2019-2020学年福建省泉州市高二上学期期末教学质量跟踪监测数学答案

泉州市普通高中2019-2020学年度上学期教学质量跟踪检测

数学（空间坐标系、空间向量、直线与圆、圆锥曲线、数列）

参及评分标准

一、单项选择题：本大题共10小题，每小题5分，共50分，在每小题给出的四个选项中，只有一项是符

合题目要求的．请将答案填在答题卡对应题号的位置上．题号123456710答案

C

B

C

D

A

C

C

D

D

B

8．选D ．解析：因为00()P x y ，到焦点的距离02d y =+，则0022y y +=，解得02y =．9．选D .

解析：因为直线0=+y x 与直线02=++y x 互相平行，

所以两直线之间的距离21

1|02|2

2

=+-=

d ，

由题意，圆C 与两直线相交，四个交点围成的四边形为正方形，则两平行线之间的距离即为正方形的边长，正方形的对角线即圆的直径．设圆的半径为r ，有222)2()2()2(+=r ，解得1=r ，

本题考查直线与圆的综合应用问题，属于中档题．

10．选B ．

解析：当α与底面趋于平行时，τ几乎成为一个圆，

因此离心率可以充分接近0．

当α与底面的夹角最大时，τ的离心率达到最大，下面求解这一最大值．

如图，A ，B 为长轴，F 为焦点时，

e 最大．2a c BF BG +===，易知1b =，

所以5434

a c ⎧=⎪⎪⎨⎪=⎪⎩，35

e =．则离心率的取值范围是305⎛⎤ ⎥⎝⎦，．

二、多项选择题：

题号1112答案

AC

BCD

11．解析：设等差数列}{

n a 的公差为d ，则1113(4)721a a d a d ++=+，解得13a d =-，

所以1(1)(4)n a a n d n d =+-=-，所以40a =，故A 正确；因为61450S S a -==，所以16S S =，故C 正确；

由于d 的正负不清楚，故3S 可能为最大值或最小值，故B 不正确；因为35420a a a +==，所以35a a =-，即35a a =，故D 错误．

12．解析：对于选项A ，()

