數(shù)學(xué)新高考-高一數(shù)學(xué)期中試卷參考答案

Word版本見：高考高中資料無水印無廣告word群559164877新高考資料全科總?cè)?32599440；高考數(shù)學(xué)高中數(shù)學(xué)探究群5622984952022~2023學(xué)年第一學(xué)期期中測試卷 高一數(shù)學(xué)參考答案 2022.11一、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)椴坏仁降慕饧癁椋?，······························································································?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，又過點(diǎn)，因此設(shè)··································································1分因?yàn)榉匠逃袃蓚€(gè)相等實(shí)根，所以，可得.····················································································3分（2）=1\*GB3①在單調(diào)遞減，在單調(diào)遞增.證明：任取，則·························································4分當(dāng)時(shí)，，，則，在單調(diào)遞增；當(dāng)時(shí)，，，則，在單調(diào)遞減.因此在單調(diào)遞減，在單調(diào)遞增.············································6分=2\*GB3②令，則.因?yàn)?，所以，?dāng)且僅當(dāng)時(shí)取等號(hào)，所以.設(shè)，······································································7分1）當(dāng)時(shí)，，在上單調(diào)遞增，······································