双基限时练12

双基限时练(十二)

1．下列各组数成等比数列的是( )

①1，－2,4，－8；②－2，2，－22，4；③x ，x 2，x 3，x 4；④a －1，a －2，a －3，a －4.

A ．①②

B ．①②③

C ．①②④

D ．①②③④

解析 由等比数列的定义，知①、②、④是等比数列．③中当x ＝0时，不是等比数列．

答案 C

2．已知等比数列{a n }中，a 1＝32，公比q ＝－1

2，则a 6等于( ) A ．1 B ．－1 C ．2

D.12

解析 a 6＝a 1q 5

＝32×? ??

??－125

＝－1.

答案 B

3．在等比数列{a n }中，a n >0，且a 2＝1－a 1，a 4＝9－a 3，则a 4＋a 5的值为( )

A ．16

B ．27

C ．36

D ．81

解析 由已知，得?

????

a 1＋a 2＝1，

a 3＋a 4＝9.

∴q 2(a 1＋a 2)＝9，∴q 2＝9. ∵a n >0，∴q ＝3.

∴a 4＋a 5＝q (a 3＋a 4)＝3×9＝27.

答案 B

4．在数列{a n }中，对任意n ∈N *，都有a n ＋1－2a n ＝0(a n ≠0)，则2a 1＋a 2

2a 3＋a 4

等于( )

A ．1 B.12 C.13

D.14

解析 由a n ＋1－2a n ＝0，得a n ＋1

a n

＝2，∴{a n }为等比数列，且公比

q ＝2，∴2a 1＋a 22a 3＋a 4＝a 1(2＋q )a 3(2＋q )＝a 1a 1q 2＝1

4

.

答案 D

5．已知等比数列{a n }满足a 1＋a 2＝3，a 2＋a 3＝6，则a 7等于( ) A ．64 B ．81 C ．128

D ．243

解析 ∵{a n }为等比数列，∴a 2＋a 3a 1＋a 2＝q ＝2.

又a 1＋a 2＝3，∴a 1＝1.故a 7＝a 1q 6＝64. 答案 A

6．已知x,2x ＋2,3x ＋3是一个等比数列的前3项，则第4项为____________．

解析 由(2x ＋2)2＝x (3x ＋3)，∵x ＋1≠0，∴4(x ＋1)＝3x ，∴x ＝－4，∴公比q ＝2x ＋2x ＝－6－4

＝3

2.

∴第4项为xq 3

＝－4×(32)3＝－27

2.

答案 －27

2

7．2＋3与2－3的等比中项是________． 答案 ±1

8．已知数列1，a 1，a 2,4成等差数列，1，b 1，b 2，b 3,4成等比数列，则a 1＋a 2

b 2

＝________.

解析 根据题意得a 1＋a 2＝5，b 2

2＝b 1b 3＝1×4＝4，又b 2>0，

∴b 2＝2，∴a 1＋a 2b 2

＝52.

答案 52

9．已知{a n }为等比数列，a 3＝2，a 2＋a 4＝20

3，则{a n }的通项公式为________．

解析 设等比数列的公比为q ，则q ≠0， a 2＝a 3q ＝2

q ，a 4＝a 3q ＝2q ， ∴2q ＋2q ＝203.解得q 1＝1

3，q 2＝3.

当q ＝1

3时，a 1＝18，∴a n ＝18×? ??

??13n －1＝2×33－n .

当q ＝3时，a 1＝29，∴a n ＝2

9×3n －1＝2×3n －3. 答案 a n ＝2×33－n 或a n ＝2×3n －3

10．已知数列{lg a n }是等差数列，求证：{a n }是等比数列． 证明：设数列{lg a n }的公差为d ，根据等差数列定义，得lg a n ＋1

－lg a n ＝d ，∴lg a n ＋1a n ＝d ，∴a n ＋1

a n ＝10d (常数)，∴{a n }是一个以10d 为公

比的等比数列．

11．已知三个数成等比数列，它们的和为13，它们的积为27，

求这三个数．

解 根据题意，设这三个数依次为a

q ，a ，aq (aq ≠0)，则?????

a q ·a ·aq ＝27，a q ＋a ＋aq ＝13，

解得?????

a ＝3，

q ＝3，

或?

??

a ＝3，q ＝1

3

.

∴所求三个数依次为1,3,9或9,3,1.

12．设数列{a n }的前n 项和为S n ，且a n ≠0(n ∈N *)，S 1，S 2，…，S n ，…，成等比数列，试问数列a 2，a 3，a 4，…，a n 成等比数列吗？证明你的结论．

解 设a 1＝a ，则S 1＝a 1＝a ，∵{S n }成等比数列，设其公比为q ，则由等比数列的通项公式有S n ＝S 1·q n －1＝aq n －1.

