2011级二诊理数

数学练习

第Ⅰ卷 （选择题 共60分）

一.选择题：本小题共12小题，每小题5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．计算

1?i的值是 i3Ａ．?1?i Ｂ．1?i

Ｃ．1?i Ｄ．?1?i

12．已知集合A?{x|()x?1},B?{x|x2?x?2?0}，则A?B?

2 Ａ．{x|?1?x?1} Ｂ．{x|1?x?2} Ｃ．{x|0?x?1}

Ｄ．{x|0?x?2}

3．样本中共有5个个体，其值分别为a，0，1，2，3，若该样本的平均值为1，则样本的方差为

Ａ．2

Ｂ．

6 5Ｃ．2 Ｄ．6 54．若tan??3，tan??

Ａ．－3

4，则tan(???)等于 31Ｂ．?

3Ｃ．3

Ｄ．

1 35．下列命题中，正确的是

n2?2nA． lim?1

n? ?n?1?3x?1(x?0)B． 若f(x)??，则limf(x)?2

x? 0?2(x?0)?x?1(x?0)C． 若f(x)??2在x?0处连续，则a?2

x?a(x?0)??2x?3 , x?1f(x)?2 D． 若f(x)??，则limx? 1 ?2 , x?1?6．对于函数f(x)?sin(2x?)?1，下列命题中正确的是

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A． 函数f(x)的图象可以由y?sin2x?1的图象向左平移B． 函数f(x)图象的一条对称轴方程是x?C． 函数f(x)图象的一个对称中心是(?D． 函数f(x)在(???个单位长度得到 6?3

7?,?1) 12??,)上是增函数 447．已知函数f(x)?cosxsinx(x?R)，给出下列四个命题：

①若f(x1)??f(x2),则x1??x2; ②f(x)的最小正周期是2?；

1

③f(x)在区间[?⑤当x???

??3? ④f(x)的图象关于直线x?对称； ,]上是增函数；

444?33??,??.其中正确的命题为

4??4

????时，

f(x)的值域为,??63?B．③④⑤

（ ） A．①②④

C．②③ D．③④

8．用0到9这10个数字，可以组成没有重复数字的三位偶数的个数为

A． 324

B． 360

C． 328

D． 648

9．已知等比数列{am}中，各项都是正数，且a1，

a?a1a3,2a2成等差数列，则910?

a7?a82

D．3?22 D． 5

A.1?2

A． 3

B. 1?2 B． 2

C. 3?22

C． 4

??????????????????????????10．已知△ABC，若点M满足MA?MB?MC?0，且存在m使得AB?AC?mAM成立，则m=

11．定义在R上的偶函数f(x)满足f(2?x)?f(x)，且在[－3，－2]上是减函数，?,?是

钝角三角形的两个锐角，则下列不等式关系中正确的是 （A）f(sin?)?f(cos?) （B）f(cos?)?f(cos?) （C）f(cos?)?f(cos?)

( D）f(sin?)?f(cos?)

12．已知f(x)是定义在(0??上的可导函数，对任意x?(0,??)都有f(x)?0，且,)x?(x)ln，则f(?)与f(e)ln?的大小关系是 f(x)?fxA．f(?)?f(e)ln?

B．f(?)?f(e)ln? C．f(?)?f(e)ln? D．不能确定

数 学（ 理工农医类）

（非选择题 共90分）

注意事项: 1．用钢笔或圆珠笔直接答在试题卷中.2．答题前将密封线内的项目填写清楚. 二、填空题：本大题共4小题，每小题4分，共16分，把正确答案填在题中横线上.

113. 二项式(x2?)6的展开式中，常数项为 . 得分 评卷人 x第Ⅱ卷

14．公差不为零的等差数列{an}的前n项和为Sn，若a4是a3与a7的等比

中项, a1??3，则S10等于 .

15．知函数f(x)是R 上的偶函数，且在（0，+?）上有f?（x）> 0，若f（－1）= 0，那

么关于x的不等式x f（x）< 0 的解集是____________． 16．已知y?f(x)（x?D）同时满足下列两个条件：（1）函数f(x)在D内单调递增或单调递

减；（2）如果存在区间[a,b]?D，使函数f(x)在区间[a,b]上的值域为[a,b]，那么称y?f(x)（x?D）为闭函数.给出下列命题：

1① 函数f(x)?2x（x?R）不是闭函数； ② 函数f(x)?x?(x?(0,??))是闭函数；

x3③ 闭函数y??x符合条件（2）的区间为[?1,1]；

1④ 若y?k?x(k?0)是闭函数，则实数k的取值范围是(?,0).

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2

其中是真命题的有 （请将正确答案的序号填在题中横线上）.

