2022—2023 学年上学期佛山市普通高中教学质量检测高二期末考试数学试卷参考答案

1 C 2 A 13 3 D 4 C 5 A 14 6 C 7 B 8 B 15 9 BC 10 AD 11 BCD 16 12 BC 3 417 18 （1）2xy30 （1）p1abc 2y3x （2）3 （2）4x2y10 3 431 9619 x2y2（1）1 43（1）（2）过定点(4,0)，证明过程详见解析 20 1 2（2）AB6 221 （1）以D为原点建立坐标系，圆弧AB的方程为x2(y6)2100(0≤y≤4) （2）此货车无法通过该洞口 22 （1）(x2)y4(3x≤4) 22（2）33423342 ,551．【答案】C

【解析】由图可知，直线l的倾斜角2．【答案】A

3． 4【解析】由ab，可得ab41(2)53x63x0，解得x2．

3．【答案】D

【解析】圆心坐标为(3,4)，半径r4．【答案】C

122

PP2，所以此圆的标准方程是(x3)(y4)2． 1221ab13【解析】b在a上的投影向量是2a(1,0,3),0,4． 44a5．【答案】A

2

C6301

【解析】依题意，取出的2个球都是红球的概率为P2，

Cn6(n6)(n5)3

(n6)(n5)90109，解得n4．

6．【答案】C

【解析】由l1//l2，1(a)2a(3a1)，且3a11，解得a0或7．【答案】B

【答案】第 1 页 共 10 页

1． 6【解析】由圆锥曲线的性质可知，kOMkABe1，即21e1，解得e

22

3．

8．【答案】B 【解析】如图，作AG//AF，且AGAF，连接FG，则四边形AAGF是平行四边形，由AAAE，AAAG，AEAGA，可得AA平面AEG，又FG//AA，FG平面AEG，

FGEG，EGEF2FG2EF2AA219，又EAG即为异面直线a，b所成的角

（或其补角），在△EAG中，由余弦定理可得cosEAG面直线a，b所成的角为

491912，EAG，所以异

223233．

EaAG9．【答案】BC

AbF 【解析】由n(AB)n(A)n(B)n(AB)，可得n(AB)3，所以事件A与事件B不互斥，A错误；

n(AB)n()n(AB)n()(n(A)n(AB))18(93)12，

P(AB)P(A)n(AB)2，B正确；

n()3111，P(B)，P(AB)，P(AB)P(A)P(B)，所以事件A与事件B相互独立，从而23621事件A与事件B相互独立，C正确；P(AB)1P(AB)1，D错误．

3310．【答案】AD

【解析】选项A，当25k9k，即k8时，曲线C:x2y217，表示半径为17的圆，A正确； 若曲线C为焦点在x轴上的椭圆，则其长轴长为225k，所以B错误；若曲线C为等轴双曲线，则

25k(9k)，该方程无解，所以C错误；若曲线C为焦点在y轴上的双曲线，则a29k，b2k25，其焦距为2c2a2b222k16，所以D正确．

11．【答案】BCD

p22【解析】设A(x1,y1)，B(x2,y2)，则由抛物线的性质可知x1x2，y1y2p，

4p23OAOBp2p20，AOB为钝角，所以A错误；

44

【答案】第 2 页 共 10 页

由AFx1正确；

px115，可得x14，代入抛物线方程并结合点A在x轴上方，可得y14，故B2ppp2p设AFx，则AF，BF，由AF2FB，可得，解得

1cos1cos1cos1cos122cos，sin，则直线AB的斜率为tan22，C正确；

33ppp

A,y1，B,y2，F,0，则FAFB(p,y1)(p,y2)p2y1y2p2p20，

222

FAFB，所以点F在以AB为直径的圆的圆上，设其圆心为D，则DADF，AAAF，AD为公共边，△AAD≌△AFD，AFDAAD90，即DFAB，所以以AB为直径的圆与直线AB相切于F．D正确．

ADB

12．【答案】BC

【解析】易证得B1D平面A1BC1，而平面A1BC1与平面A1EF相交，所以B1D与平面A1EF不垂直，A错误；因为平面ACD1//平面A1BC1，而平面A1BC1与平面A1EF相交，所以平面ACD1与平面A1EF也相交，所以B正确；以D为坐标原点建立如图所示空间直角坐标系，则A1(1,0,1)，E

11

,1,0，F0,,1，

22

111111

O1,,，A1E,1,1，A1F1,,0，AO0,,，则点O到直线A1E的距离为 1

222222

22AO1421A1E．C正确；设平面A1EF的法向量n(x,y,z)， AO1A1E2961nAExyz01A1On12则，可取n(2,4,3)，所以点O到平面A1EF的距离为，所1229nnAFxy0