当n ≥2时，a n ＝S n －S n －1＝aq n －1－aq n －2＝aq n －2(q －1)． a n ＋1＝S n ＋1－S n ＝aq n －aq n －1＝aq n －1(q －1)．

当q ＝1时，{S n }为常数列，此时a n ＝0与题设条件a n ≠0矛盾，故q ≠1.

又a n ＋1a n

＝aq n －1(q －1)

aq n －2(q －1)＝q (n ≥2)，

故数列a 2，a 3，a 4，…，a n ，…成等比数列．

人教新课标版语文高一人教版必修三双基限时练 12动物游戏之谜

双基限时练(十二)动物游戏之谜 一、基础测试 1．下列词语中，加点字的注音完全正确的一项是() A．缅．甸(miàn)聒．噪(ɡuō) 潜．游(qián) 嬉．闹(xī) B．脊．背(jǐ) 挨．挤(ái) 倒．立(dào) 消耗．(hào) C．汲．取(jí) 安抚．(fǔ) 尾鳍．(qí) 格．斗(ɡé) D．反馈．(kuì) 模．式(mó) 捉．摸(zuó) 嚼．烂(jiáo) 解析A．缅miǎn；B.挨āi；D.捉zhuō。 答案 C 2．下列词语中，没有错别字的一项是() A．伙伴兴高彩烈平衡史无前例 B．默契得意扬扬厮打销声匿迹 C．跳越猩猩相惜消融铁打金刚 D．元气自娱自乐调剂杂草老藤 解析A．兴高彩．烈—采；B.得意扬扬 ．．—洋洋；C.跳越．—跃，猩猩 ．．相惜—惺惺。 答案 D 3．下列加点的成语使用不正确的一项是() A．在动物身上，无论从形态结构、生理过程，还是行为方面去分析，尽可能节省能量 的例子几乎俯拾皆是 ．．．．。 B．单独游戏时，动物常常兴高采烈 ．．．．地独自奔跑、跳跃，在原地打圈子。 C．每当刮起大风时，成群的露脊鲸把尾鳍高高举出水面，正对着大风，以便像船帆似 的，让大风推着它们，得意洋洋 ．．．．地“驶”向海岸。 D．研究者们各执己见，众说纷纭 ．．．．。 解析A项中应是“比比皆是”。“比比皆是”和“俯拾皆是”都表示相同的事物很多，到处都是。前者侧重表示多得很，到处都是；后者侧重表示容易得到。 答案 A 4．下列各句没有语病的一句是() A．动物的游戏行为，是动物行为研究中被认为最复杂、最难以捉摸、引起争论最多的

行为。 B．在动物身上，无论从形态结构、生理过程和行为方面去分析，尽可能节省能量的例子比比皆是。 C．地震发生后，当地政府及解放军部队全力救助，目前灾区群众已住进了临时帐篷，防止余震再次发生。 D．有人认为科学家终日埋头科研，不问家事，有点不近人情，然而事实却是对这种偏见的最好说明。 解析B项中应将“和”改为“还是”；C项不合逻辑，应为“预防余震发生”；D项中“事实是说明”正好说反了，依句意应是“事实正是对上述说法的反驳”。 答案 A 5．下列句子中标点符号的使用，不正确的一项是() A．马驹常常欢快地连续扬起前蹄，轻盈地蹦跳；猴类喜欢在地上翻滚，拉着树枝荡秋千……单独游戏时动物显得自由自在。 B．这些科学家认为，动物游戏是为了“自我娱乐”，而自我娱乐是动物天性的表现，正像捕食、逃避敌害，繁殖行为等是动物的天性一样。 C．这几种假说，哪一种更有道理？动物的游戏，究竟是为了“演习”，为了“自娱”，为了“学习”，还是为了“锻炼”？研究者们各执己见，众说纷纭。 D．电视剧《水浒传》标明作者为施耐庵、罗贯中，引起许多观众的疑惑：《水浒传》的作者不是施耐庵吗？怎么又多出个罗贯中？ 解析B项中，第二个“自我娱乐”也应加引号，“逃避敌害”后的逗号应改为顿号，因为三者处在同一等级的并列上。 答案 B 6．下列语句排列正确的一项是() ①否则，我就只能相信，这些曲调是真正的音乐。 ②那些歌也许是有关航行，或有关浮游节肢动物的来源，或有关领地界限的简单而实打实的叙述和声明。 ③当然还有其他方法来解释鲸鱼之歌。 ④除非有一天有人证明，这些长长的、缭绕的、执著的曲调，被不同的歌唱者重复着，又加上了它们各自的修饰，这不过是为了向海面下数百英里之外传递像“鲸鱼在这儿”之类寻常的信息。 ⑤但迄今证据还没有得到。 A．③①②⑤④B．②③①④⑤ C．⑤②③①④D．③②⑤④① 解析注意“否则”“除非”“但”等词。

