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高二数学参考答案第1页三明一中2020-2021学年高二下开学考数学参考答案一、单选题题号12345678答案DBCADCCA二、多选题9.AB10.ABD11.AC12.BC三、填空题13.3−14.②15.答案不唯一,只需满足20mn=+,或者0,20mn+
即可16.5;94四、解答题17.解:(1):26px−,................................................................................1分∵
p是q的充分条件,∴2,6−是2,2mm−+的子集,022426mmmm−−+,......................4分∴m的取值范围是)4,+........................................................
.....................5分(2)由题意可知,当5m=时,,pq一真一假,..........................................6分p真q假时,26,37,xxx−−或≤≤x,高二数学参考
答案第2页p假q真时,37,26,xxx−−或≤≤得)(3,26,7x−−,..............................9分所以实数x的取值范围是)(3,26,7−............................................
........10分18.解:(1)设圆的方程为,其中2240EFD+−.因为圆经过,,,所以....................................................................2分解得,,,.....
...........................................................5分所以所求圆的方程为......................................6分(
2)由(1)可得圆的圆心,半径................................8分所以圆心到轴的距离,所以轴被圆截得的弦长为..................12分19.解:(1)证明:因为
,分别是,的中点,所以DEPC∥...................................................................................................2分因为平面,PC平面,所以DE∥平
面......................................................................................4分(2)如图,在△中,过作于,则点为点在底面的正投影.................................
................................................................................5分M220xyDxEyF++++=M()2,3A−()1,6B−()6,7C49230,13660,
3649670,DEFDEFDEF+−++=+−++=++++=6D=−6E=−7F=−M226670xyxy+−−−=M()3,3M=5rMx3Mdy==xM22222598rd−=−=DEBCPBDEPACPACPA
CPADPPOAD⊥OOPABC高二数学参考答案第3页...................................................................................6分理由如下:因为,,是的中点,所以,.因为PDADD=,
PD平面PAD,AD平面PAD,所以平面....................................................................................8分因为平面,所以BC
PO⊥,.................................................................................................9分又,BC平
面ABC,AD平面ABC,BCADD=所以平面,...................................................................................11分故点为点在
底面的正投影...........................................................12分20.解法一:(1)设切点坐标为200(,)2xx,因为22yx=,yx=,所以曲线C
在点200(,)2xx处的切线方程为2000()2xyxxx−=−,..............2分又切线过点1,12P−,所以200011()22xxx−−=−,即20020xx−−=,解得0=2x或0=1x−,..
.................................................4分所以切点分别为(2,2)A,1(1,)2B−,PBPC=ABAC=DBCPDBC⊥ADBC⊥BC⊥PADPOPADPOAD⊥PO⊥ABCOPABC高二数学参考答案第4页所以
直线AB的方程为1222(2)21yx−−=−+,整理得220xy−+=.........6分(2)由(1)知曲线C在点200(,)2xx处的切线方程为2000()2xyxxx−=−,若切线过点(),1Pt−,则有20001()2xxtx−−=−,整
理得200220xtx−−=,................................................................................7分设切点211(,)2xAx,222(,
)2xBx,则1x,2x为方程2220xtx−−=的两根,所以有122xxt+=,122xx=−,.....................................................................9分又
直线AB的方程为22221121122()2xxxyxxxx−−=−−,即212122xxxxyx+=−,..............................................................
....................10分所以直线AB:1ytx=+,............................................................................11分所
以直线AB过定点(0,1)................................................................................12分解法二:(1
)依题意过点1,12P−与抛物线C相切的直线斜率一定存在,设其方程为:11()2ykx+=−,高二数学参考答案第5页由211(2)2ykxxy+=−=,,得222+0xxkk−+=,..........
.......................................2分由24(2)40kk−+==可得2k=或1k=−,...................................
........3分代回方程222+0xxkk−+=可解得2x=或1x=−,..................................4分故(2,2)A,1(1,)2B−,所以直线AB的方程为1222(2)21yx−−=−+,整理得220x
y−+=.........6分(2)依题意过点(),1Pt−与抛物线C相切的直线斜率一定存在,设PA:11()ykxt+=−,PB:21()ykxt+=−,由12)21(ykxtyx+==−,得21122+20
kkxxt−+=,①2114(22)40kkt−=+=得211=22kkt+,方程①即为221120xkxk−+=,解得1xk=,.....................................
..........7分即切点A的横坐标为1k,故211(,)2kAk,同理可得222(,)2kBk,.......................................................
.................................8分高二数学参考答案第6页故直线AB的方程为22221121122()2kkkyxkkk−−=−−,即212122kkkkyx+=−,9分因为211=
22kkt+,同理222=2+2kkt,即12,kk满足方程2220xtx−−=,由韦达定理得:121222kktkk+==−,........10分所以直线AB:1ytx=+,......................................................
......................11分所以直线AB过定点(0,1)...............................................................
.................12分21.解:(1)由1BC⊥平面ABC,AB平面ABC,得1ABBC⊥,.........1分又ABAC⊥,1CBACC=,故AB⊥平面1ABC,............
...................3分因为AB平面11ABBA,故平面11ABBA⊥平面1ABC...............................4分(2)以C为原点,CA为x轴,1CB为z轴,建立如图所示空间直角坐标系,则()0,0,0C,()1,
0,0A,()1,1,0B,...........................5分又2BC=,113BBAA==,故11CB=,()10,0,1B,10,0,2E,...............................
.................6分所以11,0,2AE=−,()111,1,1AABB==−−,()1,0,0CA=,........................................................
...........................................8分高二数学参考答案第7页设平面11AACC的一个法向量为(),,nxyz=,则100nCAnAA==,即00xxyz=−−+=,令1y=,则1z=,所以()0,1,1n=,
..............................................................................................10分设直线AE
与平面11AACC所成的角为,则1102sin101214nAEnAE===+,即直线AE与平面11AACC所成角的正弦值为1010......................................12分22.解:(1)因为与关于轴对
称,由题意知在上,..........................................................................1分当在上时,,.........................................
.....................................2分,,.......................................................................
...............3分当在上时,,,∴与矛盾,.............................................................................5分33,233,2−
x33,2M()2,0C2a=233144b+=23b=()2,3−M22231ab+=223314ab+=2249ba=ab高二数学参考答案第8页故椭圆的方程为...............................................
......................6分(2)由224143xmyxy=++=,,得:22(34)24360mymy+++=,........................7分2220(24)436(34)04mmm−+由,........
.........................8分设11)(,Axy,22)(,Bxy,则1212222436,3434myyyymm+=−=++,2121221144OAOBxxyymymyyy=+=+++(())()()21212=141
6myymyy++++2222236+369616343434mmmmm−+=+++()2212100=34mm−++................................................................10分2
116434m=−++,..........................................................11分∵24m∴23416m+,∴134,4OAOB−
,∴OAC的取值范围是134,4−.................................................................12分M22143xy+=