1

2

以D错误，故选BC．

【答案】第 3 页 共 10 页

zD1A1ODAEBxCADFB1C1A1OEBD1FB1CyC1

13．【答案】

3 43

【解析】从长度为4，6，8，10的4条线段中任取3条，事件总数为C44，其中只有4，6，10不能构成三角形，所以这三条线段能构成一个三角形的概率为

114．【答案】abc

211【解析】BMBBBAAMAAABACabc．

2215．【答案】y3x

【解析】设双曲线的左焦点为F1，则AF1AF，PFPF12a，所以△APF的周长为

3． 4APAFPFAPAF12aPF≥2AF12a212a232a2a2152a，

当且仅当点P为线段AF1与双曲线的交点（如图）时等号成立，则由题可得2a152a10，

2a2155a，两边平方，解得a1，所以双曲线的渐近线方程为ybx，即y3x． a

16．【答案】 4x2y10

【解析】设P(x0,y0)(x00,y00)，设F1PF22，则tan22tan4，整理得：

1tan2312tan23tan20，(2tan1)(tan2)0，显然为锐角，故tan．由椭圆焦点三角形

2【答案】第 4 页 共 10 页

的面积公式可得：S△F1PF2btancy0，解得y0线方程为

2

33

，从而x01，P1,，过点P的椭圆的切22

xy11，其斜率为．又F1PF2的角平分线与点P的椭圆的切线垂直，所以角平分线的斜4223率k2，所以角平分线所在直线方程为y2(x1)，整理得：4x2y10．

2512··························3分 17．【解析】（1）由A(1,2)，B(3,0)可得中点M(2,1)，故kCM42因此边AB上的中线CM所在直线的方程为y12(x2)，整理得2xy30．·············5分 （2）由（1）知直线CM的方程为2xy30， 因此，点B到CM的距离为d230322(1)2

335，·············································7分 55又CM

(42)2(51)22025．································································8分

1135dCM253．··········································10分 225 所以△BCM的面积为S

18． 【解析】 （1） 设A1，A2分别表示甲两轮猜对1题，2题的事件，B1，B2分别表示乙两轮猜对1

21442

题，2题的事件．根据独立性假定，得P(A1)2，P(A2)．··············1分

33939P(B1)2p(1p)2p2p2，P(B2)p2．·······························································2分

设C“甲、乙两人在两轮竞答活动中答对3题”，

则CA1B2A2B1，且A1B2与A2B1互斥，A1与B2，A2与B1分别互相独立，

2

55，即P(A1B2)P(A2B1)P(A1)P(B2)P(A2)P(B1)，·······················4分 1212445322即(2p2p)p，结合0≤p≤1，可得p．······································6分 99124所以P(C)（2）设E1，E2，E3分别表示甲三轮猜对1题，2题，3题的事件，F1，F2，F3分别表示乙三轮猜对1题，2题，3题的事件．根据独立性假定，得

2112221482P(E1)3，P(E2)3，P(E3)．·····················7分

33393339327311933127327

P(F1)3，P(F2)3，P(F3)．·················8分

4446444464464

设D“甲、乙两人在三轮竞答活动中答对4题”，则DE1F3E2F2E3F1，···················9分

【答案】第 5 页 共 10 页

3

3

且E1F3，E2F2与E3F1两两互斥，E1与F3，E2与F2，E3与F1分别互相独立，所以

P(D)P(E1F3)P(E2F2)P(E3F1)P(E1)P(F3)P(E2)P(F2)P(E3)P(F1)

2274278931．·············································································11分 96496427649631因此，甲、乙两人在三轮竞答活动中答对4题的概率为．··············································12分

9619．【解析】（1）由于P3，P4两点关于x轴对称，故由题设知C经过P3，P4两点， 又由

1119知，C不经过点P1，所以点P2在C上，·······································1分 2222aba4b3

1,a24b2

因此解得2，···············································································3分

19b31,224bax2y2

故C的方程为1．························································································4分

43

（2）由（1）知F(1,0)，设直线l的方程为xmy1，A(x1,y1)，B(x2,y2)，则D(x1,y1)，(5分)

x2y26m9122

yyyy由4，得，所以，，(7分) (3m4)y6my903121222

3m43m4xmy1



因为直线BD的斜率kBD

y1y2yy2

，所以直线BD的方程为yy21(xx2),·············8分

x2x1x2x1

令y0，则y2

y1y2xyxy

(xx2)，即x2112，·················································9分

x2x1y1y2

有

(my11)y2(my21)y12my1y2

14，······························································11分

y1y2y1y2

所以直线BD恒过x轴上定点(4,0)．············································································12分 20．【解析】（1）因为平面ABC平面ACDE，平面ABC平面ACDEAC，ABAC，AB平面ABC，得AB平面ACDE．连接DA，则BDA即为BD与平面ACDE所成角．·········2分 在平面ACDE内，过D作DFAC于F，则CF313，DF，AF． 222因为平面ABC平面ACDE，平面ABC平面ACDEAC，DFAC，DF平面ACDE，得

DF平面ABC．····································································································3分

连接BF，由于BF平面ABC，有DFBF，

【答案】第 6 页 共 10 页

3322222．···············································4分 则BDABAFDF122AB11，即BD与平面ACDE所成角的正弦值为．··6分 因此，在Rt△ABD中，sinBDABD2222zDExCFByA