双基限时练26

双基限时练(二十六) 1．已知下列四个等式： ①sin(α＋β)＝sin αcos β＋cos αsin β； ②cos(α＋β)＝cos αcos β－sin αsin β； ③cos ? ?? ??π2＋α＝－sin α； ④tan(α－β)＝tan α－tan β1＋tan αtan β . 其中恒成立的等式有( ) A ．2个 B ．3个 C ．4个 D ．5个 解析 ①，②，③对任意角α，β恒成立，④中的α，β还要使正 切函数有意义． 答案 B 2.1－tan15°1＋tan15° 的值为( ) A. 3 B.33 C ．1 D ．－ 3 解析 原式＝tan45°－tan15°1＋tan45°tan15° ＝tan(45°－15°)＝tan30°＝33. 答案 B 3．设tan(α＋β)＝25，tan ? ????β－π4＝14，则tan ? ?? ??α＋π4等于( ) A.1328 B.1322 C.322 D.16 3．已知α，β为锐角，cos α＝45，tan(α－β)＝－13，则tan β的值为 ( )

A.13 B.139 C.1315 D.59 答案 B 4．已知tan α＋tan β＝2，tan(α＋β)＝4，则tan αtan β等于( ) A ．2 B ．1 C.12 D ．4 解析 因为tan(α＋β)＝tan α＋tan β 1－tan αtan β＝21－tan αtan β ＝4，所以tan αtan β＝12. 答案 C 5．若0<α<π2，0<β<π2，且tan α＝17，tan β＝34，则α＋β等于( ) A.π6 B.π4 C.π3 D.3π4 解析 由已知可求得tan(α＋β)＝1. 又0<α＋β<π，∴α＋β＝π4. 答案 B 6．已知tan α和tan ? ????π4－α是方程ax 2＋bx ＋c ＝0的两个根，则a ，b ，c 的关系是( ) A ．b ＝a ＋c B ．2b ＝a ＋c C ．c ＝b ＋a D ．c ＝ab 解析 由韦达定理可知tan α＋tan ? ????π4－α＝－b a 且tan αtan ? ?? ??π4－a ＝

高二数学 双基限时练3

双基限时练(三) 1．设f ′(x 0)＝0，则曲线y ＝f (x )在点(x 0，f (x 0))处的切线( ) A ．不存在 B ．与x 轴垂直 C ．与x 轴平行 D ．与x 轴平行或重合 答案 D 2．一木块沿某一斜面自由下滑，测得下滑的水平距离s 与时间t 之间的函数关系为s ＝18t 2，则当t ＝2时，此木块在水平方向的瞬时速 度为( ) A. 2 B. 1 C.12 D .14 解析 s ′＝lim Δt →0 Δs Δt ＝lim Δt →0 18(t ＋Δt )2－18t 2Δt ＝lim Δt →0 14tΔt ＋18(Δt )2Δt ＝lim Δt →0 (14t ＋18Δt )＝14t . ∴当t ＝2时，s ′＝12. 答案 C 3．若曲线y ＝h (x )在点P (a ，h (a ))处切线方程为2x ＋y ＋1＝0，则 ( ) A ．h ′(a )<0 B ．h ′(a )>0 C ．h ′(a )＝0 D ．h ′(a )的符号不定 解析 由2x ＋y ＋1＝0，得h ′(a )＝－2<0. ∴h ′(a )<0. 答案 A

4．曲线y ＝9x 在点(3,3)处的切线方程的倾斜角α等于( ) A ．45° B ．60° C ．135° D ．120° 解析 k ＝y ′＝lim Δx →0 Δy Δx ＝lim Δx →0 9x ＋Δx －9x Δx ＝lim Δx →0 －9x (x ＋Δx )＝－9x 2. ∴当x ＝3时，tan α＝－1.∴α＝135°. 答案 C 5．在曲线y ＝x 2 上切线倾斜角为π4的点是( ) A ．(0,0) B ．(2,4) C ．(14，116) D ．(12，14) 解析 y ′＝lim Δx →0 Δy Δx ＝lim Δx →0 (x ＋Δx )2－x 2Δx ＝lim Δx →0 2xΔx ＋(Δx )2Δx ＝lim Δx →0 (2x ＋Δx )＝2x . 令2x ＝tan π4＝1，∴x ＝12，y ＝14. 故所求的点是(12，14)． 答案 D 6．已知曲线y ＝2x 2上一点A (2,8)，则过点A 的切线的斜率为________． 解析 k ＝f ′(2)＝lim Δx →0 2(2＋Δx )2－2×22Δx

