山东省某重点校2022-2023学年高三数学上学期期末考试试卷（PDF版含答案）

2022-2023学年度第一学期期末学业水平检测高三数学试题本试题卷共6页，22题。全卷满分150分。考试用时120分钟。注意事项：1．答卷前，考生务必将自己的姓名、考生号等填写在答题卡和试卷指定位置上，并将准考证号条形码粘贴在答题卡上的指定位置。2．回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。如需要改动，用橡皮擦干净后，再选涂其它答案标号。回答非选择题时，将答案写在答题卡上，写在本试卷上无效。3．考试结束后，将本试卷和答题卡一并交回。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。i21．复数的虚部为1i3331A．iB．C．D．22223422．若(ax)(ax)的展开式中含有x项的系数为18，则a333A．2B．C．或2D．或2222223．已知集合A{(x，y)|xy2x0}，B{(x，y)|yk(x1)}．若AB，则33A．kB．3k33333C．k或kD．k3或k3334．“阿基米德多面体”也称为半正多面体，是由边数不全相同的正多边形为面围成的多面体，它体现了数学的对称美．如图，将一个正方体沿交于一顶点的三条棱的中点截去一个三棱锥，共可截去八个三棱锥，得到八个面为正三角形，六个面为正方形的“阿基米德多面体”，则该多面体中具有公共顶点的两个正三角形所在平面的夹角正切值为2A．2B．1C．2D．22高三数学试题第1页（共6页） x2m5．“m1”是“函数f(x)为奇函数”的x2mA．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6．已知函数f(x)2sin(x)(0π)的部分图象如下图所示，将f(x)的图象向左平移πx个单位后得到函数yg(x)的图象，则函数yg(x)g()的最小值为122yA．429B．47Oπ13πxC．3124D．07．为了解甲、乙两个班级学生的物理学习情况，从两个班学生的物理成绩（均为整数）中各随机抽查20个，得到如图所示的数据图（用频率分布直方图估计总体平均数时，每个区间的值均取该区间的中点值），关于甲、乙两个班级的物理成绩，下列结论正确的是频数频率7组距60.03050.02540.0203210.00505758596768697987888998分数05060708090100分数甲班物理成绩乙班物理成绩A．甲班众数小于乙班众数B．乙班成绩的75百分位数为79C．甲班的中位数为74D．甲班平均数大于乙班平均数估计值8．已知定义域为[0，1]的“类康托尔函数”f(x)满足：①0xx1，f(x)f(x)；1212x1②f(x)2f()；③f(x)f(1x)1．则f()320231111A．B．C．D．3264128256高三数学试题第2页（共6页） 二、多项选择题：本题共4小题，每小题5分，共20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5分，部分选对的得2分，有选错的得0分。29．通过长期调查知，人类汗液中A指标的值X服从正态分布N(10，2.5)．则A．估计100人中汗液A指标的值超过10的人数约为50B．估计100人中汗液A指标的值超过12.5的人数约为16C．估计100人中汗液A指标的值不超过15的人数约为955D．随机抽检5人中汗液A指标的值恰有2人超过10的概率为16参考数据：2若XN(，)，则P(X)0.6827；P(2X2)0.9545．10．已知对任意平面向量AB(x，y)，把AB绕其起点沿逆时针方向旋转角得到向量AP(xcosysin，xsinycos)，叫做把点B绕点A沿逆时针方向旋转角得到点P．已知平面内点A(2，1)，点B(2t，1t)，|AB|22，ABOA0，点B绕点A沿逆时π针方向旋转角得到点P，则3A．|BP|22B．AB(2，2)C．B的坐标为(4，1)D．P的坐标为(33，3)226xy11．已知O为坐标原点，离心率为的椭圆C:1(a0，b0)的左，右焦点分别223ab为F，F，C与曲线ycosx恰有三个交点，则12A．椭圆C的长轴长为3B．C的内接正方形面积等于3C．点W在C上，WFWF，则WFF的面积等于11212D．曲线C与曲线y2x4lnx2ln21没有交点3b3ann12．已知数列{a}和{b}满足a1，b0，aa1，2bb1．则nn11n1nn1n42241A．a2b222B．数列{a2b}是等比数列nnC．数列{a2b}是等差数列nnD．aan1n高三数学试题第3页（共6页） 三、填空题：本题共4个小题，每小题5分，共20分。13．已知sinsin1，coscos2，则cos()．14．将8块完全相同的巧克力分配给A，B，C，D四人，每人至少分到1块且最多分到3块，则不同的分配方案共有种（用数字作答）．215．已知O为坐标原点，抛物线C:y2px(p0)的焦点为F，过F的直线交C于A，B两点，A，B中点D在x轴上方且其横坐标为1，|AB|3，则直线AB的斜率为．16．已知球O的半径为2，圆锥W的顶点和底面圆周上的点均在球O上，记球心O到圆锥W底面的距离为h，圆锥W的底面半径为r．则（1）hr的最大值为；（2）圆锥W体积的最大值为．（本题第一空2分，第二空3分）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．（10分）在ABC中，sinBsinCcosA2sinAsinCcosB3sinAsinBcosC，内角A，B，C的对边分别记为a，b，c．22a2b（1）求的值；2c（2）求cosC的最小值．18．（12分）如图1所示，在ABC中，点E，F在线段AB上，点D在线段BC上，AEEFFB1，CE2，DF1，CEAB．将ACE，BDF分别沿CE，DF折起至点A，B重合为点G，形成如图2所示的几何体W，在几何体W中作答下面的问题．（1）证明：平面EFG平面CEFD；（2）求点D到平面CFG的距离．CCDDEFAEFB图1G图2高三数学试题第4页（共6页） 19.（12分）记数列{a}的前n项和为S，a1，．给出下列两个条件：nn1nn1条件①：数列{a}和数列{Sa}均为等比数列；条件②：2a2a2ana．nn112nn1试在上面的两个条件中任选一个，补充在上面的横线上，完成下列两问的解答：（1）求数列{a}的通项公式；n2ni（2）记正项数列{bn}的前n项和为Tn，b1a2，b2a3，4Tnbnbn1，求[(1)bibi1]．i1注：如果选择多个条件分别解答，按第一个解答计分．20．（12分）由mn个小正方形构成长方形网格有m行和n列．每次将一个小球放到一个小正方形内，放满为止，记为一轮．每次放白球的概率为p，放红球的概率为q，pq1．1（1）若m2，pq，记y表示100轮放球实验中“每一列至少一个红球”的轮数，统2计数据如下：n12345y7656423026ˆˆ求y关于n的回归方程lnybna，并预测n10时，y的值（精确到1）；12（2）若m2，n2，p，q，记在每列都有白球的条件下，含红球的行数为随机33变量X，求X的分布列和数学期望；mnnm（3）求事件“不是每一列都至少一个红球”发生的概率，并证明：(1p)(1q)1．