0s4B-4c0s2B+155cos'B4c0s2B-2=4c0s2B+4cos2B2≥2、/4c0s2B.91391314cos2B 2当且仅当4cos2B=,94B时等号成立，山由B为镜角，则cB=号所以当8=君时取最小值-212分21解四圆+a>b>0路a路为5即二另长轴长为4,0=2，c1,b三3，故椭圆的方程为+？=1，…(2)设M(x1,y1),N(x2,y2),[y=hx +m联立x2y2,得(3+42)x2+8mx+4(m2-3)=0,+兮则△=(8m)2-16(3+4k2)(m2-3)=48(4k2+3-m2)>0,-8m_4(m2-3)名，+=3+4…5分3+4k2’所以6oyko=-（,+m)(,+m）_x3+mk(x1+)+mX1X2X1X2X1X2_4(m2-3)-82m2+m(3+4)_3(m-4)=-1,4(m2-3)4(m2-3)..7m2=12k2+12,………….1MW1=V1+F1x1-1=V1+k45V42+3-m3+42:原点0到l的距离d=1m√1+k2IMNI.d2W3√4k2+3-m21ml-12/16k4+25k2+9∴.SAOMN=23+427V16k4+24k2+9宣城市高三数学参考答案第3页（共4页）

由(1)可知，AC⊥面ABB1A1,所以AC为面范围：（Ⅱ）若h(x)有2个极值点x1,,(x10,f(x)单调递增：当x∈(1，(2)根据直线与辆圆的位置关系→联立十o)时，f'(x)<0,f(x)单调递减.y=k1(x-1),所以f(x)在x=1处取得极大值，也是最大值·求解直线过定点→进而求得MG与即f(x)x=f(1)=一1.+y2-1所以当a=1时，函数f(x)的最大值为一1.(4分)N心共线.(2(I)h(x)=lr+受r一ar,则h(x)的定义域解：(1)由椭圆的定义可知，|QF,|+|QF,|=2a=4,解得a=2因为离心率e=子-受所以(=5，由0为(0，十o∞).h'(x)=r2-4x+若h(x)在定义域上单调递增，则h'(x)≥0在(0，+o∞)b2十c2可得b3=1.上恒成立，即ax2一a.r十1≥0在(0，+∞)上恒成立，所以椭圆C的方程为号+y=1当a=0时，1≥0恒成立，满足题意.(4分)当a<0时，二次函数y=ax2一ar+1的图象开口向下，(2)证明：由题意知11，l:的斜率必存在.设1的斜率为不满足a.x2一a.x十1≥0，赦不成立，k1,l2的斜率为k2,A(x1,y1),B(x2y2.当a>0时，二次函数y=ax2一ax十1图象的对称轴为y=k(x-1),则11的方程为y=k1(x一1)，联立x=之，且当x=0时，y=1,要满足ar2-ar+1≥0x消去在(0，十o∞)恒成立，只需满足△≤0，即（一a)2一4a≤0，解y可得(1十4k7)x3-8k1x十4k一4=0.得0a≤4.8k1显然△>0恒成立，则x1十工=十妆1x:4k7-4综上述，实数a的取值范围是[0,4]，(7分)1+4k7(I)证明：因为(x)=二士，若(x)有2个极4k一k1所以w=1十hyw=,红M一)1+依(6分)值点x1,x:(x10,kx1+x2=1,同理可得xy11+4k9k+4则a>4,又x1x2=>0,x1

绝密★启用前2023高考考前冲刺押题卷（六）·数学[满分150分·用时120分仲]1答题前，考生务必特自已的学校、灶名、班级、准考证号填写在答题卡上加应的位置注意事项：2全部答案在答题卡上完成，答在本试卷上无效。点回答选排题时，选出每小题答案后，用2B铅笔把答题十上对应题1的答案标号涂不。如劣改动，用粮度擦千净后，再选涂其他答案标号。回答非选择道时，将谷案川0.5飞米及以上黑色笔迹签字笔写在答题卡上。4考试结来后，将本试卷和答题卡一并交回参考公式：锥体的体积公式：V=号h(其中：为锥休的底面积，h为维休的高)一、单项选择题（本题共8小题，每小题5分共0分在每小题给出的四个选项中，只有一项是符合题日要求的）1.已知集合A=(xx2-3.x<0),B=(1,2,3,1,5),则(CRA)∩B=(A.D.{2.3,4,5》A.1.2B.{4.5)C.(3.4,52已知复数：满足：-21-3十=0，则：的共轭复数：=2=1+乞(6A.1+1B.1-iC.5+iD.5-i3.已知某样本的容量为50，均数为36，方差为48，现发现在收集这些数据时，其中的两个数据记录有误，一个错将24记录为34，另一个错将48记录为38.在对错误的数据进行更正后，重新求得样本的均数为x,方差为2，则A.x=36,s2<48B.x=36,s2>48C.x>36,s2<48D.x<36,s2>484.已知三角形中三边长为a,b,c,若lga,lgb,lgc成等差数列，则直线ax+by=a与直线bx十cy=b的位置关系为(A.行B.相交但不垂直C.垂直D.重合5.设函数f(x)=sin(x十)(00成立，则2的最大值为A日B.1C.D【名校之约系列2023高考考前冲刺押题卷（六）·数学第1页（共4页）】