1112

AD A A B A BC =-+

，选项A 错误；对于选项B ，过点D 作1AA 的平行线交11A C 于点1D ．

以D 为坐标原点，DA ，DB ，1DD

分别为x ，y ，z 轴的正方向建立空间直角坐标系Oxyz ．设棱柱底面边长为a ，侧棱长为b ，则002a A ⎛⎫

⎪⎝⎭，，002B a ⎛⎫ ⎪ ⎪⎝⎭，，102B a b ⎛⎫ ⎪ ⎪⎝⎭，，102a C b ⎛

⎫

- ⎪⎝⎭，所以122a BC b ⎛⎫=-- ⎪ ⎪⎝⎭ ，，122a AB b ⎛⎫=- ⎪ ⎪⎝⎭

，．11BC AB ⊥ ，110BC AB ∴⋅=，即2

2

2

3+=022a a b ⎛⎫⎛⎫-- ⎪ ⎪ ⎪⎝⎭⎝⎭

，解得22b a =．因为DE 平面11ABB

A ，则动点E 的轨迹的长度等于1B

B =

．选项B 正确．对于选项C ，在选项A 的基础上，002a A ⎛⎫

⎪⎝⎭，，3002B a ⎛⎫ ⎪ ⎪⎝⎭，，()000D ，，1022a C a ⎛⎫- ⎪ ⎪⎝⎭，所以002a DA ⎛⎫

= ⎪⎝⎭

，，1222a BC a ⎛⎫=- ⎪ ⎪⎝⎭ ，．因

为2

1112cos 6a BC DA BC DA BC DA ⎛⎫ ⎪⋅<>==

，．选项C 正确．

对于选项D ，点E 的轨迹为抛物线的一部分．选项D 正确．

三、填空题：本大题共4小题，每小题5分，共20分．每小题有难度明显差异的两空，第一空占2分，

第二空占3分．请将答案填在第4页对应题号的位置上．

13．(01)-，

，1y =．14．2，63．

153y =或3430x y -+=．

第二空解析：设切线长为L ，则L =

，所以当切线长L 取最小值时，PC 取最小值，

过圆心()02C ，作直线l 的垂线，则点P 为垂足点，此时，直线PC 的方程为063=+-y x ，

联立⎩⎨⎧=+-=-+0630123y x y x ，得3

3x y =⎧⎨=⎩

，点P 的坐标为()33，

．由题意可得

11

|

332|2=+-+-k k ，解得0=k 或4

3

=

k ，此时，所求切线的方程为3=y 或0343=+-y x ．

本题考查过点的圆的切线方程的求解，考查圆的切线长相关问题，在过点引圆的切线问题时，要对直线的斜率是否存在进行分类讨论，另外就是将直线与圆相切转化为圆心到直线的距离等于半径长，考查分析问题与解决问题的能力，属于中等题．16．376，760．

解析：集合4

3

2

1

443210{|22222}A n n a a a a a ==⨯+⨯+⨯+⨯+⨯，

当4012310a a a a a =====，时，16n =；当012341a a a a a =====时，31n =；

所以4{16171831}A = ，

，，共有16个元素，故激活码为16(1631)

3762

⨯+=；

结合二进制表示，当5k =时，{}n c 的各项可以看成首位为1的六位二进制数，对于41a =，符合条件()1f x =的有8个数，

同理对于32101111a a a a ====，，时，符合条件的也分别是8个数，故激活码为(

)5

43210

162822222

760⨯+++++=．

四、解答题：本大题共6小题，共70分．解答应 出文字 明， 明过程或演算步 ．

17．（本小题满分10分）解：（1）设{}n a 的公差为d ，

因为3a 是1a 与7a 的等比中项，

所以2317a a a =⋅．························································································1分因为12a =，

所以2

(22)2(26)d d +=⋅+．·········································································2分解得1d =，··································································································4分所以1n a n =+．····························································································5分（2）因为(21)(3)

22

n n n n n S +++=

=

，··································································7分令

(3)

272

n n +=，·························································································8分则2

3540n n +-=，······················································································9分解得6n =．

所以存在n 值为6，使得n S 的值为27．·····························································10分（注：第（2）问用其它方法求得6n =同样给分）

18．（本小题满分12分）解：（1）法一：

线段AB 中垂线为2x =，···············································································2分又圆心C 在直线3y =上，

故圆心C 为(23)，．·······················································································4分

半径r =

=·································································5分

所以圆C 方程为22(2)(3)5x y -+-=．····························································6分

法二：设圆心(3)C a ，圆C 方程：222()(3)x a y r -+-=．········································1分

由(42)A ，(02)B ，在C 上有

222

222

(4)(23)(23)a r a r ⎧-+-=⎪⎨+-=⎪⎩，

，

···············································································3分

解得2a r =⎧⎪⎨=⎪⎩，································································································5分

∴圆C 的方程为2

2

(2)(3)5x y -+-=．·························································6分

（2）由已知得圆心C 到直线l

距离为1d ==．···················································8分

①当l 斜率不存在时，直线方程为1x =，此时圆心到直线l 距离为1，符合题意．········9分②当l 斜率存在时，设l ：1(1)y k x -=-，即10kx y k --+=．

圆心C 到直线l

距离1d =

=，解得3

4

k =

．·····························11分故l 方程为3

1(1)4

y x -=

-，即3410x y -+=．综合①②有，所求直线方程为1x =或3410x y -+=．·······································12分

19．（本小题满分12分）

思路一

（1）解：设所求抛物线方程为2

2y px =，································································2分

依题意得36p =，···················································································3分解得2p =，···························································································4分故所求抛物线方程为24y x =.····································································5分

