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永安一中2020-2021学年上学期期中考试高二数学试题答案和解析一、选择题123456789101112BDCDCCABABCBCACDABC二、填空题13.xxx11),,1[214.]65,4[15.216.112三、解答题17.解:
(1)设球半径为r,∴圆柱的高为r2,底面圆半径为r,334rV球,3222rrrV圆柱....................................4分∴图案中球与圆柱的体积比为:32圆柱球VV.................................
....5分(2)球半径cmr10,由题意得,3232000231cmrrV锥..................7分圆锥的母线长cmrrrl5105)2(22..............
.................8分22511005100100cmrlrS)(表.....................10分18、解:(1)当1a时,06522aaxx,解得:6
1x,............1分31x,解得:42x...........................................2分由p∧q为真命题,则41,4261xxx解得:................
........4分(2)由)(006522aaaxx;可得:axa6.......................5分p是q的必要不充分条件,则aaa6420........
.............................8分解得:2a.............................................................10分19.(1)证
明:∵直线01ymxl:经过定点)1,0(D.........................2分点D到圆心)2,0(的距离等于1小于圆的半径5,.............................4分故定点)1
,0(在圆的内部,故直线l与圆C相交.................................5分(2)设中点M的坐标为),(yx,则由直线和圆相交的性质可得CMAB.........7分由于定点)
1,0(D、圆心C、点M构成直角三角形,由勾股定理得222CDDMCM,22222)12()1()2(yxyx................9分0462222yyx,即41)23(22
yx.............................11分此圆在圆C:5)2(22yx的内部,故点M的轨迹方程为:41)23(22yx.................................12分20.证明:(1)BDC
AADB平面BDADB平面CA..................................1分又D是圆周上一点,AB是圆O的直径,DBDA...........................3分ACDCDBBCDBDACDBDACADACADDACADCABDCA
BDDA面面面面面面........................6分(2)连结OD,作,,DEEBCOE连结于ADBCA平面,,ABCCA平面ADBABC平面平面
..................................................7分ABODADBD,,又ADBOD平面,ABADBABC平面平面,,ABCOD平面.......
................................................8分ODBCABCBC,面又ODEBCEDEOEOEBC平面,,,DEBC的平面角为二面角DBCAOED..
......................10分又ADBDADAC,21,1,1230,63,42DEOEOD,所以二面角DBCA的余弦值为510...................................12分21.(1)证明:如图,连接CB1交1
BC于点O,连接OD,则点O为CB1的中点,又D是AC中点,ODAB//1.....................................2分DBCABDBCABDBCOD11111//面面面
.......................................4分(2)解:在三棱柱111CBAABC中,各个侧面均是边长为2的正方形,则CACBCBCCCACCCACBCAB,,211且BDCCCC11ABCBDAB
C面面D为线段AC的中点,BCAB,ACBD,又CACCC1,11AACCBD面............................6分BC1在面11AACC上的射影为DC1,BDC1即为直线BC1与平面11AACC所成的角,在,90,2,121111
DCCCCACCDCDCRt中,52211CDCCDC,同理:410225cos,221111BCDCBDCBC;........................8分(3)解:如图在DBC1内的平面区域(包括边界)存在一点E,使DMCE,此时点E是在线段DC1上,....
.......................................................10分证明如下:过C作DCCE1交线段DC1于E,由(2)可知11AACCBD面,而11AACCCE面,DBCCED
DCBDCEDCCEBD111面.....................11分又DBCDM1面,所以DMCE...........................................................12分22、
解:(1)设点00,Mxy,因为圆M与圆N关于直线0:20lxy对称,且1,2N,根据直线MN与直线0l垂直,,MN中点在直线0l上,得000021,11220,22yxxy解得000,1,xy
即0,1M,...........................3分所以543MP,3r...............................................4分所以圆22:19Mxye,圆22:129Nxye.........
.............6分(2)由题可知1:70lxy,设点,7Aaa,设,BC中点为Q假设23BC,则3BQ又∵3BM,90BQM,∴936MQ,∵BMQ与AMB相似,∴MQBMBMAM,.........................
........9分∴293626BMAMMQ,...........................................11分∴22360712aa,整理得24521202aa
,∵45144421602,所以方程无解,假设23BC不成立,所以32||BC...................................14分