中职学业诊断考试数学试卷及参考答案

一、选择题（本大题共15个小题，每小题4分，共60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）

1．已知集合A{5,7}，B{x|3x≤7}，则下列说法正确的是（ ） A．BA

B．AB

C．AB

D．BA

2．“xy”是“x2y2”的（ A．充分不必要条件 C．充要条件

）

B．必要不充分条件 D．既不充分又不必要条件

3．下列函数中，定义域为(0,)的函数是（ A．yx2

B．y ）

D．y2x

x

C．ylog2x

4．设f(x)是定义在R上的奇函数，当x0时，f(x)2x1，则f(2)（ A．5

B．5

C．

）

5 4

D．5 45．下列等式正确的是（

2A．222

1 ）

412

B．lg2lg510 C．93

D．lne12e

6．某地为了抑制某种寄生生物的肆意增长，引入一种以该寄生生物为食物的特殊动物．已知该动物繁殖后的数量y（只）与引入时间t（年）的关系为ymlog3(t2).若该动物在引入1年后的数量为1000只，则7年后它们可以发展到（ ） A．1000只

B．1500只

C．2000只

D．2500只

7．若象限角满足A．一或二

sin=1，则所在的象限是（ ） sinB．二或三

C．三或四

D．一或四

1

8．右图是正弦型函数ysin(2x)一个周期的图象，则可以是（ A．0 C．

B．D．

）

 12 3 69．等比数列{an}中，a23，则该数列前3项之积是（ A．27

B．－27

C．9

D．－9

）

10．如图，点E、F、G、H分别是正方形ABCD各边的中点，且EG，HF交于点O，则向量AE的负向量的个数是（ A．4个 C．6个

B．5个 D．7个

）

11．已知直线l与直线3x6y40垂直，且两直线在y轴上的截距相等，则直线l的方程可以是（

）

B．6x3y20 D．3x6y40

A．6x3y20 C．3x6y40

12．如图，某公司准备在办公楼前修建一个椭圆形的喷泉池，椭圆的长轴长为12m，离心率为

2．现要求两个喷泉口安放在该椭圆的焦点F1、F2处，则喷泉口位置F1与顶点A的距离是3（ ）

A．2m C．4m

B．3m D．5m

13．如图，一个天然形成的石臼，由高为1m、底面直径为6m的圆柱和直径为6m的半球组成，则该石臼能盛雨水约（ A．9m3

3C．27m

）

B．18m3

3D．36m

14．某校组织6名女生和4名男生参加市冬运会，从中任选3名学生参加开幕式，则所选3人中恰有1名男生的概率是（ A．0.3

）

C．0.6

D．0.7

B．0.5

2

15．右图是样本容量为400的频率分布直方图，则样本数据落在8,18内的频数为（ A．120 B．160 C．200 D．280

二、填空题（本大题共5个小题，每小题4分，共20分） 16．不等式52x3≥0的解集是 ． 17．若tan

）

1sin2cos，则 ． 24sincos18．埃及一金字塔共有40级石阶，石阶高度从下往上成等差数列，最下面一级石阶高1m，第二级石阶高0.98m，则该金字塔石阶的总高度是 ．

19．已知抛物线顶点在原点，焦点在y轴上．抛物线上一点Q的纵坐标是3，到焦点的距离为5．则此抛物线方程为 ．

20．用0，1，2，3组成没有重复数字的4位数，其中偶数共有 个．

三、解答题（本大题共6个小题，共70分。解答应写出文字说明、证明过程或推演步骤） 21．（本小题满分10分）已知ABC顶点坐标分别为A(2,2)，B(2,0)，． C(1,a)（a0）（1）用坐标表示向量

1AB； 2（2）若C90，求实数a的值．

22．（本小题满分10分）已知等差数列an的前n项和为Sn，且S42a42，a62a32． 求数列an通项公式．

23．（本小题满分12分）在ABC中，角A,B,C所对应的边分别为a,b,c，且满足

2cos2AcosA2． 2（1）求角A的值；

（2）若a2，bc4，求ABC的面积．

24．（本小题满分12分）最近，网络小说作者达康写了一部小说，三生、人和两家出版社正在竞争这部小说的版权．三生出版社给出的稿酬方案为：发行量前3000册每册支付6%的版权费用；超过3000册的部分每册支付8%的版权费用，且每册另加2元的稿酬．人和出版社给出的

3

稿酬方案为：发行量前4000册不支付版权费用，但超过4000册的部分每册将支付10%的版权费用，且每册另加3元的稿酬．假定每册的定价为30元．

（1）分别写出两家出版社给出的方案中稿酬y（元）与发行量x（册）的函数关系式． （2）作者达康将在两家出版社中选择一家出版社合作.请你比较两种稿酬方案，为作者达康给出合理性建议．

