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厦门市2009~2010学年(下)高一质量检测数学参考答案一、选择题CBBDADCBCD二、填空题11.12.13.114.三、解答题15.(本小题满分10分)解:(Ⅰ)该几何体的直观图如图:┅┅┅┅┅┅┅┅┅5分(Ⅱ)该几何体是四棱锥,其底面的面积,┅┅┅┅┅┅┅┅┅7分高,K^S*5U.C#O%下┅┅┅┅┅┅┅┅┅┅┅┅┅┅┅┅┅┅┅8分则体积(体积单位)┅┅┅┅┅┅┅┅┅┅┅┅┅10分16.(本小题满分12分)K^S*5U.C#O%下解:(Ⅰ)∵不等式的解集为,∴和是方程,¥高#考#资%源*网则得,∴.┅┅┅┅┅┅┅┅┅┅┅┅┅6分(Ⅱ)函数┅┅┅┅┅┅┅┅┅┅┅┅┅┅8分,┅┅┅┅┅┅┅┅┅┅┅┅10分当且仅当时取“=”号.注意到,┅┅┅┅┅┅┅┅┅┅11分所以函数的最小值是,对应x的值是.┅┅┅┅┅┅12分17.(本小题满分12分)解:(Ⅰ)中,∵,∴,∴┅┅┅┅┅┅┅┅┅┅┅┅┅┅4分.┅┅┅┅┅┅┅┅┅┅┅┅┅┅┅┅┅┅┅┅┅┅6分(Ⅱ),又,∴,得,┅┅┅┅┅┅┅┅┅┅┅┅┅┅9分∴∴.┅┅┅┅┅┅┅┅┅┅┅┅┅┅12分B卷(共50分)甲卷四、填空题18.62519.20.21.21五、解答题22.(本小题满分10分)解:(Ⅰ)经过1小时后,甲船到达M点,乙船到达N点,,,,┅┅┅┅┅┅┅┅┅┅┅┅┅2分A岛B岛北EF∴,∴.┅┅┅┅┅┅┅┅┅┅┅┅┅┅4分(Ⅱ)设经过t()小时小船甲处于小船乙的正东方向.则甲船与A距离为海里,乙船与A距离为海里,,,┅┅┅5分则由正弦定理得,K^S*5U.C#O%下即,┅┅┅┅┅┅┅┅┅┅┅┅┅┅7分ABCDEPGM第23题O.┅┅┅┅┅┅┅┅9分答:经过小时小船甲处于小船乙的正东方向.┅┅┅┅┅┅┅┅┅10分23.(本题满分12分)(Ⅰ)证明:∵底面,∴,┅┅┅┅┅┅┅┅┅┅┅┅┅┅2分∵底面为正方形,∴,┅┅┅┅┅┅┅┅┅┅┅┅┅┅3分∵,∴平面.┅┅┅┅┅┅┅┅┅┅5分(Ⅱ)解:连结,取中点,连结.∵,平面平面,∴,┅┅┅┅┅┅┅┅┅┅┅┅┅┅8分在中,E为的中点,所以点O为AC的中点,在正方形中,是中点,则是MG中点,,,┅┅┅┅┅┅┅┅┅┅┅┅┅┅10分而,,所以.┅┅┅┅┅┅┅┅┅┅┅┅┅┅12分24.(本小题满分12分)解:(Ⅰ)∵(),,∴,┅┅┅┅┅┅┅┅┅┅┅┅┅┅2分即,即(),∴数列是以为首项、以2为公比的等比数列┅┅┅┅┅┅4分(Ⅱ)由(Ⅰ)知,所以,∴,┅┅┅┅┅┅┅┅┅┅┅┅┅┅5分∴┅┅┅┅┅┅┅┅┅┅┅┅┅7分记┅┅┅┅┅┅┅┅┅①则┅┅┅┅┅┅②∴①②得,∴┅┅┅┅┅┅┅11分所以.┅┅┅┅┅┅┅┅┅┅┅┅┅┅12分乙卷四、填空题18.540019.20.21.30五、解答题22.(本小题满分12分)解:仅依据所测得的数据,不能计算出山顶建筑物CD的高度.┅┅┅┅┅┅2分因为依据所测得的三个数据(,,),只能确定的形状与大小,图形中其余的量还是不确定的.┅┅┅┅┅┅┅┅┅┅┅4分例如150300BACDE第22题山坡的坡度(相对于水平面)显然是变量,┅┅┅┅┅┅┅┅┅┅┅┅┅┅6分则,在中,,所以,在中,由正弦定理得,┅┅┅┅┅┅┅┅8分∴与山坡的坡度有关,K^S*5U.C#O%下┅┅┅┅┅┅┅┅┅┅┅┅┅┅10分所以依据所测得的数据,不能计算出山顶建筑物CD的高度.说明:本题的其它角度的说理酬情给分.K^S*5U.C#O%下23.(本小题满分12分)K^S*5U.C#O%下解:(Ⅰ)∵底面,∴,┅┅┅┅┅┅┅┅┅┅┅┅┅┅1分∵底面为正方形,∴,K^S*5U.C#O%下ABCDEPG第23题OM∵,∴平面.∴是三棱锥的高.┅┅┅┅┅┅┅3分∵点G在BC边上且,∴,┅4分∵是的中点,∴,┅┅5分∴.┅┅┅┅┅┅┅┅┅┅┅┅┅┅6分(Ⅱ)在边上是否存在点M,.连结,取中点O,连结EO、GO,延长GO交于点,则.下面证明之.┅┅┅┅┅┅┅┅┅┅┅┅┅┅8分在中,E为的中点,点O为AC的中点,K^S*5U.C#O%下∴,又∵平面,平面∴.┅┅┅┅┅┅┅┅┅┅┅┅┅┅10分在正方形中,是中点,则是MG中点,,,而,,所以.┅┅┅┅┅┅┅┅┅┅┅┅┅┅12分24.(本小题满分12分)解:(Ⅰ)∵(),∴,┅┅┅┅┅┅┅┅┅┅┅┅┅┅2分即(),∴数列是以为首项、以2为公比的等比数列.┅┅┅┅┅┅3分(Ⅱ)由(Ⅰ)知,∴,K^S*5U.C#O%下∴┅┅┅┅┅┅┅┅┅┅┅┅5分∴.K^S*5U.C#O%下┅┅┅┅┅┅┅┅┅┅┅┅┅┅7分(Ⅲ),∵,∴K^S*5U.C#O%下∴即┅┅┅┅┅┅┅┅┅①┅┅┅┅┅┅┅┅8分∴┅┅②②①得┅┅┅┅┅┅┅┅┅③┅┅┅┅┅10分∴┅┅┅┅┅┅┅┅┅④K^S*5U.C#O%下④③得,┅┅┅┅┅┅11分则,即(),所以,