20.解：1)证明：连接AB,设AB0AB,=M,对A1B中：点为M,且AM⊥A1B,面ABC⊥面ABB1A1且交线为A1B,AMC故g(xmn=g(xo)=1-x0十lna十x0+1=lna+2≥面ABB1A1,AM⊥面A1BC0,解得a>己BCC面ABC,.AM⊥BC又直三棱柱ABCA1B,C1,BB1⊥BC,.2分综上所，实数a的取值范为[十.12分:AMOBB1=B1,AM,BB1C面ABB1A1,C-.BC⊥面ABB1A1,ABC面ABB1A1,.AB⊥BC.4分22.解：(1)由题意可得{411(2)连接MC,由(1)知AM⊥面A1BC5分a2+=1，2分a2=b2+c2故直线AC与面A1BC所成的角为∠ACM=晋解得u2=8,b2=2.…3分不坊设AB=2,AM=√2，AC=2√2，则BC=所以C的方指为管+-1..4分√AC2-AB2=2,(2)①设B(x0,y),因为O=1OA+μOB=(2十ux0,以B为原点，BA,BC,BB所在方向分别为x,y,z轴的λ+4y0),正方向建立空间直角坐标系，所以点P的坐标为(2入十μx0,入十μy0),则A(2,0,0),C(0,2,0),E(1,1,1),又因为点P在椭圆C上，设面ABE的法向量为n=(x,y,2),7分所以②λ+)2+Q+o)2P=12|n·BA=2x=0,n·BE=x十y+z=0,故n=(0,1,-1),..9分化商可得2+2（货+9）+2经+%=1,6分设面CBE的法向量为m=(x1,y1,21）,因为管+-1且+-1，1m·BC=2y1=0,所以2受+00=0，m·BE=x1十y1十z1=0,故m因为B,P为C上不与A重合的两点，(1,0,-1),.-11分设面ABE与面BCE所成锐二所以μ≠0，y0=一220,面角为日即直线0B的斜来k-公28分∴.cos0=n·m=故9=晋1nm...12分②设1的方程为)y=号+m(m≠0)，Mn21解：1)当a=0,fx)=1-1nx,x>0,1y=-2x+m,令f(x)=二2+nx=0,解得工=e2,N(x2,y2),由消去y,..2分r2则当x∈(0，e2)时，f(x)<0,f(x)单调递减，可得x2-2mx+2m2-4=0,当x∈(e2,十∞)时，f(.x)>0,f(x)单调递增由△=4m2-4(22-4)>0,可得-20,1MN√+a--号√/（x1+x2)2-4x1x2今g(x)=axe+1-lnx-x,则g(x)=(x+1)·=√5(4-m2),(ae-）=(z+1).ac.5分x点A到直线1的距离d=lm一2=2(2-m)+5当a≤0时，g'(x)<0,g(1)=ae≤0，则x>1时，g(x)0时，设h(x)=aer·x-1,x>0,h(日)=ae÷.}-1=e2-1>0,h0)=-1<0,√(4-m2)(2-n)2=√(2+n)(2-n)3.10分令f(m)=(2+m)(2-m)3(-20,所以h(x)8分当∫(m)>0,解得-20,所以f(m)在(-2，-1)上单调递增，在（一1,0），(0,2)则当x∈(0，x0),g(x)<0;当x∈(x0,+∞)，g'(x)>0,上单调递减，则g(x)在(0，x)上单调递减，在(x0,十∞)上单调递增，所以f(m)的最大值为f(一1)=27，.-11分所以g(x)mn=g(x0)=azoe-x0-lhxo十1.10分所以S≤3√3，12分因为ae3,=上，所以ae5,=1,即lna十z=一ln0即△AMN面积的最大值为3√3.

