蘇州市2022-2023學(xué)年高一上學(xué)期期中調(diào)研測試數(shù)學(xué)試卷（含答案）

蘇州市2022～2023學(xué)年第一學(xué)期期中測試卷高一數(shù)學(xué)2022.11注注意事項學(xué)生在答題前請認(rèn)真閱讀本注意事項及各題答題要求：1．本卷共4頁，包含單項選擇題（第1題~第8題）、多項選擇題（第9題~第12題）、填空題（第13題~第16題）、解答題（第17題~第22題）．本卷滿分150分，答題時間為120分鐘．答題結(jié)束后，請將答題卡交回．2．答題前，請您務(wù)必將自己的姓名、調(diào)研序列號用0.5毫米黑色墨水的簽字筆填寫在答題卡的規(guī)定位置．3．請在答題卡上按照順序在對應(yīng)的答題區(qū)域內(nèi)作答，在其他位置作答一律無效．作答必須用0.5毫米黑色墨水的簽字筆．請注意字體工整，筆跡清楚．4．請保持答題卡卡面清潔，不要折疊、破損．一律不準(zhǔn)使用膠帶紙、修正液、可擦洗的圓珠筆．一、單項選擇題：本題共8小題，每小題5分，共40分．在每小題給出的四個選項中，只有一項是符合題目要求的．1.已知集合，則A. B． C． D．2.命題“存在一個素數(shù)，它的平方是偶數(shù)”的否定是A．任意一個素數(shù)，它的平方是偶數(shù) B．任意一個素數(shù)，它的平方不是偶數(shù)C．存在一個素數(shù)，它的平方是素數(shù) D．存在一個素數(shù)，它的平方不是偶數(shù)3.若集合A的子集個數(shù)有4個，則集合A中的元素個數(shù)是A.2 B.4 C.8 D.164.已知是定義在上的增函數(shù)，則A.函數(shù)為奇函數(shù)，且在上單調(diào)遞增B.函數(shù)為偶函數(shù)，且在上單調(diào)遞減C.函數(shù)為奇函數(shù)，且在上單調(diào)遞增D.函數(shù)為偶函數(shù)，且在上單調(diào)遞減5.已知冪函數(shù)為偶函數(shù)，則關(guān)于函數(shù)的下列四個結(jié)論中正確的是A．的圖象關(guān)于原點對稱 B．的值域為C．在上單調(diào)遞減 D．6.若函數(shù)在區(qū)間上的最大值是，最小值是，則A．與有關(guān)，且與有關(guān) B．與有關(guān)，但與無關(guān)C．與無關(guān)，且與無關(guān) D．與無關(guān)，但與有關(guān)7.已知函數(shù)的圖象關(guān)于點成中心對稱圖形的充要條件是函數(shù)為奇函數(shù).利用該結(jié)論，則函數(shù)圖象的對稱中心是A. B． C． D．8.若將有限集合A的元素個數(shù)記為card(A)，對于集合，，下列說法正確的是A.若，則B.若，則或C.若，則D.存在實數(shù)，使得二、多項選擇題：本題共4小題，每小題5分，共20分．在每小題給出的四個選項中，有多項符合題目要求．全部選對的得5分，部分選對的得2分，有選錯的得0分．

9.下列命題為真命題的是A．是的必要不充分條件B．或為有理數(shù)是為有理數(shù)的既不充分又不必要條件C．是的充分不必要條件D．的充要條件是10.函數(shù)滿足條件：①對于定義域內(nèi)任意不相等的實數(shù)恒有；②對于定義域內(nèi)的任意兩個不相等的實數(shù)都有成立，則稱其為函數(shù).下列函數(shù)為函數(shù)的是A. B.C. D.11.函數(shù)是定義在上的函數(shù)，則A.若，則函數(shù)的值域為B．若，則函數(shù)的值域為C．若函數(shù)單調(diào)遞增，則的取值范圍是D．若函數(shù)單調(diào)遞增，則的取值范圍是12.下列說法正確的是A．函數(shù)，與函數(shù)，是同一個函數(shù)B．直線與函數(shù)的圖象至多有一個公共點C．滿足“值域相同，對應(yīng)關(guān)系相同，但定義域不同”的函數(shù)組不存在D．滿足“定義域相同，值域相同，但對應(yīng)關(guān)系不同”的函數(shù)有無數(shù)個三、填空題：本題共4小題，每小題5分，共20分．13.若，則的取值范圍是▲．14.若函數(shù)為奇函數(shù)，則▲．15.已知正數(shù)滿足，若不等式恒成立，則實數(shù)的最大值是▲．16.若函數(shù)的定義域為，對任意的，都有，且，則不等式的解集是▲．四、解答題：本大題共6小題，共70分．解答應(yīng)寫出文字說明、證明過程或演算步驟．17.（10分）已知函數(shù)的定義域是，集合.（1）若，求，；（2）若命題“，”是真命題，求實數(shù)的取值范圍．▲▲▲18.（12分）已知函數(shù).（1）若關(guān)于的不等式的解集為，求實數(shù)的值；（2）若關(guān)于的不等式的解集為，求實數(shù)的取值范圍.▲▲▲19.（12分）閱讀：序數(shù)屬性是自然數(shù)的基本屬性之一，它反映了記數(shù)的順序性，回答了“第幾個”的問題.在教材中有如下順序公理：=1\*GB3①如果，那么；=2\*GB3②如果，那么.（1）請運用上述公理=1\*GB3①=2\*GB3②證明：“如果，那么.”（2）求證：▲▲▲20.（12分）某地區(qū)上年度電價為0.8元/（kW·h），年用電量為akW·h，本年度計劃將電價下降到0.55元/（kW·h）至0.75元/（kW·h）之間，而用戶期望電價為0.4元/（kW·h）.經(jīng)測算，下調(diào)電價后新增用電量和實際電價與用戶的期望電價的差成反比（比例系數(shù)為）.該地區(qū)的電力成本價為0.3元/（kW·h）.記本年度電價下調(diào)后電力部門的收益為（單位：元），實際電價為（單位：元/（kW·h））.（收益=實際電量（實際電價—成本價））（1）當(dāng)時，實際電價最低定為多少時，仍可保證電力部門的收益比上年至少增長20%？（2）當(dāng)時，求收益的最小值.▲▲▲21.（12分）已知函數(shù)，.（1）當(dāng)時，，用表示，中的較大者，記為，求的最小值；（2）若不等式對任意，（）恒成立，求實數(shù)的取值范圍.▲▲▲22.（12分）已知二次函數(shù)的圖象經(jīng)過點，且=，方程有兩個相等的實根.（1）求的解析式；（2）設(shè)，=1\*GB3①判斷函數(shù)的單調(diào)性，并證明；=2\*GB3②已知，求函數(shù)的最小值.▲▲▲

