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福建省德化一中2012届最后一卷(数学理)

2024-01-05 来源:汇智旅游网
德化一中2012届毕业班考前适应性考试数学(理)试卷

德化一中2012届毕业班考前适应性考试

数学(理)试卷

第Ⅰ卷(选择题 共50分)

一、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. 1.sin225的值是(★★★).

A.

2 2

B. 32 C. 

22 D.3 2x1,2.已知变量x、y满足条件y2, 则xy的最小值是(★★★).

xy0, A. 4

B. 3

C. 2

D.1

3.在递减等差数列{an}中,若a1a50,则Sn取最大值时n等于(★★★).

A. 2 B. 3 C. 4 D.2或3 4.下列命题中,真命题的个数有(★★★).

1①xR,x2x≥0;

4

②x21的充分条件是x1;

③函数y2x是单调递增函数;④yx3和ylog3x互为反函数.

A. 0个 B. 1个 C. 2个 D.3个 5.某流程图如右图所示,现输入如下四个函数,则可以输出函数的是(★★★).

A. f(x)x

2

B. f(x)1 xC. f(x)lnx2x6 D.f(x)sinx

6.由两个完全相同的正四棱锥组合而成的空间几何体的正(主)视图、侧(左)视图、俯视图都相同,其图形如右图所示,其中视图中四边形ABCD是边长为1的正方形,则该几何体的体积为(★★★). A.

2 B.

22 C. 322 D. 36x2y27.双曲线221(a0,b0)的两条渐近线将平面划分为“上、下、左、右”四个区域(不

ab德化一中2012届毕业班考前适应性考试数学(理)试卷 第 - 1 - 页 共 9 页

德化一中2012届毕业班考前适应性考试数学(理)试卷

含边界),若点(1,2)在“上”区域内,则双曲线离心率e的取值范围是(★★★). A. (3,) B. (5,) C. (1,3) D.(1,5) 8.若某同学连续三次考试的名次(第一名为1,第二名为2,以此类推且没有并列名次情况)不超过3,则称该同学为班级的尖子生.根据甲、乙、丙、丁四位同学过去连续3次考试名次数据,推断一定不是尖子生的是(★★★). ..A. 甲同学:均值为2,中位数为2 B. 乙同学:均值为2,方差小于1 C. 丙同学:中位数为2,众数为2 D. 丁同学:众数为2,方差大于1

P到9.如图,在正方体ABCDA1B1C1D1中,若平面A1BCD1上一动点

AB1和BC的距离相等,则点P的轨迹为(★★★).

A.椭圆的一部分 B.圆的一部分

C.一条线段 D.抛物线的一部分

10.将方程xtanx0的正根从小到大地依次排列为a1,a2,,an,,给出以下不等式: ①0an1anA1D1C1B1DAPCB22an1an2an; ③

②an1an;

22an1an2an; ④

第9题图

其中,正确的判断是(★★★).

A. ①③ B. ①④ C. ②③ D. ②④

第Ⅱ卷(非选择题 共100分)

二、填空题:本大题共5小题,每小题4分,共20分.把答案填在答题卡相应位置. 11.已知复数zabi(其中i为虚数单位,a、bR),若|a|1且|b|1,则|z|1的

概率为★★★★★.

12.已知(1x)5a0a1xa2x2a3x3a4x4a5x5,则a1a2a3a4a5的值等于

★★★★★.

13.在△ABC中,A60,b1,SABC3,则ABAC等于★★★★★. 214.已知函数f(x)x3ax2bx(a、bR)的图象如图所示,

它与x轴在原点相切,且x轴与函数图象所围成的区域(如图阴影部分)的面积为

1,则a等于★★★★★. 1215.类比数学归纳法的证题思路,如果要证明对于任意的nZ-(Z-表示负整数集),命题p(n)都成立,可先证明命题p(1)成立,然后在假设命题

p(k)(kZ-)成立的基础上,证明命题★★★★★成立即可.