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(2)现有甲、乙两所高校的笔试环节都设有三门考试科目，某考生参加每门科日考试是否通过相互独立。若该考生报考甲高校，每门笔试科日通过的概率均为？：该考生报考乙高校，每门笔试利目道过的概率根次为m写手其中00.75时，认为两个变量24-20-列之间存在较强的线性相关关系。21.(12分)已知4，B为椭圆E:兮+y=1上两点，点P1,0满足而=历，过点A与点Q0,2》的直线与直线x=5交于点M.(1)当AB⊥x轴且A在x轴上方时，求直线BM的斜率：(2)已知1∈，2]，记△AP的面积先9，△ABM的面积为S,求三的取值范围.22.(12分)已知函数f(x)=(x+1-2a)n(x-a)(1)当a=2时，求函数f(x)的极值：(2)当x≥a+1时，f(x)≥x-1恒成立，求实数a的取值范围.数学试题第4页共4页

答案专期2022一2023学年聊城专版九年级第9~12期分数学用报MATHEMATICS WEEKLY因为小王到达B地后即以15千米/时的速度返回A地，所以当3

教全国@0所名校高三月考卷记22.(12分)已知函数f(x)=芒与g(x)=azlnx的最大值相等。(1)求a的值；(2)判断函数y=f(x)一g(x)的零点个数；(3)证明：存在直线y=b,其与两条曲线y=f(x)和y=g(x)共有三个不同的交点，且从左到右三个交点的横坐标成等比数列，【解题分析】1)f()=一2，易知函数f(x)在（一0,1）上单调递增，在(1，十∞)上单调递减，f(x)≤f1)=eg(x)=alnx+1),显然a≠0，若a>0,则当0<<时，g(x)<0,当>是时g)>0,函敦g()在1=是处取得授小值，不合题高：若a<0,则当0<<是时g()>0,当>时，g(x)<0,函数g()在x=上处取得极大值，且为最大值，ee则g)=-8=日仑，解得11.4e(2)y=fw)-g(x)=芒(1+elnx设(x)=1+elhx,则'(x)=e(lnx+子).设(x)=h叶子，则及()=会，易知百数()在(0.1D上单调递减，在1，十)上单调递特，)≥k1)1,则()>0，函数()是增面数，h(日)=1-e<0,h(1)=1>0,函数hx)在（日，1》上有唯一零点，又总>0西数y=f)一gx)有准一零点，…8分(3)函数f(x),g(x)的部分图象大致如图，由(2)知，当0xxo时，f(x)g(x),X3当x>xo时，f(x)>g(.x),且f(xo)1,则0

8.二进制是计算技术中广泛采用的一种数制，由8世纪德国数理哲学大师菜布尼兹发现，二进制数据是用0和1两个数码来表示的数，如01,10分别表示不同的二进制数，在有一个0，两个1组成的二进制数中，两个1和邻的概率是120分钟，请合理分配时间。题。A.3收答题卷。D9.在边长为8的正方形ACD中，E为AB边上一点，AE=3E,连接DE,G为DE中点，若点M在正方形5CD的边上，且MG=5,则满足条件的点M的个数是A、3个B.4个C.5个D,6个题目要求的10、已知、二次函数y=a2+(2a-1)x+1的对称轴为y轴，将此函数向下移3个单位，若点M为二次函数图象在(-1≤x≤1)部分上任意一点，O为坐标原点，连接OM,则OMD君长度的最小值是完全自主设计建造的首熊A.5位一写B.2温回健于成c.3D行装置，满载排水量800002二、填空题（本大题共4小题，每小题5分，满分20分）♪，个，个人方隆D.0.8x1011、64的立方根是12.如图，用半径为9，圆心角为120°的扇形围成一个圆锥的侧面，则这个圆锥底面半径D.（-2a2b)2=4a02为视图是13.如图，在面直角坐标系中，反比例函数y=”（之0，m为常数)的图象与一次函数y=女+b的图象交于点A(2,a)和点B,过点A、B分别作x、y轴的垂线，交x轴于D点C,交y轴于点D,AC与BD交于点E,若点E恰为AC中点，三角形ADC的面积为4,则k的值为义赠没食玉个前，示深图世《orD.44A120°发津5B)E2B图第12题图第13题图第14题图能是I4.如图，△ABC中，∠ABC=90,AB=BC,点D是边AC上一点，CD=2AD,连接D,后D.-1过点C作CE⊥BD于点E,连按AE,拓晨超甲F=2,FG=4,则菱(1)∠AEC=_10)发(2)若BC=35,则AE=一一·()量风n单人P钟D.322023年九年级质量调研（三）数学试题第2页共6页C⑤扫描全能王于前全日