kxiyikxy5附：经验回归方程系数：bˆi1，aybx；nlny53,lny3.8．kii22i1xikxi1高三数学试题第5页（共6页） 21．（12分）22y已知O为坐标原点，动直线l:ykxm(km0)与双曲线C:x1(b0)的渐近线交2bx222于A，B两点，与椭圆D:y1交于E，F两点．当k10时，2(OAOB)3(OEOF)．2（1）求双曲线C的方程；（2）若动直线l与C相切，证明：OAB的面积为定值．22．（12分）1已知函数f(x)xlnx的最小值和g(x)ln(1x)ax的最大值相等．e（1）求a；x2（2）证明：lnxe；ex11m12m2（3）已知m是正整数，证明：[1]e．2m(m1)高三数学试题第6页（共6页） 2022-2023学年度第一学期期末学业水平检测高三数学评分标准一、单项选择题：本题共8小题，每小题5分，共40分。1--8：BCADABDC二、多项选择题：本题共4小题，每小题5分，共20分。9．ABD10．ACD11．BCD12．BCD三、填空题：本题共4个小题，每小题5分，共20分。1256π13．；14．19；15．2；16．（1）2；（2）．281四、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.（10分）解：（1）由正弦定理得：bccosA2accosB3abcosC··································2分222222222bcaacbabc由余弦定理得：bc()2ac()3ab()··········4分2bc2ac2ab222222222所以(bca)2(acb)3(abc)22222a2b化简得a2b3c，所以3·······················································5分2c22122222ab(a2b)abc3(2)由余弦定理：cosC···························6分2ab2ab2212ab3312ab()······················7分2ab6ba12ab22······························9分6ba3当仅当b2a（即a:b:c3:6:5）时取等号2所以cosC的最小值为···········································································10分318.（12分）解：（1）由题知：因为CEEG,CEEF,EGEFE，所以CE平面EFG····················································································3分又因为CE平面CEFD，平面EFG平面CEFD·········································4分（2）在平面EFG内过点F做EF的垂线FH，因为平面EFG平面CEFD，所以FH平面CEFD······································5分C如图，以F为坐标原点，直线FE，FH，FD分别为x，y，z轴，建立空间直角坐标系··············································6分z13D则F(0,0,0),C(1,0,2),G(,,0),D(0,0,1)···········7分22uuur13E所以FD0,0,1，FG(,,0),FC(1,0,2)········8分F22xGHy高三数学答案第1页（共5页） 设平面CFG的法向量为n(a,b,c)，a3bnFG00因为，即22，取b2，从而n(23,2,3)···············11分nFC0a2c0|FDn|357所以D到平面CFG的距离为h···································12分|n|191919.（12分）解：（1）方案一：选条件①．因为数列{Sa}为等比数列n122所以(Sa)(Sa)(Sa)，即(2aa)2a(2aaa)·················2分211131121123设等比数列{a}的公比为q，因为a1n122所以(2q)2(2qq)，解得q2或q0（舍）·····································5分n1n1*所以aaq2(nN)······································································6分n1方案二：（1）选条件②．nn1*当n2时，因为2a2a2ana(nN)①12nn1n1n2所以2a2a2a(n1)a12n1nnn12所以2a2a2a2(n1)a②···············································3分12n1nan+1①②得2ana2(n1)a，即=2(n2)·········································4分nn1nana2当n1时，2aa，=2适合上式····························································5分12a1所以数列{a}是首项为1，公比为2的等比数列nn1n1*所以aaq2(nN)······································································6分n1（2）由题知：4Tbb,4Tbbnnn1n1n1n2两式做差得：4(TT)bbbbn1nn1n2nn1所以4bb(bb)，得bb4······················································8分n1n1n2nn2n*所以{b}(kN)为首项b4，公差等于4的等差数列，2k2所以b4(k1)44k2k*同理：{b}(kN)为首项b2，公差等于4的等差数列，2k11所以b2(k1)44k22k1所以b2nn2n2nii所以[(1)bibi1]4(1)i(i1)i1i12ni2所以4(1)i(i1)4[(26)(1220)4n]8n8n····················12分i1高三数学答案第2页（共5页） 20.