三、解答题：本大题共6小题，共74分，解答应写出文字说明，证明过程或演算步骤．

?17．（本小题满分12分）在△ABC中，角A、B、C的对边分别为a、b、c，C?，b?5，△ABC

3的面积为103. （Ⅰ）求a、c的值；（Ⅱ）求sin(A?)的值.

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18．（本小题满分12分）在某次抽奖活动中，一个口袋里装有5个白球和5个黑球，所有球除颜色外无任何不同，每次从中摸出2个球，观察颜色后放回，若为同色，则中奖.

（Ⅰ）求摸球一次中奖的概率；

（Ⅱ）记连续3次摸球中奖的次数为?，求?的分布列及期望.

???19已知?ABC中的内角A,B,C的对边分别为a,b,c，定义向量m?2sinB,?3，

????????2Bn??cos2B,2cos?1?且m//n．

2??（Ⅰ）求函数f?x??sin2xcosB?cos2xsinB的单调递增区间； （Ⅱ）如果b?2，求?ABC的面积的最大值

20．（本小题满分12分）

已知f(x)?(x2?a)ex，其中无理数e=2.71828…. （Ⅰ） 若a?3，求函数f(x)的单调递减区间；

3（Ⅱ）已知x1、x2是f(x)的两个不同的极值点，且|x1?x2|?|x1x2|，若3f(a)?a3?a2?3a?b2恒成立，求实数b的取值范围．

3

得分 评卷人 21．已知数列?an?是等差数列，cn2??an?an?1?n?N?

2 （1）判断数列?cn?是否是等差数列，并说明理由；

（2）如果a1?a3???a25?130,a2?a4???a26?143?13kk为常数，试写出

数列?cn?的通项公式；

（3）在（2）的条件下，若数列?cn?得前n项和为Sn，问是否存在这样的实数k，使Sn当且仅当n?12时取得最大值。若存在，求出k的取值范围；若不存在，说明理

由。

??得分 评卷人 22．（本小题满分14分）已知在数列{an}中，a1?2,a2?4，曲线f(x)?an?1x3?3(an?1?3an)x?1(n?2)在点(2,f(2))处的切线方程平行于

直线y=0．

（Ⅰ）证明数列{an?1?an}是等比数列，并求数列{an}的通项公式；

（Ⅱ）当1?i?j?n（i,j,n均为正整数）时，求ai和aj的所有可能的乘积aiaj之和Tn； （Ⅲ）设函数g(x)?2x，M是数列{

13g(n)}前n项之和，求证：?M?． Tn24

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数 学(理工农医类)参考答案及评分意见

评分说明：

1． 本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比

照评分参考制订相应的评分细则．

2． 对计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度．可视

影响的程度决定后继部分的给分，但不得超过该部分正确解答应得分数的一半；如果后继部分的解答有较严重的错误，就不再给分．

3． 解答右侧所注分数，表示考生正确做到这一步应得的累加分数． 4． 只给整数分数，选择题和填空题不给中间分．

一、选择题 题号 答案 1 B 2 D 3 A 4 D 5 B 6 C 7 D 8 C 9 C 10 A 11 D 12 B 二、填空题

13. 15 14. 60 15. 答案 (??,?1)?(0,1)， 16. （3）（4）

三、解答题

17．解：（Ⅰ）∵C??3，b?5，

? S?ABC?absinC ， 即 103?a?5sin，

3∴a?8， ················································································································ 2分 由余弦定理可得:c2?64?25?80cos1212??3?49,

∴c?7. ················································································································· 4分 （Ⅱ）由（Ⅰ）有cosA?49?25?641·································································· 6分 ?, ·

70743, ··········································· 8分 7∵A是三角形的内角，∴sinA?1?cos2A? ∴sin(A?)?sinAcos?cosAsin ····························································· 10分

666????43311??? 7272?13. ······················································································································ 12分 1422C5418．解：（Ⅰ）设摸球一次中奖的概率为p1,则p1=2=， ················································ 4分

C109 5

（Ⅱ）?的取值可以是0，1，2，3， ············································································· 5分

P(??0)=(1-p1)3=

125, ························································································· 6分 729300100=, ···································································· 7分 72924324080=, ··································································· 8分 7292432P(??1)=C13(1?p1)p1?P(??2)=C32(1?p1)p12?P(??3)=p13=

64， ····························································································· 9分 729所以?的分布列如下表

? P 0 1 2 3 125 729100 24380 24364 729 ································································································································ 10分 ············································································································ 11分 ????. ·

43??2B19解：19 答案 解：（Ⅰ）?m//n ?2sinB(2cos?1)??3cos2B

2?sin2B??3cos2B 即 tan2B??3

又?B为锐角 ?2B??0,??