河北省高三年级2月联考英语参考答案听力部分录音材料听力部分。该部分分为第一、第二两节。注意：回答听力部分时，请先将答案标在试卷上。听力部分结束前，你将有两分钟的时间将你的答案转涂到答题卡上。停顿00'10现在是听力试音时间。M:Hello.International Friends Club.Can I help you?W:Oh,hello.I read about your club in the paper today and I thought I'd phone to find out a bit more.M:Yes,certainly.Well,we are a sort of social club for people from different countries.It's quite a new clubwe have about 50 members at the moment,but we are growing all the time.W:That sounds interesting.I'm British actually,and I came to Washington about three months ago.I'm lookingfor ways to meet people.Er,what kinds of events do you organize?M:Well,we have social get-togethers,and sports events,and we also have language evenings.W:Could you tell me something about the language evenings?M:Yes.Every day except Thursday we have a language evening.People can come and practice theirlanguages-you know,over a drink or something.We have different languages on different evenings.Mon-day-Spanish;Tuesday-Italian;Wednesday-German;and Friday-French.On Thursday we usually havea meal in a restaurant for anyone who wants to come.W:Well,that sounds great.I really need to practice my French.M:OK.Well,if you can just give me your name and address,I'll send you the form and some more informa-tion.If you join now you can have the first month free.试音到此结束。听力考试正式开始。停顿00'10”请看听力部分第一节。第一节听下面5段对话。每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项。听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。每段对话仅读一遍。停顿00'02”例如：现在你有5秒钟的时间看试卷上的例题。停顿00'05”你将听到以下内容：M:Excuse me.Can you tell me how much the shirt is?W:Yes,it's nine fifteen.停顿00'02你将有5秒钟的时间将正确答案标在试卷上。停顿00'05”衬衫的价格为9镑15便士，所以你选择C项，并将其标在试卷上。【高三英语·参考答案第1页（共8页）】·23-302C·

位-小，。-1=号数列日-1是以号为首项，写为公比的等比数列，放A正确：3对于C,易知止系+1关于a单两递减数列日}是遮减散列，又心>0数列口为造拍敦列，放心错误；时安侍明京小层-哈，1-3n=n-+10,∴.{an}是递减数列，且当1≤n≤6时，an>0,当n≥7时，an<0,∴.当Sn取最大值时，n的值为6.15.答案2.5%命题意图本题考查全概率公式的应用.解析设A车间的次品率为之，则B车间的次品率为，C车间的次品率为号，则0.2×号+0.3×号+0.5×5=0.3x=0.015,解得x=0.05,所以A车间的次品率为0.025=2.5%.16.答案(-a命题意图本题考查利用导数研究函数性质。CULTURE解析关于x的方程公=-2+*+1有三个不等实数根等价于函数y=(-2+x+1)。的图象与直线y=m有三个公共点，设g(x)=(-x2+x+1)e,则g'(x)=-(x-1)(x+2)e,当x∈(-2,1)时，g'(x)>0,g(x)单湖道烟，在(-，-2和1，+)g()<0,8)单调递减6(-2)=三，当<1≥5或>2时，g(x)<0,且当x→-0时，g(x)0,当x→+∞时，g(x)→-∞.画出g(x)=（-x2+x+1)e的大致图象如图，要使g(x)的图象与直线y=m有三个交点，需g(-2)

8:16l令61数学511A答案(1).pdf【高一数学·参考答案第5页（共7页）】·23-511A·解法二：如图，连接BD,BE,AB,过E作EF⊥AB,垂足为F..A1D1∥BC,且A1D1=BC,∴.四边形A1BCD1是行四边形，BB1⊥底面ABCD,BB1⊥BC,BC⊥AB,AB∩BB=B,∴.BC⊥面ABB1A1.EFC面ABB1A1,∴.BC⊥EF.又A1B∩BC=B,.EF⊥面ABCD1.…6分sin-格AB=VAs+A=25,∴EF=AB·AE25AB…7分5面ABBA1∥面CDDC,∴.BC⊥面CDDC1,∴.BC⊥CD.CD=AB=2/5S6m,=2BC…CD,=25.…8分ED,=√A1E+A1D=2√2，CE=√AB+BC+AE=2√5，ED+CE=A1B,∴CE⊥EDSAcD,=2CE·ED=26.…9分设点B到面CED1的距离为h.:Vgm=3Sm,·h=Vem=}Sm·EF,…10分h二S△D,·EF=S即点B到面CED,的距离为SACED…12分评分细则：第(1)问，先证明BD⊥面ACCA1,再证明面BDDB1⊥面ACCA1,按照相应步骤给分2.解：(1)由f0)=2c0s9=1,得c0s9=号，因为-5<9<0，所以9=-5.…1分由/3)=2cos(5。一3)=1，得0（骨。哥）=7则肾。青=晋+2kr或-晋+2kx从∈Z),…2分得w=2十6k或w=6k(k∈Z).…3分又0+经当f(x)在[a,a十]上单调时，e+<+经…5分【高一数学·参考答案第6页（共7页）】·23-511A·所以b-a=fa)-fa+5)1=2cos(2a-号)-2cos(2a+号)1=|cos2a十√3sin2a-cos2a十√3sin2a=2W3sin2a.…6分又登+k≤2a<+kmh∈D,所以sm2a∈停.1.所以ba=25sn2a∈[3,2v3…7分

