安徽省宿州市泗县2023-2024学年度第二学期七年级期末质量检测试题(数学)

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安徽省宿州市泗县2023-2024学年度第二学期七年级期末质量检测试题(数学)试卷答案

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MyTeacherMr.MooreThere'sateacherMr.Moore.Whoislovelyandthirty-four.Alwaysencouragingustotry.Heleadsustoaworldof“why”.Wealladmirehimmoreandmore.Theauthorofthispoemisunknown,butitremindsmeofthecountlessteacherswhohavetaughtme.WheneverIreadthispoem,Icouldn'thelpthinkingofmymathteacher,Mrs.Davies.Ifirmlybelievethatshewassuchateacherdescribedinthepoem.Inhighschool,Mrs.Davieswasmymathteacher,whotookherjobseriouslyandexpec-tedherstudentstoputforththesamedegreeofcommitment.Thoughshewas30,shelookedveryyoungforherage.Andshewasoneofthemostpopularteachersinourschool.Comparedwithotherteachers,Mrs.Daviespaidmoreattentiontoherwayofteaching.Shetriedvariouswaystomakeherclasseslivelyandinteresting.Strictbutfair,sheheldourattentionwithhertime-provenapproach.Oneofherclasseswasforstudentswhoweremoreadvancedacademically.Ichosetotakeitandsoonrealizedthatthecoursewasabitovermyhead.Istruggledwiththecomplexproblemsthatothersseemedtolearnwithoutmucheffort.OneFridayweweretestedonourabilitytouseatheorem(thatnooneintheclassthoroughlyunderstood.Theteacherwouldgradeourpapersbasedonourabilitytoprogressthroughtheproblem.Iwasabsolutelylost.Finally,Ididn'tusetherequiredtheorem.Instead,Idecidedtousemorefamiliartheoremstoarriveattheanswer.IknewIhadfailedbecauseIhadn'tdonetherequiredassignment.Ibecameresignedtomyfate.注意：1.续写词数应为150左右；2.请按如下格式在答题卡的相应位置作答

OurtestpaperswerehandedbackthefollowingMonday.ThoughIdidn'tusetherequiredtheorem,Mrs.Daviesseemedgenuinelyproudofme.【高三英语第10页（共10页）】·23-93C·

分析对于函数${f_1}（x）=\frac{{{2^x}-1}}{{{2^x}+1}}$=1-$\frac{2}{{2}^{x}+1}$，定义域为R，由2^x＞0，可得f₁（x）∈（-1，1），满足①，又f₁（-x）=-f₁（x），函数f₁（x）是奇函数，关于原点中心对称，即可判断出结论．同理即可判断出f₂（x）是否是“P-函数”．

解答解：①存在M＞0，使得对任意的x₁，x₂∈D，都有|f（x₁）-f（x₂）|＜M?函数f（x）在D上是“有界函数”．
对于函数${f_1}（x）=\frac{{{2^x}-1}}{{{2^x}+1}}$=1-$\frac{2}{{2}^{x}+1}$，定义域为R，∵2^x＞0，∴0＜$\frac{1}{{2}^{x}+1}$＜1，∴f₁（x）∈（-1，1），∴满足①，又f₁（-x）=$\frac{{2}^{-x}-1}{{2}^{-x}+1}$=-$\frac{{2}^{x}-1}{{2}^{x}+1}$=-f₁（x），∴函数f₁（x）是奇函数，关于原点中心对称．∴f₁（x）是“P-函数”．
${f_2}（x）=lg（\sqrt{{x^2}+1}-x）$，定义域为R，令x=tanα$（α∈（-\frac{π}{2}，\frac{π}{2}））$，则f₂（x）=lg$（\frac{1}{cosα}-tanα）$=lg$\frac{1-sinα}{cosα}$，∵$\frac{1-sinα}{cosα}$∈（0，+∞），∴f₂（x）不满足①，因此，f₂（x）不是“P-函数”．
故选：B．

点评本题考查了函数的有界性、奇偶性、新定义函数、简易逻辑的判定方法，考查了推理能力与计算能力，属于中档题．