44答：仅一次摸球中奖的概率为，连续3次摸球中奖的期望为. ················ 12分

93?2B?2?? ?B? 33?? f?x??sin2xcosB?cos2xsinB?sin?2x???? 3?2k???2?2x??3?2k?????2

∴函数的单调递增区间是?k???12,k??5??． 7分 12??a2?c2?b2得 （Ⅱ）?B?,b?2，由余弦定理cosB?32ac?a2?c2?ac?4?0

又?a?c?2ac 代入上式得：ac?4（当且仅当 a?c?2时等号成立．）

22 6

S?ABC?13acsinB?ac?3（当且仅当 a?c?2时等号成立．） 24. ·································································································································· 12分

20．解：（Ⅰ）∵f(x)?(x2?3)ex，

∴f?(x)?(x2?2x?3)ex， ····················································································· 1分 当f?(x)?(x2?2x?3)ex?0时，x?(?3,1)， ······················································ 2分 ∴f(x)的单调递减区间为(?3,1)， ······································································ 3分 （Ⅱ）∵f?(x)?(x2?2x?a)ex，

∴设f?(x)?(x2?2x?a)ex?0即x2?2x?a?0， ·············································· 4分 ∵x1,x2是f(x)的两个不同的极值点， ∴x1,x2是方程x2?2x?a?0的两实根，

且x1?x2??2,x1x2??a， ····················································································· 5分 ∵|x1?x2|?|x1x2|，

∴?2?a?2， ········································································································ 6分 又??4?4a?0，

∴ ?1?a?2， ······································································································ 7分

33记g(a)?3f(a)?a3?a2?3a?3(a2?a)ea?a3?a2?3a， ··························· 8分

22∴g?(a)?3(a2?a?1)ea?3a2?3a?3， 令3(a2?a?1)ea?3a2?3a?3?0， ∴3(a2?a?1)ea?3a2?3a?3?0，

?1?5∴a?0,a?， ···························································································· 9分

2 · ································································································································································· a

g?(a)(?1,0) g?(a)?0递增(0,5?1)2(5?1,2)2g?(a)?0 递增 g?(a)?0 递减g(a) ∵g(0)?0,g(2)?6e2?8， ················································································· 10分 ∴g(a)有最大值为6e2?8， ··············································································· 11分 ∴b?6e2?8. ········································································································· 12分

21答案．解：（1）设{an}的公差为d，则

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2222cn?1?cn?(an?1?an?2)?(an?an?1) 222?2an?(a?d)?(a?d) ?1n?1n?1??2d2

?数列{cn}是以?2d2为公差的等差数列…………4分

（2）?a1?a3???a25?130

a2?a4???a26?143?13k

?两式相减：13d?13?13k ?d?1?k…………6分

13(13?1)?13a1??2d?130

2?a3??2?12k…………8分

?an?a1?(n?1)d?(1?kn?(13k?3))

22?cn?an?an?1?(an?an?1)(an?an?1)

?26k2?32?6?(2n?1)(1?k2)

??2(1?k)2?n?25k?30k?5…………10分

（3）因为当且仅当n?12时Sn最大

?有c12?0,c13?0…………12分

22????24(1?k)?25k?30k?5?0?k?18k?19?0??2即? 22????36(1?k)?25k?30k?5?0?k?22k?21?0

?k?1或k??19???k??19或k?21…………15分 ?k?21或k?122解: （Ⅰ）∵f(x)?an?1x3?3(an?1?3an)x?1(n?2)，

∴f?(x)?3an?1x2?3(an?1?3an)， ········································································· 1分 ∵曲线f(x)?an?1x3?3(an?1?3an)x?1(n?2)在点(2,f(2))处的切线方程为y=0． ∴f?(2)?0，即2an?1?3an?an?1?0， ····························································· 2分 ∴an?1?an?2(an?an?1)，

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∵a1?2,a2?4,?an?an?1?0,

?an?1?an?2,

an?an?1∴数列{an?1?an}是以2为首项，2为公比的等比数列， ·································· 3分

an?(an?an?1)?(an?1?an?2)??(a2?a1)?a1 ?(2?22?…+2n?1)?2

2(1?2n?1)??2

1?2·········································································································· 4分 ?2n(n?2)，

··························································· 5分 ?当n?1时，上式也成立，∴an?2n， ·

（Ⅱ）∵an?2n，

由ai和aj的所有可能乘积aiaj?2i?j（1?i?j?n）可构成下表：

21?1，21?2，21?3，…，21?(n?1)，21?n， 22?2，22?3，…，22?(n?1)，22?n， 23?3，…，23?(n?1)，23?n，

……

2n?n，

构造如下n行n列的数表： 21?1，21?2，21?3，…，21?(n?1)，21?n， 22?1，22?2，22?3，…，22?(n?1)，22?n， 23?1，23?2，23?3，…，23?(n?1)，23?n，

……

2n?1，2n?2，2n?3，…，2n?(n?1)，2n?n，

设上表第一行的和为T4(1?2n)，则T?············································· 7分 ?4(2n?1).