高一數(shù)學(xué)參考答案一、單項選擇題：本大題共8小題，每小題5分，共計40分．題號12345678答案ABACDBCC二、多選選擇題：本大題共4小題，每小題5分，共計20分．題號9101112答案BDBCBDABD三、填空題：本大題共4小題，每小題5分，共計20分．13．14．315．16．四、解答題：本大題共6小題，共計70分．17．（10分）由解得，故.··························································2分若，則.，.·····································································4分若命題“”是真命題，則.·····················································6分·············································································8分故實數(shù)的取值范圍是.········································································10分18．（12分）解：（1）法一：因為不等式的解集為，所以，·······························································································2分且方程的兩不等根為和1（）由韋達(dá)定理得，·····················································4分所以.···············································································6分法二：因為不等式的解集為，所以，，··················································································2分且即························································4分所以.················································································6分當(dāng)時，不等式的解集為，不滿足題意；································8分當(dāng)時，由，可得的解集為所以即··············································································10分所以.···························································································12分19.（12分）解：（1），················································2分同理，························································································3分.·································································································5分（2）法一：當(dāng)同號時，，.當(dāng)異號時，，，.····························································································9分綜上可知，的取值范圍為，的取值范圍為····················································10分且，······································································11分由（1）中的結(jié)論可知：.······································································································12分法二：令，則關(guān)于的函數(shù)在區(qū)間和上單調(diào)遞增，在和上單調(diào)遞減，的值域為.令，則的取值范圍為，···········································9分令函數(shù)，則在上單調(diào)遞減，在上單調(diào)遞增.所以函數(shù)的值域為，······················································································11分所以，故.·········································12分法三：令，則，令，則的取值范圍為，·········································7分又，所以.因為·································8分當(dāng)時，；當(dāng)時，.······························10分所以，又，所以，原命題即證.···························12分20．（12分）由題意知，下調(diào)電價后新增用電量為.故電力部門的收益，.（1）當(dāng)時，.················2分由題意知且.············3分化簡得.解得.或又.·····················································································5分答：實際電價最低定為時，仍可保證電力部門的收益比上年至少增長20%.·····································································································6分（2）當(dāng)時，.令，，.······································8分，···················10分當(dāng)且僅當(dāng)時取等號.故收益的最小值.·······································································12分21．（12分）解：（1）當(dāng)時，，當(dāng)即時，；··········1分②當(dāng)即時，；····················2分所以在上單調(diào)遞減，在上單調(diào)遞增，所以.···························································4分（2）記函數(shù)，由題意，當(dāng)時，都有，即在區(qū)間上單調(diào)遞增，··························································6分的對稱軸為，=1\*GB3①當(dāng)即時，要使得在區(qū)間上單調(diào)遞增，則需，解得，所以；························································································8分=2\*GB3②當(dāng)即時，在區(qū)間上不可能單調(diào)；···········9分=3\*GB3③當(dāng)即時，要使得在區(qū)間上單調(diào)遞增，則需，解得，所以；·······················································································11分綜上：或.···········································································12分22．（12分）（1）（法一）設(shè)，則，由得，化簡得恒成立，則，即；········································1分因為方程有兩個相等實根，所以，可得，..·····················································································3分（法二）由可得對稱軸為，又過點，因此設(shè)······················································