蘇州市2022-2023學(xué)年高一上學(xué)期期中調(diào)研測(cè)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(cè)試卷高一數(shù)學(xué)2022.11注注意事項(xiàng)學(xué)生在答題前請(qǐng)認(rèn)真閱讀本注意事項(xiàng)及各題答題要求：1．本卷共4頁(yè)，包含單項(xiàng)選擇題（第1題~第8題）、多項(xiàng)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(shí)間為120分鐘．答題結(jié)束后，請(qǐng)將答題卡交回．2．答題前，請(qǐng)您務(wù)必將自己的姓名、調(diào)研序列號(hào)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(qǐng)?jiān)诖痤}卡上按照順序在對(duì)應(yīng)的答題區(qū)域內(nèi)作答，在其他位置作答一律無(wú)效．作答必須用0.5毫米黑色墨水的簽字筆．請(qǐng)注意字體工整，筆跡清楚．4．請(qǐng)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(xiàng)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(gè)選項(xiàng)中，只有一項(xiàng)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(gè)素?cái)?shù)，它的平方是偶數(shù)”的否定是A．任意一個(gè)素?cái)?shù)，它的平方是偶數(shù) B．任意一個(gè)素?cái)?shù)，它的平方不是偶數(shù)C．存在一個(gè)素?cái)?shù)，它的平方是素?cái)?shù) D．存在一個(gè)素?cái)?shù)，它的平方不是偶數(shù)3.若集合A的子集個(gè)數(shù)有4個(gè)，則集合A中的元素個(gè)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(gè)結(jié)論中正確的是A．的圖象關(guān)于原點(diǎn)對(duì)稱 B．的值域?yàn)镃．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(wú)關(guān)C．與無(wú)關(guān)，且與無(wú)關(guān) D．與無(wú)關(guān)，但與有關(guān)7.已知函數(shù)的圖象關(guān)于點(diǎn)成中心對(duì)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(duì)稱中心是A. B． C． D．8.若將有限集合A的元素個(gè)數(shù)記為card(A)，對(duì)于集合，，下列說(shuō)法正確的是A.若，則B.若，則或C.若，則D.存在實(shí)數(shù)，使得二、多項(xiàng)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(gè)選項(xiàng)中，有多項(xiàng)符合題目要求．全部選對(duì)的得5分，部分選對(duì)的得2分，有選錯(cuò)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?yàn)橛欣頂?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(duì)于定義域內(nèi)任意不相等的實(shí)數(shù)恒有；②對(duì)于定義域內(nèi)的任意兩個(gè)不相等的實(shí)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?yàn)锽．若，則函數(shù)的值域?yàn)镃．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(shuō)法正確的是A．函數(shù)，與函數(shù)，是同一個(gè)函數(shù)B．直線與函數(shù)的圖象至多有一個(gè)公共點(diǎn)C．滿足“值域相同，對(duì)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(duì)應(yīng)關(guān)系不同”的函數(shù)有無(wú)數(shù)個(gè)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(shí)數(shù)的最大值是▲．16.若函數(shù)的定義域?yàn)?，?duì)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(shuō)明、證明過(guò)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(shí)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(shí)數(shù)的值；（2）若關(guān)于的不等式的解集為，求實(shí)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(gè)”的問(wèn)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(qǐng)運(yùn)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(jià)為0.8元/（kW·h），年用電量為akW·h，本年度計(jì)劃將電價(jià)下降到0.55元/（kW·h）至0.75元/（kW·h）之間，而用戶期望電價(jià)為0.4元/（kW·h）.經(jīng)測(cè)算，下調(diào)電價(jià)后新增用電量和實(shí)際電價(jià)與用戶的期望電價(jià)的差成反比（比例系數(shù)為）.該地區(qū)的電力成本價(jià)為0.3元/（kW·h）.記本年度電價(jià)下調(diào)后電力部門的收益為（單位：元），實(shí)際電價(jià)為（單位：元/（kW·h））.（收益=實(shí)際電量（實(shí)際電價(jià)—成本價(jià)））（1）當(dāng)時(shí)，實(shí)際電價(jià)最低定為多少時(shí)，仍可保證電力部門的收益比上年至少增長(zhǎng)20%？（2）當(dāng)時(shí)，求收益的最小值.▲▲▲21.（12分）已知函數(shù)，.（1）當(dāng)時(shí)，，用表示，中的較大者，記為，求的最小值；（2）若不等式對(duì)任意，（）恒成立，求實(shí)數(shù)的取值范圍.▲▲▲22.（12分）已知二次函數(shù)的圖象經(jīng)過(guò)點(diǎn)，且=，方程有兩個(gè)相等的實(shí)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(xiàng)選擇題：本大題共8小題，每小題5分，共計(jì)40分．題號(hào)12345678答案ABACDBCC二、多選選擇題：本大題共4小題，每小題5分，共計(jì)20分．題號(hào)9101112答案BDBCBDABD三、填空題：本大題共4小題，每小題5分，共計(jì)20分．13．14．315．16．四、解答題：本大題共6小題，共計(jì)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(shí)數(shù)的取值范圍是.········································································10分18．（12分）解：（1）法一：因?yàn)椴坏仁降慕饧癁?，所以，······························································································?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?yàn)椴坏仁降慕饧癁椋?，，·················································································?分且即························································4分所以.················································································6分當(dāng)時(shí)，不等式的解集為，不滿足題意；································8分當(dāng)時(shí)，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(hào)時(shí)，，.當(dāng)異號(hào)時(shí)，，，.····························································································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?yàn)?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?yàn)椋ぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁ?1分所以，故.·········································12分法三：令，則，令，則的取值范圍為，·········································7分又，所以.因?yàn)椤ぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁ?分當(dāng)時(shí)，；當(dāng)時(shí)，.······························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(jià)后新增用電量為.故電力部門的收益，.（1）當(dāng)時(shí)，.················2分由題意知且.············3分化簡(jiǎn)得.解得.或又.·····················································································5分答：實(shí)際電價(jià)最低定為時(shí)，仍可保證電力部門的收益比上年至少增長(zhǎng)20%.·····································································································6分（2）當(dāng)時(shí)，.令，，.······································8分，···················10分當(dāng)且僅當(dāng)時(shí)取等號(hào).故收益的最小值.·······································································12分21．（12分）解：（1）當(dāng)時(shí)，，當(dāng)即時(shí)，；··········1分②當(dāng)即時(shí)，；····················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(shí)，都有，即在區(qū)間上單調(diào)遞增，··························································6分的對(duì)稱軸為，=1\*GB3①當(dāng)即時(shí)，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(shí)，在區(qū)間上不可能單調(diào)；···········9分=3\*GB3③當(dāng)即時(shí)，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(jiǎn)得恒成立，則，即；········································1分因?yàn)榉匠逃袃蓚€(gè)相等實(shí)根，所以，可得，..·····················································································3分（法二）由可得對(duì)稱軸為，又過(guò)點(diǎn)，因此設(shè)······················································