7.What does the woman advise the man to do?A.Save some money.B.Check the booking.C.Book the plane tickets ahead.听第7段材料，回答第8、9题。8.Where does the conversation probably take place?A.In a hospital.B.In a school.C.In a restaurant.9.Who will drive the man home?A.The woman.B.His wife.C.His friend.听第8段材料，回答第10至12题。10.How was the man's job interview?A.Competitive.B.Easy.C.Informal.11.What do we know about the man?A.He got the job.B.He failed the interview.C.He hasn't got the result.12.What are the speakers going to do?A.Enjoy a meal.B.Go to a party.C.Have a drink.听第9段材料，回答第13至16题。13.What did the woman do to what the man said?A.She repeated it.B.She wrote it down.C.She typed it out.14.When did the accident happen according to the man?A.At 7:20 a.m.today.B.At 7:20 p.m.today.C.At 7:20 a.m.yesterday.15.What was the girl probably doing when the accident happened?A.Riding a bike.B.Listening to music.C.Getting on a car.16.What does the man have to do in the end?A.Sign his name.B.Copy his ID card.C.Record their conversation.听第10段材料，回答第17至20题。17.How did the speaker get to Miami?A.By ship.B.By plane.C.By train.18.Why did the speaker go on holiday in Miami?A.It was warmer.B.It was nearer.C.It was cheaper.19.Where did the speaker eat breakfast?A.Outdoors near the pool.B.In the dining hall.C.In the hotel room.20.When did the speaker go dancing?A.In the morning.B.In the afternoon.C.In the evening.【高一英语第2页（共10页）】·23-311A·

Be(0,）B-号∈(-号，）B-牙=晋，…(1分)B=牙，即△ABC是直角三角形.…(12分)21.命题意图本题考查空间几何体的结构特征以及相关计算解析（I)001=5dm,.P0,=2dm。…(2分)六玻璃罩的容积V=弓×62×2+62×5=24+180=204(dm3)=204(L).…(4分)(Ⅱ)连接A101,设P0,=xdm(0

2022-2023学年度下期高2023届三诊模拟考试英语参考答案及评分标准第一部分听力（满分30分）1~5 BCAAC6~10ABCAB11~15 ABCCC16~20 BABCC评分标准：1~20小题，每小题1.5分。第二部分第一节阅读理解（满分30分）21~23BBC24~27 CDBD28-31 AAAD32~35 BBAB评分标准：2135小题，每小题2分。第二节（满分10分）36-40 FGEBA评分标准：36~40小题，每小题2分。第三部分第一节完形填空（满分30分）41~45 BACCC46~50 ADDDA51~55 CBBBD56~60 DDCDC评分标准：41~60小题，每小题1.5分。第二节（满分15分）61.have completed 62.on63.who64.formerly65.a66.technologies 67.as68.included69.and70.to be used评分标准：61~70小题，每小题1.5分。有任何错误，包括用词错误、单词拼写错误（含大小写)或语法形式错误，均不给分。第四部分第一节短文改错（满分10分）My favourite inventor is Alexander Graham Bell.At a young age,Bell started to show muchinterest A helping deaf people communicate,which led to their invention of the microphone.inhisBeside,he also invented the telephone in 1876.However,at the beginning,what he actually triedBesides学使to design was not a telephone but a multiple telegraph.During his search to improve the telegraph,Bell invents the first telephone.One of his most famous saying is:"Leave the beaten trackinventedsayingsoccasional and dive into the woods.Every time you do so you will be certain to find somethingoccasionallywhich is completely new in your life."Indeed,what it was his acute curiosity that made his success.thatBell was a honourable inventor all his life.He made his first invention at 11 but his last invention at 75.anand评分标准：有任何错误，包括用词错误、单词拼写错误（含大小写）或语法形式错误，均不给分。第1页（共5页）第二节书面表达（满分25分）一、各档次语言要点配分参考标准档次要点数要点分语言要，点表达情况划档根据第五档521-25语言基本无误，行文连贯，表达清楚第四档16-20语言有少量错误，行文基本连贯，表达基本清楚第三档311~15语言有一些错误，尚能表达第二档26-10语言错误很多，影响表达第一档10-5只能写出与要求内容有关的一些单词