德化一中2012届毕业班考前适应性考试数学(理)试卷 第 - 2 - 页 共 9 页

德化一中2012届毕业班考前适应性考试数学(理)试卷

三、解答题:本大题共6小题,共80分.解答应写出文字说明,证明过程或演算步骤. 16.(本小题满分13分)

如图,在底面为平行四边形的四棱锥PABCD中,ABAC,PA面ABCD,

APAB3,AD5,点E是PD的中点.

(Ⅰ)求证:PB∥平面AEC;

(Ⅱ)求直线AB与平面EAC所成角大小.

17.(本小题满分13分)

D

P

E

A C

B

ABC中,角A、B、C的对边分别为a、b、c,若函数f(x)x2mx数,且f(cos1为偶函4B)0. 2(Ⅰ)求角B的大小; (Ⅱ)若△ABC的面积为18.(本小题满分13分)

随机变量X的分布列如下表如示,若数列pn是以p1为首项,以q为公比的等比数列,则称随机变量X服从等比分布,记为Q(p1,q).现随机变量XQ(15373,其外接圆半径为,求△ABC的周长. 431,2). 15  1 2 n X

 pn p1 p2 P

(Ⅰ)求n的值并求随机变量X的数学期望EX;

(Ⅱ)甲乙两人举行乒乓球比赛,已知甲赢得每一局比赛的概率都等于P(X2),比赛采用三局两胜制(即在三局比赛中,只要有一方赢得两局比赛,就取得胜利,比赛也就随之结束了),求甲在比赛中赢的局数比输的局数多的概率. 19.(本小题满分13分)

已知抛物线y2px(p0)的焦点为F,A是抛物线上横坐标为4且位于x轴上方的点,A到抛物线准线的距离等于5.过A作AB垂直于y轴,垂足为B,OB的中点为M.

(Ⅰ)求抛物线方程;

(Ⅱ)过M作MNFA,垂足为N,求点N的坐标;

2,0)(Ⅲ)以M为圆心,MB为半径作圆M,当K(m判断直线AK与圆M的位置关系.

是x轴上一动点时,

德化一中2012届毕业班考前适应性考试数学(理)试卷 第 - 3 - 页 共 9 页

德化一中2012届毕业班考前适应性考试数学(理)试卷

20.(本小题满分14分)

某公司有价值a万元的一条流水线,要提高该流水线的生产能力,就要对其进行技术改造,从而提高产品附加值,改造需要投入,假设附加值y(万元)与技术改造投入x(万元)之间的关系满足:①y与ax和x的乘积成正比;②x③0a2时,ya;2xt,其中t为常数,且t[0,1].

2(ax)(Ⅰ)设yf(x),求f(x)表达式,并求yf(x)的定义域; (Ⅱ)求出附加值y的最大值,并求出此时的技术改造投入.

21.本题有(1)、(2)、(3)三个选答题,每题7分,请考生任选2题作答,满分14分.如果多做,则按所做的前两题记分.作答时,先用2B铅笔在答题卡上把所选题目对应的题号涂黑,并将所选题号填入括号中. (1)(本小题满分7分)选修4-2:矩阵与变换

10已知矩阵M.

21(Ⅰ)请写出矩阵M对应的变换f的变换公式;

(Ⅱ)从变换的角度说明矩阵M可逆吗?如果可逆,请用求逆变换的方式求出对应的逆矩阵M

(2)(本小题满分7分) 选修4—4:极坐标与参数方程

在直角坐标平面内,以坐标原点O为极点,x轴的非负半轴为极轴建立极坐标系. 已知点A、B的极坐标分别为(1,0)、(1,参数,r0).

(Ⅰ)求直线AB的直角坐标方程;

(Ⅱ)若直线AB和曲线C只有一个交点,求r的值.

(3)(本小题满分7分)选修4—5: 不等式证明选讲

已知正实数a、b、c满足条件abc3, (Ⅰ) 求证:abc3;

(Ⅱ)若cab,求c的最大值.

1.

xrcos,),曲线C的参数方程为(为2yrsin德化一中2012届毕业班考前适应性考试数学(理)试卷 第 - 4 - 页 共 9 页

德化一中2012届毕业班考前适应性考试数学(理)试卷

德化一中2012届毕业班考前适应性考试

数学(理)试卷参考解答及评分标准

一、选择题:

1.B 2.C 3.D 4.C 5.D 6.C 7.D 8.D 9.D 10.D 二、填空题 11.