（2）以A为原点，建立空间直角坐标系Axyz如图所示，设ABt，

3133则B(t,0,0)，C(2,0,0)，D,0,，， 2E2,0,22

131333ED(1,0,0)，，，，··················8分 BE,t,BD,t,CD,0,222222

13z10BEm0x1ty1设平面BDE的法向量为m(x1,y1,z1)，则，即2， 2x0EDm0

1

令y13，得m(0,3,2t)，··················································································9分

33xtyz2022BDn022设平面BCD的法向量为n(x2,y2,z2)，则，即， 

1x3z0CDn0

2222令y223，得n(3t,23,t)，·············································································10分

mn62t22由于平面BDE与平面BCD的夹角为，所以cos，·11分 2244mn234t124t解得t

66，因此AB的长为．·············································································12分 22【答案】第 7 页 共 10 页



21．【解析】法一：（1）如图，以圆弧AB所在圆的圆心为坐标原点，AB方向为x轴建立平面直角坐标系，

······························································································································1分

B(8,R4)，C(0,R)，D(0,R4) 设圆弧AB所在圆的半径为R，则AB16，则A(8,R4)，CD4，

······························································································································2分 在Rt△BOD中，ODBDOB，即(R4)282R2，则R10，··························4分 故C(0,10)，D(0,6)，从而圆弧AB的方程为xy100(6≤y≤10)．·························6分

2

2

2

2

2

（2）如图，由题意知，隔墙EM2，车宽MP2．·····················································7分 过P作x轴的垂线交圆弧AB于Q，垂足为H，则OH3．············································8分 连接OQ，在Rt△OQH中，HQ2OQ2OH2，即HQ210232，则HQ故PQHQHP91．·······9分

9163.6，·············································································11分

从而，此货车无法通过该洞口．····················································································12分 法二：（1）如图，以点D为坐标原点，线段AB所在直线为x轴建立平面直角坐标系，···········1分 由题知，点C，B坐标分别为(0,4)，(8,0)，设圆心坐标为(0,b)，圆的半径为r，则圆弧AB所在圆的方程为x(yb)r，··························································································3分

2

2

2

02(4b)2r22

r100，从而圆弧AB所在圆的方程是因为点C，B都在圆上，有2，得，b622

8(0b)rx2(y6)2100，·································································································5分

故圆弧AB的方程为x(y6)100(0≤y≤4)．·····················································6分

2

2

【答案】第 8 页 共 10 页

（2）此货车不能通过该路口．······················································································7分 由题知，隔墙在y轴右侧1米，车宽2米，车高3.6米，所以货车右侧的最高点坐标为(3,3.6)，(9分) 代入x2(y6)2100(0≤y≤4)后得32(3.66)2100，

因此，此货车无法通过该洞口．····················································································12分 法三：（1）如图，以点A为坐标原点，线段AB所在直线为x轴建立平面直角坐标系，············1分 由题意，点A，C坐标分别为(0,0)，(8,4)，设圆心坐标为(8,b)，圆的半径为r，则圆弧AB所在圆的方程是(x8)2(yb)2r2，

(08)2(0b)2r22因为点A，C都在圆上，有，得，r100， b6222

(88)(4b)r从而圆弧AB所在圆的方程是(x8)2(y6)2100，···················································5分 故圆弧AB的方程是(x8)2(y6)2100(0≤y≤4)．··············································6分

（2）此货车不能通过该路口．······················································································7分 由题知，隔墙在道路中间右侧1米，车宽2米，车高3.6米，

所以货车左侧的最高点坐标为(5,3.6)，··········································································9分 代入(x8)2(y6)2100(0≤y≤4)后得32(3.66)2100，

因此，此货车无法通过该洞口．····················································································12分 22．【解析】（1）由题意，圆C:x2y28x120的圆心为(4,0)，半径为2．··················1分 若直线l1的斜率为0，线段AB的中点为M(4,0)；···························································2分 若直线l1的斜率不为0，设M(x,y)，则kABkMC1，···················································3分 即

yy1，得(x2)2y24，········································································4分 xx4设与圆C:x2y28x120与(x2)2y24交于点E和点F，则

(x2)2y24由2，得E(3,3)，F(3,3)，·····················································5分 2

xy8x120【答案】第 9 页 共 10 页

因此，线段AB的中点M的轨迹的方程为(x2)2y24(3x≤4)．··························6分 （2）由（1）知，轨迹即为劣弧EF，

若直线l2:ykx上存在点P，使得以点P为圆心，2为半径的圆与有公共点，则

当k0时，存在圆P:(x2)y4满足条件；·························································7分 当k0时，只需点E(3,3)到l2:ykx的距离d2，

2

2

即3k3k212，得0k

3342；·····································································9分 5当k0时，点F(3,3)到l2:ykx的距离d2，

即3k3k212，得3342k0；···································································11分

5因此，k的取值范围为33423342．······················································12分 ,55

【答案】第 10 页 共 10 页