20.12分)在菱形AB5C中，AB=2,∠CAB=60,将△ABC沿对角线BC翻折至△DBC的位置，使得AD=3(1)证明BCL AD(2)求三棱锥A-BDE的体积0BCLAV2L.(12分)已知在正三棱柱ABC-A,B,C,中，AA,=AB,点D为AMA,的中点，点E在CC,的延长线上，且GE=}CE(1)证明：AC,∥面BDE:(2)求二面角B-DE-C的正切值，biinA asmcC22.(12分)已知△ABC中，内角A,B,C所对的边分别为a,b,c,且si4-asin C_csin B152135(1)求C;(2)若c=14,点E,F分别在AC,BC上，下面两个条代中选)个，求EF的最小值①△EFC的周长为△ABC周长的；②△EFC的面积为△ABC面积的)注：如选择多个条件分别解答，按第一个解答计分5mA,5bGin1-50数学第4页（共4页）

athes aiferent and that his classmateswill laugh at him because of his deafness,"she said.Rebecca and Mason were heart-broken and worried.After chattingM:(温馨提示：答熟词生义和长难句分析见本期B3版ENGLISH CORNER)(听力材料见下期中缝、参考答案见下期B4版)上期B1-B4版w1-5ABCBA6-10 CBBCA11-15 CAABAcould not have made it.16-20 BBCCA31-35 AABDC21-25 DCBDD26-30 ACDABScenes of reciting one of Shakespeare's poemsOne possible version:46-50 DABAC36-40 EGDFA41-45 DBCADbefore hundreds of students and six serious-lookingSoon,Jane reached the school canteen.She57.producer51-55 DADBCjudges still appear vividly before me.It was kind ofbought an icy drink and a cheese sandwich.Then,she58.largest56.has achieved59.combinedyou to set aside half an hour every day from your busycomplaining about what a bad day she was having,-started her car,driving to Sally.Just when Sally was60.the61.to occupyschedule to help me from choosing the right poem tow.63.as64.who62.respeetivelycorrecting my pronunciation.M:写作65.boostingThanks again for your generous help.Jane pulled up behind her and got out.Smiling,第一节Jane approached Sally and introduced herself.SallyYours.【写作提示】also briefly introduced herself and talked about herLi Hua本写作要求给外教Mr Smith写一封电子邮件(范文中的画线句子可作为佳句背诵)problem.After comforting her,Jane returned to her car感谢他在你参加英语诗歌朗诵大赛期间的帮助。时第二节and showed her the drink and food she had bought for写作提示】her.态应以一般现在时和一般过去时为主，人称应以第W:人称和第三人称为主。审题可知，第一段应告知外个炎热的夏日，Saly开车去修理F修车，中Then,she invited Sally into her car to enjoy the教Mr Smith你获奖的消息，并表达感谢；第二段回顾途车抛锚了，在等待拖车时她又渴又饿。附近一所学food.Sally gladly accepted.Once inside the car,Sally外教在哪些方面帮助了你；第三段再次表达感谢校的老师Jane看到这一幕后，决定给她送一瓶冰饮felt cool since Jane had turned on her air conditioner.M:One possible version:料和食物。根据续写第一段的首句可知，Jane来到了Finally,the tow truck arrived.Sally said goodbye toW:Dear Mr Smith,学校食堂。她给Sally买了什么？接下来她会和SallyJane and thanked her sincerely.She was grateful for如何互动？本段可据此展开叙述。根据续写第二段the way Jane touched her life.Although Sally didn'tW:I am writing to inform you that I won first prizein the English poetry recitation contest held in our的首句可知，Jane邀请Sally?到自己的车里享用食物.meet Jane again,she tried to be kind,like her,doingschool last Friday.Thanks for your help.Without yourSally接受Jane的好意了吗？她向Jane道谢了吗？这acts of kindness to a person in need.M:件事对Sally产生了什么影响？本段可据此展开叙patient guidance and continuous encouragement,I述本版主编杨娟芳责编李冬妮W:本报地址山西省太原市小店区龙兴街190号，邮政编码030032联系电话0351-27776792777680（发行中心）订阅查询0351-27776482777649印别音中开发区置臻卸业有限公司（晋中开发区汇通产业园区南六堡段恒易通公司院内）订阅方式为方便订阅订户可通过当地发行工作站订购订阅时请核验其授权书护”“工作证”或直接向本报发行中心咨询以防假冒盗版行为激光排版本报激光照排中心

8.已知函数f(x)=sin,甲、乙、丙、丁四名同学研究f(x)在（一∞，一a)上的零-2x2-4x十3-a,x≤0，点分布情况，各得出一个结论：甲：若f(x)在（一∞，一a)上有0个零点，则实数a的取值范围为(5，十∞）；乙：若f(x)在(-∞，一a)上有1个零点，则实数a的取值范围为(0,1)；丙：若f(x)在(-∞，一a)上有3个零点，则实数a的取值范围为[一6，一4)；丁：若f(x)在（一∞，一a)上有5个零点，则实数a的取值范围为[一10，一8）.则这四名同学中得出正确结论的是A.甲、丙、丁B.甲、乙C.乙、丙D.丙、丁二、选择题：本题共4小题，每小题5分，共20分。在每小题给出的选项中，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.已知函数f(x）=2cos(3x-),则Af(x)的图像的两条相邻对称轴之间的距离为晋B.f(x)的图像关于点(，0)对称Cfx)在[0，受]上不单调D.f)在[-晋，号]上的最小值为-510.下列物体中，能够整体放入半径为1（单位：）的球体容器（容器壁厚度忽略不计）内的有A.棱长为1m的正方体B.底面棱长为1m,侧棱长为√3m的正六棱锥C.底面直径为1.1m,高为W3m的圆柱体D.底面直径为0.8m,高为1.8m的圆柱体11.已知函数y=xf(x)是R上的偶函数，f(x一1)+f(x十3)=0，当x∈[一2,0]时，f(x)=22-x十x,则A.f(x)的图像关于直线x=2对称B.4是f(x)的一个周期C.f(x)在(0,2]上单调递增D.f2023)