（2）联立方程组244y x y x ⎧=⎨=-⎩，

消去x ，得2

4160y y --=，①·······························6分

设()11A x y ，()22B x y ，由方程①得：1216y y =-，································7分

又2112

2244y x y x ⎧=⎪⎨=⎪⎩，

则2212121616y y x x ==，·····················································9分所以()()112212120OA OB x y x y x x y y ⋅=⋅=+=

，，·······································11分

所以OA OB ⊥.······················································································12分

思路二：（1）同思路一

（2）联立方程组244y x y x ⎧=⎨

=-⎩，

，

消去y ，得2

12160x x -+=，①·······························································6分设()11A x y ，()22B x y ，由方程①可得：12121216x x x x +=⎧⎨

⋅=⎩，

，

·········································································7分

由()11OA x y = ，()22OB x y =

，

故1212OA OB x x y y ⋅=+

，·············································································8分

又114y x =-，224y x =-，·········································································9分故()12121241616y y x x x x =-++=-，··························································10分

则0OA OB ⋅=

，························································································11分

所以OA OB ⊥．·························································································12分

20．（本小题满分12分）

解：[解法一]（1）因为1(1)1n n na n a +-+=，两边同除以(1)n n +，得111(1)

n n a a n n n n +-=++．1分令n n a b n =，则11(1)

n n b b n n +-=+．由：213243112123134

b b b b b b -=

⨯-=⨯-=⨯ ，，11(1)n n b b n n

--=-·····································································3分得11111122334(1)n b b n n

-=++++⨯⨯⨯-⨯ 1111111(1)(()()223341n n

=-+-+-++-- 11n =-．···········5分又11

11b a ==，所以12n b n =-．························································6分（2）21n n a nb n ==-，令2(21)2n n n n C a n =⋅=-⋅，·································7分

所以1231123252(23)2(21)2n n n S n n -=⋅+⋅+⋅++-⋅+-⋅ ，·········8分

23412123252(23)2(21)2n n n S n n +=⋅+⋅+⋅++-⋅+-⋅ ，·············9分

①—②得231(12)22(222)(21)2n n n S n +-=++++--⋅ ，············10分

11

11

14(12)22(21)212

28(21)(21)2(23)26n n n n n n S n n n -+-++--=+⨯--⋅-=+---⋅=--⋅-，

所以1(23)26n n S n +=-⋅+．···························································12分

[解法二]

（1）因为1(1)1n n na n a +-+=，

所以11(1)(1)n n n n na n a n a na +--+=--，····················································1分

所以112n n n na na na +-=-，·······································································2分

所以112n n n a a a +-=+，

所以112n n n a a a +-=+，

所以{}n a 为等差数列，·············································································3分

又当1n =时，2121a a -=且11

a =所以23a =，··························································································4分

所以212d a a =-=，

所以21n a n =-，····················································································5分所以12n n a b n n =

=-．··············································································6分（2）同解法1

（注：第（1）问通过求123b b b ，发现12n b n

=-，给2分）

21.（本小题满分12分）

解法一：

明：（1）连接DO ．设4EH a =，则EF =，翻折后的4BD DE FB a =+=．1分

在SAC △中，SA SC ==，4AC a =，O 为AC 的中点，2SO a ∴=．········2分又 在SOB △中，2BS a =，SP BO ⊥，P ∴为BO 的中点，·····························3分SP DO ∴ ．·······························································································4分SP ⊄ 平面ACD ，DO ⊂平面ACD ，

SP ∴ 平面ACD ．······················································································5分

解：（2）B DMN D BMN V V --= 且三棱锥D BMN -的高为定值，

BMN S ∴△最大时，三棱锥B DMN -的体积取得最大值．········································6分

设AM BN x ==（0x ≤≤），

()22111

sin )sin 3sin 222BMN S BM BN MBN x x MBN x a MBN ⎡⎤∴=⋅⋅∠=-∠=--+∠⎢⎥⎣⎦

△又sin MBN ∠ 为定值，

∴当x =时，BMN S △最大，即三棱锥B DMN -的体积最大．

此时M ，N 分别是AB ，BC 上的中点．···························································7分由（1）可得SP DO ，SP BO ⊥，DO BO ∴⊥．