25．（本小题满分13分）如图所示，正方形AA1D1D与矩形ABCD所在平面互相垂直，点E为AB的中点．

（1）求证：A1D平面AD1B； （2）求证：BD1//平面A1DE．

x2y226．（本小题满分13分）已知双曲线221ab的焦半距c6a，过A(a,0)，B(0,b)的直223． 3线到原点的距离是（1）求此双曲线的标准方程；

（2）若直线ykx1(k0)与双曲线交于不同的点C，D，且C，D都在以M(0,1)为圆心的圆上，求k的值．

参考答案及评分标准

一、选择题（本大题共15个小题，每小题4分，共60分） 题号 答案 1. B 2. D 3. C 4. A 5. A 6. C 7. C 8. D 9. B 10. 11. 12. 13. 14. 15. C A A C B D 二、填空题（本大题共5个小题，每小题4分，共20分） 题号 答案 16. 17. 18. 19. 20. [1,4] 5 224.4m x28y 10 三、解答题(本大题共6个小题，共70分) 21．（本小题满分10分）

4

解：（1）AB(2,0)(2,2)(4,2),

2分

··························································

1AB(2,1)． ··········································································· 4分 2（2）由C90可知向量ACBC，

···························································· 5分 AC(1,a)(2,2)(3,a2)， ·

································································· 6分 BC(1,a)(2,0)(1,a)， ·则ACBC(3,a2)(1,a)

3a(a2)

a22a30． ····················································································· 8分

解得a3或a1（舍去）． ········································································· 9分 故a3． ·································································································· 10分 22.（本小题满分l0分） 解：由题意可知S44(a1a4)2a12a42a42， 2所以a11． ····························································································· 4分 设等差数列an公差为d，

则a3a12d2d1，a6a15d5d1， ··············································· 6分

a62a35d12(2d1)d12，

即d1． ··································································································· 8分 所以数列an通项公式为：ana1(n1)d

············································································· 10分 1(n1)1n2． ·23.（本小题满分l2分） 解：（1）由2cos2AcosA2， 2得1cosAcosA2，即cosA∵0A， ∴A1． ··························································· 2分 23． ································································································· 4分

222（2）∵abc2bccosA， ··································································· 5分

5

∴a(bc)2bc2bc221 2(bc)23bc． ························································································ 6分

又∵a2，bc4，

∴bc4． ································································································· 8分 故SABC1bcsinA ···················································································· 10分 214sin3． ················································································· 12分 2324.（本小题满分l2分）

（1）解：三生出版社给出的稿酬方案为：

1.8x, 0x3000y14.4x7800, x3000 ·································································· 3分

人和出版社给出的稿酬方案为：

0, 0x4000y26x24000, x4000 ································································· 6分

（2）当0x4000时，显然y1y2，所以选择三生出版社． ………………….. 7分 当x4000时，令y1y2，即4.4x78006x24000，

解得x10125． ························································································· 8分 当4000x10125时，y1y2， 当x10125时，y1y2，

当x10125时，y1y2，……………………………………………………………. 10分 综上所述

当发行量0x10125时，建议选择三生出版社； 当发行量x10125时，可以任选一家；

当发行量x10125时，建议选择人和出版社． ················································· 12分 25.（本小题满分l3分） 证明：

（1）因为ADD1A1为正方形，

所以A······················································································· 2分 1DAD1． ·因为ABCD为矩形，所以ABAD， ····························································· 3分

6

又因为平面ADD1A1平面ABCD，

所以AB平面ADD1A············································································ 4分 1， ·又A1D平面ADD1A1，

所以ABA1D， ························································································· 5分 又因为AB交AD1于A，

所以A·················································································· 7分 1D面ABD1． ·

O，则O是AD1的中点，连接OE， ·（2）设AD1交A································ 8分 1D于

又因点E为AB的中点，所以OE为ABD1的中位线．

所以OE//BD1． ························································································ 10分 又因为OE平面A1ED，BD1平面A1ED，

所以BD1//平面A············································································· 13分 1DE． ·26．（本小题满分l3分）

解：（1）原点到直线AB的距离h因为23， 311·········································································· 2分 a2b2hab， ·

22且c2a2b2， ·························································································· 3分

ab23所以， ························································································· 4分 c3因为c所以b6a， 22,a2． ····················································································· 5分

x2y21． ·故所求双曲线方程为································································· 6分 42x2y21中，得到(12k2)x24kx60 ·（2）把ykx1(k0)代入··········· 7分 42设C(x1,y1),D(x2,y2)，CD的中点E(x0,y0)，则 由x1x2

4k． ···················································································· 8分 212k7

得x012k1， ·········································· 10分 (x1x2),ykx100212k212k21， k∵CD是圆E的弦，∴EMCD，则kEM∴kEMy0(1)y011，

x00x0k112112k． ·即·················································································· 12分 2kk12k2化简得1k21．

即k22，故k2． ············································································· 13分

8