限定以若两个贝有线性相关关系的亦：的相们火性越强，刚样本相关系数”的值越接近于1.记无现数=2.71828182845904水数点后第7位上的数字是y,我们把y看作是x的函数，B.函数图象(r)是一群孤立的点记作y=(),定义域为A,值城为，其下列说法M的是A.定义域A是值城B的子集D.x也是y的函数，记作x=f(y②，如图，所有边长均为4的正三梭柱AB沁CA'BC巾，点E,F分别作线段AM和CC'上，且满尼C.f(6)=1AE=CF=1,若过B,B,F三点的面把该几何体分成上下两部分，则A.上半部分是三棱柱B.下半部分是四棱锥C.上半部分的体积是1633D.下半部分的体积是163三、填空题：本题共4小题，每小题5分，共20分.8,有3个不同的旅游景点，甲，乙人每人选2个去游玩，若在同一个某点两人必然相遇，则两人拾好在其中一个景点相遇的概率为4.柏拉图立体被称为最有规律的立体结构，其所有面都是全等的正多边形，每一个角都是由相同数量的面结合而成。数学家已证明柏拉图立体只有五种，分别为正四面体，正六面体，正八面体，正十二面体，和正二十面体，现有一个正八面体，每个面都是边长为3的正三角形，则该其体积为5.将曲线y=0sz上所有友向右移0@一0)个单位，得到函数y一osz的图象，则p的最小值一为16已知双曲线C荐-茶-1的左右焦点分别为R,F,过右焦点F且倾斜角为45的直线1交C的右支于A,B两点(A在x轴上方)，且满足A=9F克，则双曲线C的离心率是四、解答题：本题共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.(本小题满分10分)在直四棱柱ABCD-A,B,CD1,AB∥DC,AB⊥AD,AB=2,AD=3,DC=4DC(1)求证：A1B∥面DCC1D;(2)若四棱柱体积为36，求二面角A1一BD一C的正切值.B

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高一数学试卷参考答案1.C因为0,1∈N,-1N,所以NM.2.C全称量词命题的否定是存在量词命题.3.C因为ac-ad=a(c-d),a<0,c0,所以ac>ad,故A错误；取a=-4,b=-1,c=1,d=2,则a-c=-5,b-d=-3,故B错误：因为21=b4,且6-a>a b.ab0,ab>0,所以0心2>0，所以日>石故C正确：取a=-2,6-1c-1,d=2,则-1b十d=1,故D错误.4.D因为不等式x2一x一6≥0的解集为{xx≤-2或x≥3}，所以{xm-1≤x≤m十1}二{xx≤-2或x≥3}，所以m-1≥3或m+1≤-2，所以m≥4或m≤-3.5.C依题意，h=-3.6t2+28.8t=-3.6(t2-8t+16)+57.6=-3.6(t-4)2+57.6,当t=4时，烟花达到最高点，6.B因为6m=2X3,所以MUN={ss=3m,n∈Z}=M.7.B存在x∈[0,3]，使得不等式x2-4x-a≥0成立，等价于a≤(x2-4x)mmx·令y=x2-4x,x∈[0,3]，当x=0时，ymx=0,所以a≤0，8A肉为2a+6+2=[(2a+1+(2+1D]水2a+千2）-5+0+190》，所以2a2a+11+2b+6+5+·雾=0当且仅当器8群。-16-时等号成立，所以2a+2%+2≥9，即a+b≥，因为a十b>2x2-6z恒成立，所以2r2-6x<号，即112x7=(2x+1D2x-)0,解得-名号9.BDAC是全称量词命题，BD是存在量词命题，10.ABD因为A={xx2-2x-3=0}={-1,3},所以-1∈B或3∈B或B=0,所以a=-3或a=1或a=0.11.AD因为a2-ab=a(a-b)>0,所以a2>ab,因为ab-b2=b(a-b)>0,所以ab>b2,所以。>a6>心，所以A正确：因为+2+，2≥2的等号成立条件x+2=2不成立，所以B错误：因为≥（生y:=1,所以。+≥2，所以C指误：因为}+2=(x十2-20+2)=22+2+2≥2(2+2)=2，当且仅当}=2即=1时，等号成立，所以D正确：12.ABD易知m≠0且△=m2-4m>0,所以m<0或m>4,故A正确；因为x1+x2=1,C1x2【高一数学·参考答案第1页（共4页）】·24-50A·