4; 12. -1 ; 13. 1 ; 14. -1; 15.p(k1).

三、解答题

16.证明: (Ⅰ)连结BD交AC于点O,并连结EO, 四边形ABCD为平行四边形

∴O为BD的中点 又E为PD的中点 ∴在PDB中EO为中位线,EO//PB

E

A

D

C

B

P

PB面AEC,EO面AEC

∴PB//面AEC .„„„„„„„5分

(Ⅱ)以A为原点,AC,AB,AP分别为x,y,z轴建立空间直角坐标系,则

33A(0,0,0),B(0,3,0),C(4,0,0),D(4,3,0),P(0,0,3),E(2,,),

2233则AB=(0,3,0),AE(2,,),AC=(4,0,0), „„„„„„„„„„„„„„7分

2233(x,y,z)(2,,)0nAE0设平面EAC一个法向量n(x,y,z),则,即, 22nAC0(x,y,z)(4,0,0)0令y1,得x0,z1,所以n(0,1,1).„„„„„„„„„„„„„„„„„„9分

设直线AB与平面EAC所成角为,则

|ABn|32, sin|cosAB,n|2|AB||n|32又[0,2],所以求直线AB与平面EAC所成角等于

2.„„„„„„„„„„„13分 417. 解:(Ⅰ)∵f(x)xmx∴f(x)f(x),即xmx21是偶函数, 411x2mx,∴m0„„„„„„„„„„„„2分 44德化一中2012届毕业班考前适应性考试数学(理)试卷 第 - 5 - 页 共 9 页

德化一中2012届毕业班考前适应性考试数学(理)试卷

又f(cosBB11cosB1)0,∴cos2,即, 2242412∴cosB,又B(0,)∴B.„„„„„„„„„„„„„„„„„„5分

23(Ⅱ)∵△ABC的外接圆半径为73 , 3∴根据正弦定理

bb1432R得,,b7.„„„„„„„„„„7分

2sinB3sin3又SABC1153,∴ac15. „„„„„„„„„„„„„„„„9分 acsinB24249,a2c234 „„„„11分 3在△ABC中,根据余弦定理得,

b2a2c22accosB,即a2c230cos∴(ac)2a2c22ac64,∴ac8,

∴△ABC的周长等于15.„„„„„„„„„„„„„„„„„„„„„„„„„„13分

18.解:(Ⅰ)依题意得,数列pn是以

1为首项,以2为公比的等比数列, „„„1分 151(12n)所以Snp1p2pn15=1 „„„„„„„„„„„„„„„„„„„3分

12解得n=4.„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„5分

122223EXp12p23p34p41234

151515151120221322423 „„„„„„„„„„„„„„„„„„„„„„6分 1549„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„7分 15(Ⅱ)由(Ⅰ)知随机变量X的分布列为

1 X

1 P 15

2 3 4 2 154 158 15设“甲在第i(i1,2,3)局取胜”为事件Ai, 依题意,P(Ai)P(X2)121„„„„„„„„„„„„„„„„„„„„9分 15155设“甲在比赛中赢的局数比输的局数多”为事件A,

则事件A等价A „„„„„„„„„„„„„„„„„„„„11分 1A2A1A2A3A1A2A3,

德化一中2012届毕业班考前适应性考试数学(理)试卷 第 - 6 - 页 共 9 页

德化一中2012届毕业班考前适应性考试数学(理)试卷

1141114113„13分 55555555125pp219. 解:(Ⅰ)抛物线y2px的准线为x,于是45,p2.

22则P(A)P(A1A2)P(A1A2A3)P(A1A2A3)∴抛物线方程为y24x.„„„„„„„„„„„„„„„„„„„„„„„„„„ 3分 (Ⅱ)∵点A的坐标是(4,4), 由题意得B(0,4),M(0,2),

43;MNFA,kMN,„„„5分 3443则FA的方程为y(x1),MN的方程为y2x.