DA DC = ，BA BC =，DO AC ∴⊥，BO AC ⊥．········································8分

以O 为坐标原点，OA ，OB ，OD 分别为x ，y ，z 轴的正方向建立空间直角坐标系Oxyz ，

则(200)A a ，，(00)B ，，(200)C a -，，(00)D ，，(0)M a ，

()

0N a =-，

，()DM a =- ，(200)NM a = ，

，(20)DA a =- ，

，(20)AB a =- ，．··················································································9分设平面DMN 的一个法向量为1111()x y z =，n ．

00DM NM ⎧⋅=⎪∴⎨⋅=⎪⎩ ，n

n 1111020ax ax ⎧+-=⎪∴⎨=⎪

⎩，取11z =，则12y =，10x =，

∴平面DMN 的一个法向量为1(021)=，n ．····················································10分设平面DAB 的一个法向量为2222()x y z =，n ．

00DA AB ⎧⋅=⎪∴⎨⋅=⎪⎩ ，n

n 22222020ax ax ⎧-=⎪∴⎨-+=⎪⎩，

取2x =221y z ==，

∴平面DAB

的一个法向量为211)=n ．···················································11分

则12121235cos 10

⋅<>==，n n n n n n ．所以平面DAB 与平面DMN

所成锐二面角的余弦值为

10．································12分解法二：

明：（1）连接DO ．

设4EH a =

，则EF =

，DO BO ==

，AB BC ==，

翻折后的4BD DE FB a =+=．······································································1分222BD DO BO =+ ，DO BO ∴⊥．·····························································3分又 SP BO ⊥，且DO ，SP ⊂平面BOD ，SP DO ∴ ．·······························4分又因为SP ⊄平面ACD ，DO ⊂平面ACD ，SP ∴ 平面ACD ．······················5分

解：（2）B DMN D BMN V V --= 且三棱锥D BMN -的高为定值，

BMN S ∴△最大时，三棱锥B DMN -的体积取得最大值．··········································6分设AM BN x ==

（0x ≤≤）

，

()22111sin )sin 3sin 222BMN S BM BN MBN x x MBN x a MBN ⎡⎤∴=⋅⋅∠=-∠=--+∠⎢⎥⎣⎦

△又sin MBN ∠ 为定值，

∴

当x =时，BMN S △最大，即三棱锥B DMN -的体积最大．

此时M ，N 分别是AB ，BC 上的中点．···························································7分DA DC = ，BA BC =，DO AC ∴⊥，BO AC ⊥．········································8分

以O 为坐标原点，OA ，OB ，OD 分别为x ，y ，z 轴的正方向建立空间直角坐标系Oxyz ，

则(200)A a ，

，(00)B ，，(200)C a -，

，(00)D ，

，(0)M a ，

，

()

0N a =-，

，()DM a =- ，(200)NM a = ，

，(20)DA a =- ，

，(20)AB a =- ，．··················································································9分设平面DMN 的一个法向量为1111()x y z =，n ．

00DM NM ⎧⋅=⎪∴⎨⋅=⎪⎩ ，n

n 1111020ax ax ⎧+-=⎪∴⎨=⎪

⎩，取11z =，则12y =，10x =，

∴平面DMN 的一个法向量为1(021)=，n ．····················································10分设平面DAB 的一个法向量为2222()x y z =，n ．