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33.What does the author intend to do inSome tasks can wait until the right motivation.paragraph 2?When my wife drove the car to do things last week,阅A.To offer some tips for relaxing our eyes.阅it was the perfect time to oil the garage door-aB.To introduce the structure of the eyeballtask easier done with the car out of the way.C40To explain how sereen time can cause myopia.D.To explain why the shape of the lens is解important.解⑧A.It's useless to blame the task.34.What does the 2019 study mentioned inB.Don't underestimate the power of social阅paragraph 3 show?A.There are many factors that cause myopia阅connection.C.Enjoy these unexpected moments when theyB.Myopia can easily lead to permanentappear理blindness.D.But what if they are paired with somethingC.The rate of myopia among children is on therise.解enjoyable?E.If the standard approach is boring,reject it.Myopia leaves people a higher risk of eyeF.Matching the task with our emotional state canconditions.make a difference.G.But are the tasks boring or are we approachingthem with a boring mind?第一节完形填空35.What does Mojarrad suggest in the lastparagraph?Wanda Smith always knew she wanted to be aA.Children should not use electronics.语teacher.But then bad luck 41 Her motherWe'd better delay the appearance of myopia.became ill just after Smith graduated from high理C.Effective measures should be taken to treatschool and she had to put off college to 42 hermyopia.运mother.D.People should learn more about visualA husband,3 kids,and nearly 20 years later,development.B41.A.existedB.occurredC.continuedD.unfoldedC42.A.wait forB.search forC.care forD.call for第二节根据短文内容，从短文后的选项中选出。能填入空白处的最佳选项。选项中有两项为多余Smith decided it was now or 43.Even though shehad to look after her family and work to make ends阅选项。We may struggle to complete a number of chores语meet,she 44 night classes.In order to provide for her family,Smith workedthat are on our to-do lists,which makes us upset.around the 45 as a school bus monitor and a理G 36 What would happen if we were to apply a运janitor(看门人).She woke up at6 am for her job.fun mind to these tasks?Consider the simple act of doing household用D43.A.forever B.later C.everD.neverchores again.They often feel tiring,right?D 37A44.A.attendedB.preparedListening to an audio-book or some light musicC.finishedD.developedtransforms the experience,turning a tiring task intoC45.A.schoolB.town C.clockD.dayan opportunity for enjoyment or learning.Shopping is another chore that often lands on theAfter years of 46 her own dreams for the sake oftiring task list.But who says it has to be that way?others,Smith finally decided it was her 47.In阅F 38 When I go alone,I choose off-peak hours,2004,she went to Sam Houston State University.making the experience quick and stress-free.BetterHer path wasn't 48 and she faced a lot ofyet,I'll hold off until I'm in the mood for a bit ofphysical activity.言运difficulties along the way.49,Smith fought through with hard work,解B 39 I discovered years ago that finding adentist with a great team made my regular check-B46.A.doubtingB.abandoningC.breakingD.destroyingups something to look forward to.Sharing storiesD47.A.duty B.ideal C.demand D.turnabout my recent travels or workout routines withA48.A.smooth B.long C.broadD.quietmy dentist made the experience more personal andC49.A.ThereforeB.Moreoverenjoyable.C.HoweverD.Otherwise

protects the planet from the sun and other interference from outer space可知，炽热的熔融状内核对我们有一定的益处。3.B。理解词汇。根据第五段中的we would no longer have working compasses...Itwould become very similar to what Mars looks like可知，地核变冷后将产生一系列危害，导致地球不再宜居。4.C。理解主旨要义。通读全文可知，文章主要介绍了地核冷却将会发生什么，故C项最能概括文章主旨。【补充解释】fare(para.1:进展

nally had a throat operation,which was very important for her and made her go on acting for another 25 years after that.Helen Perry was famous for several famous plays.But few people know her talent in writing.We have the letters she exchanged with a writer and they show she had a great writing style.She also had quite a few degrees.But she said what really pleased her was that the audience loved her.

第一部分 听力（共两节，满分30分） 第一节（共5小题；每小题1.5分，满分7.5分） 听下面5段对话。每段对话后有一个小题，从题中所给的A、 B、C三个选项中选出最佳选项，并标在试卷的相应位置。听完每 段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。 每段对话仅读一遍。 1.How many members are there in Alice's group in the end? A.Two. B.Four. C.Six. 2.What can we learn about the man in the contest? A.He did well in spelling. B.He made some mistakes. C.He was satisfied with the result.