34又∵F(1,0), ∴kFA84xy(x1)53,得解方程组y23xy454

84N(,).„„„„8分

55(Ⅲ)由题意得,圆M的圆心是点(0,2),半径为2. „„„„„„„„„„„„„„9分 当m4时,直线AK的方程为x4,此时,直线AK与圆M相离;„„„„„„„10分 当m4时,直线AK的方程为y圆心M(0,2)到直线AK的距离d4(xm), 即为4x(4m)y4m0, 4m|2m8|16(m4)2,令d2,解得m1

∴当m1时,直线AK与圆M相离; 当m1时,直线AK与圆M相切; 当m1时,直线AK与圆M相交. „„„„„„„„„„„„„„„„„„„„„„„„„„13分 20. 解:设yk(ax)x,当xa2时,ya,可得:k4,∴y4(ax)x„„2分 2x0(1)2(ax)x由0 „„„„„„„„„„„„„„„„„„3分 t得2(ax)xt(2)2(ax)又x0所以由(1)得ax0,即0xa „„„„„„„„„„„„„„„„„„4分 所以(2)可化为x2(ax)tx因为t[0,1],所以

2at „„„„„„„„„„„„„„„„„„„5分 12t2ata „„„„„„„„„„„„„„„„„„„„„„„„„6分 12t德化一中2012届毕业班考前适应性考试数学(理)试卷 第 - 7 - 页 共 9 页

德化一中2012届毕业班考前适应性考试数学(理)试卷

综上可得函数f(x)定义域为[0,2at],其中t为常数,且t[0,1].„„„„„„„„7分 12t(2)y4(ax)x4(x)2a2 „„„„„„„„„„„„„„„„„„„„„9分

a22ata1a时,即t1,x时,ymaxa2 „„„„„„„„„„„„„„11分 12t2222ata12at当,即0t,y4(ax)x在[0,]上为增函数 12t2212t当

8a2t2at∴当x时,ymax „„„„„„„„„„„„„„„„„„„13分 2(12t)12t答:当

1at1,投入x时,附加值y最大,为a2万元; 228a2t12at当0t,投入x时,附加值y最大,为万元 „„„„„„„14分

(12t)2212t21. ⑴矩阵与变换 解:(Ⅰ)假设M10把任一点x,y变成x',y', 21x'10x则

y'21yx'x ················································································································· 3分 y'2xy(Ⅱ)从变换的角度看,变换f是可逆的.由(Ⅰ)得矩阵M对应的变换f是y轴方向上的

切变变换.因为变换f把每个点在横坐标不变的情况下,纵坐标变为原来纵坐标加上横坐标的2倍,所以它的逆变换f1应该是把每个点在横坐标不变的情况下,纵坐标变为原来纵坐标

减去横坐标的2倍. ·············································································································· 5分

x'x ····················································································································· 6分 y'2xy∴M110············································································································· 7分 . ·

21(2)极坐标与参数方程

解:(Ⅰ)∵点A、B的极坐标分别为(1,0)、(1,2),

∴点A、B的直角坐标分别为(1,0)、(0,1), ································································· 2分 ∴直线AB的直角坐标方程为xy10. ······································································ 4分

xrcos,C(Ⅱ)由曲线的参数方程化为普通方程为x2y2r2, ·· 5分 (为参数)yrsin ∵直线AB和曲线C只有一个交点,

德化一中2012届毕业班考前适应性考试数学(理)试卷 第 - 8 - 页 共 9 页

德化一中2012届毕业班考前适应性考试数学(理)试卷

∴直线AB与圆C相切 ∴半径r······························································································· 7分 2. ·

222111⑶不等式证明选讲

解:(Ⅰ)由柯西不等式得(abc)2(abc)(111) 代入已知 abc3

(abc)29

abc3

当且仅当abc1取等号. „„„„„„„„„„„„„„„„„„„„„„„„3分 (Ⅱ)由ab2ab得2abc3,若cab,则2cc3,

c3 c10,

所以c1,c1,当且仅当ab1时,c有最大值1.„„„„„„„„„„„„7分

德化一中2012届毕业班考前适应性考试数学(理)试卷 第 - 9 - 页 共 9 页

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