2022-2023學(xué)年度第一學(xué)期期中學(xué)業(yè)水平檢測高三數(shù)學(xué)評分標(biāo)準(zhǔn)一、單項(xiàng)選擇題：本大題共8小題．每小題5分，共40分．1-8：D C B A  A C D A  二、多項(xiàng)選擇題：本大題共4小題．每小題5分，共20分．9．ACD；     10．AD；     11．AB；     12．ACD．三、填空題：本大題共4小題，每小題5分，共20分．13．；           14．；           15．；     16．(1)；(2)．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）解：（1）由題知，····················································1分所以·······························································2分所以·······························································3分結(jié)合正弦定理，所以···················································4分（2）由（1）知：····················································5分所以，即，所以······················································7分解得或（舍）························································8分所以········································10分18．（12分）解：（1）由題知：····················································3分設(shè)點(diǎn)到平面的距離為，則，因?yàn)?/span> ，所以·························································5分（2）由題知：，······················································6分以為坐標(biāo)原點(diǎn)，直線，，分別為，，軸，建立空間直角坐標(biāo)系，則，·······························································7分設(shè)，則， 則直線的單位方向向量為···············································8分則點(diǎn)到直線的距離為 ··················································10分··········································11分所以的面積所以面積的取值范圍為················································12分19．（12分）解：（1）在中，由正弦定理知···········································1分所以，即····························································2分又因?yàn)?/span>，所以························································3分所以（舍）··························································4分（2）在中，，所以····················································5分又因?yàn)?/span>·····························································6分所以，·····························································7分又因?yàn)?/span>，所以························································8分在中，由余弦定理知：·················································9分所以，即···························································10分解得或（舍）·······················································11分所以，即···························································12分20．（12分）解：（1）由題知：平面，所以···········································1分因?yàn)?/span>平面平面，平面平面，平面，所以平面····························································4分因?yàn)?/span>平面，所以······················································5分（2）若選擇①因?yàn)?/span>平面，平面，平面平面所以，因此四邊形為平行四邊形，即為中點(diǎn)·································6分若選擇②因?yàn)?/span>平面，平面，所以，所以四邊形為平行四邊形，即為中點(diǎn)·······································6分所以，因?yàn)橹本€平面，所以直線與平面所成角為，所以··········································7分所以·······························································8分以為坐標(biāo)原點(diǎn)，分別以所在直線為軸建立空間直角坐標(biāo)系設(shè)，則·····························································9分，為平面的一個(gè)法向量················································10分設(shè)平面的一個(gè)法向量為，且，由，令，則，解得······························································11分設(shè)平面與平面所成銳二面角為，則 12分21．（12分）解:（1）由題知：，且·················································2分①當(dāng)時(shí)，有，所以，在上單調(diào)遞增，在上單調(diào)遞減，在上單調(diào)遞增···················································4分②當(dāng)時(shí)，有，所以在上單調(diào)遞增·····················································5分③當(dāng)時(shí)，有，所以，在上單調(diào)遞增，在上單調(diào)遞減，在上單調(diào)遞增···················································7分（2）由（1）知：若，當(dāng)時(shí)，，所以·······························································9分所以 ······························································10分······························································11分綜上，命題得證······················································12分 22．（12分）解:（1）若，則，·····················································1分所以·······························································2分令，所以當(dāng)時(shí)，，；當(dāng)時(shí)，，，；所以，對恒成立所以，在上單調(diào)遞增···················································3分又因?yàn)?/span> 所以，當(dāng)時(shí)，，在上單調(diào)遞減；當(dāng)時(shí)，，在上單調(diào)遞增；又因?yàn)?/span>，所以·······························································4分（2）若，則·························································5分由，得，·······························································6分令再令，則··························································7分若，令，則所以，當(dāng)時(shí)，，在上單調(diào)遞減；當(dāng)時(shí)，，在上單調(diào)遞增；所以，，得和則，滿足題意···················································8分若，則，不合題意·················································9分若，因?yàn)?/span>在上單調(diào)遞增，且···························································10分所以存在，使得，即，即·······················································11分所以，當(dāng)時(shí)，，在上單調(diào)遞減；當(dāng)時(shí)，，在上單調(diào)遞增；所以綜上，數(shù)的取值范圍是················································12分 

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