1?2∴2Tn?T(1?2?22???2n?1)?(22?24?26??22n)， ········································· 9分

4(1?4n)4n?4(2?1)(2?1)??(2?1)(2n?2?2)，

1?43nn4·························································································· 10分 Tn?(2n?1)(2n?1?1)，·

34（Ⅲ）由题意g(x)?2x,?g(n)?2n又Tn?(2n?1)(2n?1?1)，

3g(n)3?2n311??(n?n?1)， ·∴··············································· 11分 nn?1Tn4(2?1)(2?1)42?12?1 9

2222n∴M?????

T1T2Tn311111111?((?2)?(2?3)?(3?4)?(n?n?1)) 42?12?12?12?12?12?12?12?131······································································································ 12分 ?(1?n?1). ·

42?1∵2n?1?1?3 ， ∴即

1313······················································································ 13分 ?(1?n?1)?，·

242?1413·········································································································· 14分 ?M?. ·

24113．∵(a?0?1?2?3)?1，∴a??1，故S2?[(?1?1)2?(0?1)552?(11?)22(2?1)?(3?1)?]22?.

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45tan??tan?3?3?1， 4．tan(???)??1?tan?tan?1?3?45333?8．首先应考虑“0”是特殊元素，

当0排在末位时，有A9?9?8?72（个），

当0不排在末位时，有A4?A8?A8?4?8?8?256（个），

于是由分类计数原理，得符合题意的偶数共有72?256?328（个）.故选B.

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?????2????10．由题目条件可知，M为△ABC的重心，连接AM并延长交BC于D，则AM?AD①，因

3??????????????????????????为AD为中线则AB?AC?2AD?mAM，即2AD?mAM②，联立①②可得m=3，.

12.要比较f(?)与f(e)ln?的大小，只须比较

ln(e)ln?与的大小 f(e)f(?)1f(x)?ln(x)?f?(x)ln(x)f(x)?f?(x)?lnxxx(x?0),? F?(x)???0, ?构造函数F(x)?f(x)f2(x)xf2(x)?F(x)在(0,??)上是增函数，即F(?)?F(e),即

22ln?ln(e),即f(?)?f(e)?ln? ?f(?)f(e)14．a4?a3a7得(a1?3d)?(a1?2d)(a1?6d)得2a1?3d?0,则d?2,所以

S10?10a?190d?60,. 216．已知函数f(x)，定义：若对给定的实数a(a?0)函数f(x?a)与f?1(x?a)互为反函数，则

称f(x)满足“a和性质”；若函数y?f(ax)与y?f?1(ax)互为反函数，则称f(x)满足“a积性质”.给出下列命题：

① 函数f(x)?1?x满足“1和性质”； ② 函数f(x)?1不满足“2积性质”； x③ 函数f(x)?kx?b(k?0)满足“1和性质”的充要条件是k?1； ④不可能存在实数m使函数f(x)?x?m既满足“a和性质”又满足“a积性质”； mx?1其中是真命题的有 ①④ （请将正确答案的序号填在题中横线上）

解：据题可知f(x?1)?1?(x?1)??x,其反函数是f数f?1?1(x)??x，由f(x)?1?x知其反函

(x)?1?x，故f?1(x?1)?1?(1?x)??x，?f(x?1)与f?1(x?1)互为反函数，

所以①正确； 由f(x)?1知其反函数为fx?1(x)?111，而f(2x)?，其反函数为,?f?1(2x)?x2x2x11

y?11，故f(x)?满足“2积性质”，②不正确

x2x?1设函数f(x)?kx?b(k?0)满足“1和性质”，则f由f(x?1)?k(x?1)?b得其反函数为y(x)?x?bx?1?b， ,?f?1(x?1)?kk?x?b?1，由“1和性质”定义知必有 kx?1?bx?b??1，即x?1?b?x?b?k，?k??1，③不正确 kkx?mx?a?mx?m?1若函数f(x)?满足“a和性质”，则f(x)? ,?f?1(x?a)?mx?1mx?1mx?ma?1x?a?m(1?ma)x?a?m由f(x?a)?知其反函数为y?，

mx?ma?1mx?1x?a?m(1?ma)x?m?a2，化简得mx?(1?ma)x?m?a?0，上式不可能??mx?ma?1mx?1恒成立，所以不存在m的值使f(x)?同理若f(x)?由

x?m满足“a和性质”. mx?1?1x?m满足“a积性质”，则fmx?1(ax)?ax?m，

max?1ax?mx?m得其反函数为y?

max?1max?aax?mx?m22，化简得ma(a?m)x?(1?a)x?m(1?a)?0，上式成立的充要??max?1max?af(ax)?条件是m?0,a??1，但此时f(x)?x?m不满足“a和性质”.所以④正确 mx?1

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