（12分）5nilnyi5nlny5353.834解：（1）由题知：i1···············1分b0.4n22554510ni5ni1所以a3.80.435，·············································································2分所以线性回归方程为：lny0.4n5·····························································3分所以，估计n10时，ye3·····································································4分（2）由题知：X的取值可能为0,1,2·······························································5分记“含红球的行数为k”为事件A(k0,1,2)，记“每列都有白球”为事件B，k4P(AB)p10所以P(X0)P(A|B)=·········································6分022P(B)[1q]2513122P(AB)CpqCpq16142P(X1)P(A|B)=·······························7分122P(B)[1q]2512P(AB)C(pq)822P(X2)P(A|B)=········································8分222P(B)[1q]25所以X的分布列为：X0121168P(Xk)25252513832所以数学期望为E(X)012········································10分25252525mn（3）证明：因为每一列至少一个红球的概率为(1p)mn记“不是每一列都至少一个红球”为事件A，所以P(A)1(1p)····················11分nm记“每一行都至少一个白球”为事件B，所以P(B)(1q)mnnm显然，AB，所以P(A)P(B)，所以(1p)(1q)1······················12分21.（12分）解:（1）设A(x,y),B(x,y),E(x,y),F(x,y)112233443因为2(OAOB)3(OEOF)，所以xx(xx)·································1分12342ykxmmm由，得x1；同理可得x2········································3分ybxbkbk2km所以xx···················································································4分1222bkykxm2224km由x2，得(12k)x4kmx2m20，所以xx·····5分2342y112k2高三数学答案第3页（共5页） 2km6km222所以，即3k3b12k，解得b3222bk12k22y所以双曲线C的方程为x1·································································6分3m3mm3m（2）由（1）得A(,)，B(,)k3k3k3k3224m2m4m2m所以|OA|||，|OB|||··················8分22(k3)k3(k3)k322114m33mS|OA||OB|sinAOB||||····························9分OAB2222k32k3ykxm222由y2，得(3k)x2kmxm30···········································10分2x13因为直线l与双曲线C相切，2222所以4km4(3k)(m3)022即mk30······················································································11分23m所以S||3为定值··································································12分OAB2k322.（12分）解:（1）由题知：f(x)的定义域为(0,)，g(x)的定义域为(1,)·················1分因为f(x)lnx1·····················································································2分11所以，当x(0,)时，f(x)0，f(x)在(0,)上单调递减；ee11当x(,)时，f(x)0，f(x)在(,)上单调递增；ee1所以f(x)f()0···················································································3分e1又因为g(x)a，1x若a0，则g(x)0，g(x)在(1,+)上无最大值；···································4分11若a0，g(x)a0，解得x1；1xa11所以，当x(1,1)时，g(x)0，g(x)在(1,1)上单调递增；aa11当x(1,)时，g(x)0，g(x)在(1,)上单调递减；aa1所以，g(x)g(1)alna10a高三数学答案第4页（共5页） 1令h(a)alna1(a0)，则h(a)1a所以，h(x)在(0,1)上单调递减，在(1,)上单调递增；所以h(a)h(1)0综上，a1································································································5分x21x1（2）要证明：lnxe；只需证xlnxxe····························7分exee1由（1）知：f(x)0，当仅当x时取等号；ex1只需证：xe0（等号不同取）····························································8分ex1x设p(x)xe，则p(x)(1x)e，e所以，p(x)在(0,1)上单调递增，p(x)在(1,)上单调递减；所以，p(x)p(1)0，当仅当x1时取等号综上，命题得证····························································································9分11m12m2（3）要证明：[1]e；2m(m1)11只需证(m1)ln[1]2m(m1)2(m1)11即证ln[1]·····························································10分22m(m1)2(m1)m11设(x)ln(1x)x，x(0,)；m12m(m1)m1m11所以(x)0·················································11分1xm11x11m1所以(x)在(0,)上单调递增，(x)(0)0m111所以，()ln(1)022m(m1)2m(m1)2(m1)11m12m2综上，[1]e成立·························································12分2m(m1)高三数学答案第5页（共5页）