00DA AB ⎧⋅=⎪∴⎨⋅=⎪⎩ ，n

n 22222020ax ax ⎧-=⎪∴⎨-+=⎪⎩，①，

②

取2x =221y z ==，

∴平面DAB

的一个法向量为211)=n ．···················································11分

则12121235cos 10

⋅<>==，n n n n n n ．所以平面DAB 与平面DMN

所成锐二面角的余弦值为

10．································12分解法三：

（1） 明：连接DO ．

设4EH a =

，则EF =

，DO BO ==

，AB BC ==，

翻折后的4BD DE FB a =+=．·············································································1分DA DC = ，BA BC =，DO AC ∴⊥，BO AC ⊥．···········································2分222BD DO BO =+ ，DO BO ∴⊥．···································································4分

以O 为坐标原点，OA ，OB ，OD 分别为x ，y ，z 轴的正方向建立空间直角坐标系Oxyz ，

则(000)O ，

，(00)B ，

，(00)D ，

，(0)S

，(00)P ，．5分

∴(00)PS = ，．

取平面DAC

的一个法向量为(00)OB = ，，··················································6分

0PS OB ⋅= 且SP ⊄平面ACD ，

SP ∴ 平面ACD ． (7)

分

（2）B DMN D BMN V V --= 且三棱锥D BMN -的高为定值，

BMN S ∴△最大时，三棱锥B DMN -的体积取得最大值．···············································8分设AM BN x ==

（0x ≤≤）

，

()22111sin )sin 3sin 222BMN S BM BN MBN x x MBN x a MBN ⎡⎤∴=⋅⋅∠=-∠=--+∠⎢⎥⎣⎦△又sin MBN ∠ 为定值，

∴

当x =时，BMN S △最大，即三棱锥B DMN -的体积最大．

此时M ，N 分别是AB ，BC 上的中点．·································································9分由（1）得(200)A a ，，(200)C a -，

，(0)M a ，

，()0N a -，

，()DM a =- ，(200)NM a = ，

，(20)DA a =- ，

，(20)AB a =- ，．设平面DMN 的一个法向量为1111()x y z =，n ．

00DM NM ⎧⋅=⎪∴⎨⋅=⎪⎩ ，n

n 1111020ax ax ⎧+-=⎪∴⎨=⎪

⎩，取11z =，则12y =，10x =，

∴平面DMN 的一个法向量为1(021)=，n ．··························································10分设平面DAB 的一个法向量为2222()x y z =，n ．

00DA AB ⎧⋅=⎪∴⎨⋅=⎪⎩ ，n

n 22222020ax ax ⎧-=⎪∴⎨-+=⎪⎩，

取2x =221y z ==，

∴平面DAB

的一个法向量为211)=n ．·························································11分

则12121235cos 10

⋅<>===，n n n n n n ．所以平面DAB 与平面DMN 所成锐二面角的余弦值为

3510．······································12分

22．（本小题满分12分）

解：（1）设()P x y ，

，由已知得：()0PA y k x x -=≠

，()0PB y k x x +=≠，1分则有3332

y y x x +⋅=-，·········································································2分化简得：()22

1032

y x x +=≠，··········································································3分由椭圆的定义可知，存在定点()101F -，()201F ，使得12PF PF +为定值．············4分

（2）由于C ，D ，E 在τ上，所以AC ，AD ，AE 斜率存在，

由条件，011AD AC AD AE AE AC k k k k k k +=⎫⇒⋅=⎬⋅=-⎭

．······························································5分设:DE l y kx m =+，()11D x kx m +，()22E x kx m +，联立2212

3y kx m x y =+⎧⎪⎨+=⎪⎩消去y 得()22326x kx m ++=，()222234260k x kmx m +++-=，..................................................................7分由韦达定理12221224232623km x x k m x x k ⎧+=-⎪⎪+⎨-⎪=⎪+⎩， (8)

分

(

1212

1AD AE k k kx m kx m x x ⋅=⇔++=(

)((

)(22121210

k x x k m x x m ⇔-+-++-=

展开222222222222226426390k m m k k m k m k m k m m --+-++-++-+=··10

分

2226390

m m ⇒-++-+

=2150m ⇒-+=．

解得m =

（舍去）或．··········································································11分

所以过定点()0，．···················································································12分