2019-2020学年广东省佛山市高二上学期期末数学试题(含答案解析).pdf
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1、2019-2020 学年广东省佛山市高二上学期期末数学试题一、单选题1已知直线l经过点1,2P,且倾斜角为135,则直线l的方程为()A30 xyB10 xyC10 xyD30 xy【答案】B【解析】由倾斜角求出斜率,写出直线方程的点斜式,化成一般式.【详解】直线l倾斜角为135,则斜率为-1,且经过点1,2P,直线l方程为2(1)yx,即10 xy.故选:B【点睛】本题考查求直线方程,属于基础题.2已知命题p:,则为A,B,C,D,【答案】D【解析】根据全称命题的否定为特称命题可得答案【详解】解:命题p:,则为,故选:D【点睛】本题考查的知识点是全称命题,命题的否定,熟练掌握全特 称命题的否
2、定方法是解答的关键3已知抛物线2yx 上的点M到其焦点的距离为2,则M的横坐标是()A32B52C74D94【答案】C【解析】求出抛物线的准线方程,设点M的横坐标,利用抛物线的定义,即可求解.【详解】第 1 页,共 16 页抛物线2yx焦点1(,0)4F,准线方程为14x,设点M的横坐标为0 x,根据抛物线的定义,0017|2,44MFxx.故选:C【点睛】本题考查抛物线定义在解题中的应用,属于基础题.4圆22460 xyx与圆22460 xyy的位置关系为()A外离B相切C相交D内含【答案】C【解析】求出两圆的圆心和半径,判断圆心距和两半径和与差的绝对值的关系,即可得出结论.【详解】2246
3、0 xyx化为22(2)10 xy,圆心1(2,0)C,半径110r;22460 xyy化为22(2)10 xy,圆心2(0,2)C,半径210r,120|2 22 10C C,所以两圆相交.故选:C【点睛】本题考查两圆的位置关系,属于基础题.5过点3,2的双曲线C的渐近线方程为0 xy,则C的方程为()A221xyB225xyC221yxD225yx【答案】B【解析】根据渐近线方程,设出双曲线方程,将点3,2代入,即可求解.【详解】双曲线C的渐近线方程为0 xy,设双曲线C的方程为22(0)xy第 2 页,共 16 页文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F
4、6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6
5、HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F
6、6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6
7、HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F
8、6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6
9、HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F
10、6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9将点3,2代入,得5.故选:B【点睛】本题考查已知双曲线渐近线方程求标准方程,合理设双曲线方程是解题的关键,属于基础题.6函数21fxxa,则“0a”是“01,1x,使00fx”的()A充分不必要条件B必要不充分条件C充分必要条件D既不充分也不必要条件【答案】A【解析】先求出“01,1x,使00fx”成立时,a的取值范围,与“0a”比较,即可得出结论.【详解】01,1x,使00fx即200()10f xxa,需max1,1,()1
11、01xf xaa,“0a”是“01,1x,使00fx”的充分不必要条件.故选:A【点睛】本题考查充分不必要条件的判断,考查存在不等式成立,属于基础题.7已知m是平面的一条斜线,直线l过平面内一点A,那么下列选项中能成立的是()Al,且lmBl,且lmCl,且lmDl,且lm【答案】A【解析】将选项 BCD 一一当做条件,都会得出与题中矛盾的结论,故选项BCD 错误,选项A 得不出矛盾,选项 A 正确.【详解】解:若l,且lm,则m或m,不符合题意,选项B 错误;若l,且lm,则m,不符合题意,选项C 错误;若l,且lm,则m,不符合题意,选项D 错误.故选:A.【点睛】本题考查了空间中线面平行
12、与垂直关系的判定与性质,属于基础题.8正四棱柱1111ABCDA B C D中,12AAAB,则异面直线1AD与1B D所成角的余弦值为()第 3 页,共 16 页文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 H
13、S5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6
14、X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 H
15、S5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6
16、X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 H
17、S5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6
18、X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9A110B110C3010D3010【答案】D【解析】建立空间直角坐标系,求出11,A DD B坐标,利用空间向量法,求出11,AD DBuuuu r uuuu r所成角余弦的绝对值,即为所求.【
19、详解】设122AAAB,以D为坐标原点,以1,DA DC DD所在的直线分别为,x y z轴,建立空间直角坐标系Oxyz,则11(1,0,0),(0,0,2),(1,1,2)ADB,11(1,02),(1,1,2)ADDBu uu u ruu uu r,111111330cos,1056ADDBADDBADDBuu uu r uuu u ruuuu r u uu u ruu uu ruuuu r.因此,异面直线1AD与1DB所成角的余弦值为3010.故选:D【点睛】本题考查用空间向量法求异面直线所成的角,属于基础题.9如图,长方体1111ABCDA B C D中,4ABBC,12 2BB,点E
20、,F,M分别为11A B,11A D,11B C的中点,过点M的平面与平面AEF平行,且与长方体的面相交,则交线围成的几何图形的面积为()第 4 页,共 16 页文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS
21、5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X
22、9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS
23、5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X
24、9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS
25、5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X9文档编码:CL9H6H4K9Y6 HS5D3G9H7T7 ZZ2G3V2F6X
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