四川省九市联考2023-2024学年度高二上期期末教学质量检测数学.考卷答案

四川省九市联考2023-2024学年度高二上期期末教学质量检测数学.考卷答案试卷答案,我们目前收集并整理关于四川省九市联考2023-2024学年度高二上期期末教学质量检测数学.考卷答案得系列试题及其答案，更多试题答案请关注微信公众号：考不凡/直接访问www.kaobufan.com(考不凡)

四川省九市联考2023-2024学年度高二上期期末教学质量检测数学.考卷答案试卷答案

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WhydidShertzerandRepaskystarttoworkonaplan?A.Theywereworriedaboutworseningairportconditions.B.Theyhopedtoswitchcareerstoagriculture.C.Theywantedtokeepbeesaway.D.Theylearnedaboutthevalueofbees.4.Whatwasthefunctionofthetoolsplacedonthesideofthetarmac?A.Tokeepthetarmacworkingasusual.B.Toeasetheburdenoftheairport'sworkers.C.Toleadbeestotravelfarthertomakehomes.D.Topreventbeesfrommakinghomesinimproperplaces.35.Whatwouldbethebesttitleforthetext?A.DeclineinbeepopulationsissettocontinueB.AnairportkeepsbeestohelpthespeciesC.BeekeepershelphandlebeehivesatairportsD.Localagriculturedoessomethinggoodforanairport根据短文内容，从短文后的选项中选出能填人空白处的最佳选项

选项中有两项为多余选项

第二节（共5小题：每小题2分，满分10分）InthechildhoodmemoriesofalmosteveryMexican,thereisanexcitingscenebreakingapinata()withaclothovertheireyesandastickintheirhands.36Forexample,itisPinatascanbemadefrommaterialssuchascardboardandclaypots,withcoloredpaperstuckhardtoimagineabirthdaypartywithoutapinata.tothem.37Themosttraditionalformistheseven-pointedstar,butnowit'sverypopulartomakepinatasthatfeatureanimals,superheroesorcartooncharacters.Peopleusuallyfillthemwith38It'softenseenthatpinatasindifferentshapesandsizestraveltoothercountriesinthesweets,fruits,ormoney.bagsofmillionsoftavers.TheyalwayshopetoretutotheirowncountrieswithsomethingthatPeopleoftenconneetpinataswithMexicanfestivals.39TheancientChineseusedtostandsforMexicanculturebest.fashionaprdmaigtoaeChineseNewYear.TheyusuallyfilledthefiguresthedkdhethstickstheseedsfewouAfterwards,theremainswerebunedandtheasheswerethoughttobringgoodluckinthecomingyear.WhenthepinatawasintroducedtoMexico,itwasgivenreligiousmeaningandusedinreligiousceremoniesforcenturies.Today,thepinatahaslostmostofitsreligiouscharacter.Peopleseldomthinkaboutitsreligiousmeaning.40A.Theycomeinvariousshapesandcolors.B.Mostsimplyseethepinatatraditionasafunpastime.C.ThrowingabirthdaypartyisimportanttoMexicans.D.Now,pinatashavebecomeapopularsouvenirofMexico.E.Thereisaninterestinghistoryanddeepmeaningsbehindthepinata.F.ThepinataplaysacentralroleinmanycelebrationsinMexico.G.However,everythingpointstothefactthatChinaisthebirthplaceofthepinata.

分析（1）直线l的方程为y=x-c，则$\frac{c}{\sqrt{2}}$=$\frac{\sqrt{2}}{2}$，解得c，又$\frac{c}{a}=\frac{\sqrt{3}}{3}$，b²=a²-c²，解得a，b即可得出．
（2）由（1）可得：椭圆C的方程为$\frac{{x}^{2}}{3}+\frac{{y}^{2}}{2}$=1．假设C上存在点P，使得当l绕P转到某一位置时，有$\overrightarrow{OP}$=$\overrightarrow{OA}$+$\overrightarrow{OB}$成立．设A（x₁，y₁），B（x₂，y₂）．
设直线l的方程为my=x-1，与椭圆方程联立化为（2m²+3）y²+4my-4=0，利用根与系数的关系及其$\overrightarrow{OP}$=$\overrightarrow{OA}$+$\overrightarrow{OB}$=（x₁+x₂，y₁+y₂），可得点P的坐标（用m表示），代入椭圆的方程即可得出．

解答解：（1）直线l的方程为y=x-c，则$\frac{c}{\sqrt{2}}$=$\frac{\sqrt{2}}{2}$，解得c=1，
又$\frac{c}{a}=\frac{\sqrt{3}}{3}$，b²=a²-c²，解得$a=\sqrt{3}$，b²=2．
∴得$a=\sqrt{3}$，b=$\sqrt{2}$．
（2）由（1）可得：椭圆C的方程为$\frac{{x}^{2}}{3}+\frac{{y}^{2}}{2}$=1．
假设C上存在点P，使得当l绕P转到某一位置时，有$\overrightarrow{OP}$=$\overrightarrow{OA}$+$\overrightarrow{OB}$成立．设A（x₁，y₁），B（x₂，y₂）．
设直线l的方程为my=x-1，联立$\left\{\begin{array}{l}{my=x-1}\\{\frac{{x}^{2}}{3}+\frac{{y}^{2}}{2}=1}\end{array}\right.$，
化为（2m²+3）y²+4my-4=0，
∴y₁+y₂=$\frac{-4m}{2{m}^{2}+3}$．
∴x₁+x₂=m（y₁+y₂）+2=$\frac{6}{2{m}^{2}+3}$．
∴$\overrightarrow{OP}$=$\overrightarrow{OA}$+$\overrightarrow{OB}$=（x₁+x₂，y₁+y₂）=$（\frac{6}{2{m}^{2}+3}，\frac{-4m}{2{m}^{2}+3}）$．
代入椭圆方程可得：$\frac{36}{3（2{m}^{2}+3）^{2}}$+$\frac{16{m}^{2}}{2（2{m}^{2}+3）^{2}}$=1，
化为2m²-1=0，
解得m=$±\frac{1}{\sqrt{2}}$．
∴直线l的方程为：y=$±\sqrt{2}$（x-1）．
由方程：${y}^{2}±\sqrt{2}y$-1=0，
解得$\left\{\begin{array}{l}{x=\frac{\sqrt{3}+1}{2}}\\{y=\frac{\sqrt{6}-\sqrt{2}}{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=\frac{1-\sqrt{3}}{2}}\\{y=\frac{-\sqrt{6}-\sqrt{2}}{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=\frac{1-\sqrt{3}}{2}}\\{y=\frac{\sqrt{2}+\sqrt{6}}{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=\frac{\sqrt{3}+1}{2}}\\{y=\frac{\sqrt{2}-\sqrt{6}}{2}}\end{array}\right.$．
因此假设正确．

点评本题考查了椭圆的标准方程及其性质、直线与椭圆相交问题、一元二次方程的根与系数的关系、向量坐标运算，考查了推理能力与计算能力，属于难题．