【文档说明】福建省三明市第一中学2020-2021学年高二下学期开学考试数学试题答案.pdf,共(8)页,287.747 KB,由管理员店铺上传
转载请保留链接:https://www.doc5u.com/view-fa94fa54a6671779e36fbc6b33c6c09d.html
以下为本文档部分文字说明:
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13.3−14.②1
5.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,DEFD
EFDEF+−++=+−++=++++=6D=−6E=−7F=−M226670xyxy+−−−=M()3,3M=5rMx3Mdy==xM22222598rd−=−=DEBCPBDEPACPACPACPADPPOAD⊥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PA
D,AD平面PAD,所以平面....................................................................................8分因为平面,所以BCPO⊥,.............................
....................................................................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2
00(,)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220xy−+=.........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+20kkxxt−+=,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AB
AC⊥,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10
0nCAnAA==,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−x33,2
M()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=1416myymyy++++2222236+369616343434mmmmm−+=+++()2212100=34mm−++.........................................
.......................10分2116434m=−++,..........................................................11分∵24m∴
23416m+,∴134,4OAOB−,∴OAC的取值范围是134,4−.................................................................12分M22143xy+=