2019-2020年高中数学 双基限时练12 新人教A版必修3

2019-2020年高中数学双基限时练12 新人教A版必修3 1．问题：①有1000个乒乓球分别装在3个箱子内，其中红色箱子内有500个，蓝色箱子内有200个，黄色箱子内有300个，现从中抽取一个容量为100的样本；②从20名学生中选出3名参加座谈会． 方法：Ⅰ.简单随机抽样法Ⅱ.系统抽样法Ⅲ.分层抽样法．其中问题与方法能配对的是( ) A．①Ⅰ，②ⅡB．①Ⅲ，②Ⅰ C．①Ⅱ，②ⅢD．①Ⅲ，②Ⅱ 解析读题知①用分层抽样法，②用简单随机抽样法． 答案 B 2．一个单位有职工160人，其中有业务员104人，管理人员32人，后勤服务人员24人，要从中抽取一个容量为20的样本，用分层抽样方法抽出样本，则在20人的样本中管理人员人数为( ) A．3 B．4 C．12 D．7 解析由题意可得20 160 ×32＝4. 答案 B 3．某地区为了解居民家庭生活状况，先把居民按所在行业分为几类，然后每个行业抽1 100的居民家庭进行调查，这种抽样是( ) A．简单随机抽样B．系统抽样 C．分层抽样D．分类抽样 答案 C 4．一个总体分为A，B两层，其个体数之比为，用分层抽样方法从总体中抽取一个容量为10的样本，则A层中抽取的样本个数为( ) A．8 B．6 C．4 D．2 答案 A 5．某大学数学系共有本科生5000人，其中一、二、三、四年级的学生数之比为4:3:2:1.要用分层抽样的方法从所有本科生中抽取一个容量为200的样本，则应抽三年级的学生( ) A．80人B．40人 C．60人D．20人

解析 分层抽样应按比例抽取，所以应抽取三年级的学生人数为200×2 10＝40. 答案 B 6．一个单位共有职工200人，其中不超过45岁的有120人，超过45岁的有80人．为了调查职工的健康状况，用分层抽样的方法从全体职工中抽取一个容量为25的样本，应抽取超过45岁的职工________人． 解析 依题意得，抽取超过45岁的职工人数为25 200×80＝10. 答案 10 7．某工厂生产A ，B ，C 三种不同型号的产品，产品数量之比依次为 现用分层 抽样方法抽出一个容量为n 的样本，样本中A 种型号产品有16件，那么此样本的容量n ＝________. 解析 由题意得n ＝16×10 2＝80. 答案 80 8．课题组进行城市空气质量调查，按地域把24个城市分成甲、乙、丙三组，对应城市数分别为4,12,8.若用分层抽样抽取6个城市，则丙组中应抽取的城市数为________． 答案 2 9．某企业有三个车间，第一车间有x 人，第二车间有300人，第三车间有y 人，采用分层抽样的方法抽取一个容量为45人的样本，第一车间被抽取20人，第三车间被抽取10人，问：这个企业第一车间、第三车间各有多少人？ 解 x ＝20×30045－20－10＝400(人)，y ＝10×300 45－20－10 ＝200(人)． 10．某单位有工程师6 人，技术员12 人，技工18 人，要从这些人中抽取一个容量为 n 的样本．如果采用系统抽样和分层抽样方法抽取，都不用剔除个体；如果样本容量增加1 个，则在采用系统抽样时，需要在总体中先剔除1个个体，求样本容量n . 解 解法1：总体容量为6＋12＋18＝36(人)．当样本容量是n 时，由题意知，系统抽样的间隔为36n ，分层抽样的比例是n 36，抽取工程师人数为n 36×6＝n 6人，技术人员人数为n 36×12 ＝n 3人，技工人数为n 36×18＝n 2 人，所以n 应是6的倍数，36的约数，即n ＝6,12,18. 当样本容量为(n ＋1)时，总体容量是35 人，系统抽样的间隔为35n ＋1，因为35 n ＋1 必须是整数，所以n 只能取6，即样本容量n ＝6. 解法2：总体容量为6＋12＋18＝36(人)． 当抽取n 个个体时，不论是系统抽样还是分层抽样，都不用剔除个体，所以n 应为

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北师大版高中英语必修二双基限时练12

高中英语学习材料 （灿若寒星*制作整理） 双基限时练(十二)Unit 5Lesson 3 Ⅰ.单词拼写 1．I'd like to ________ (辞职) my job because it's too hard for me. 答案quit 2．A famous ________ (音乐家) will perform in my city theatre tomorrow evening. 答案musician 3．His sister is such a ________ (美人) that many young men want to marry her. 答案beauty 4．He still hasn't found his own ________ (才能) in painting. 答案talent 5．Your ________ (发型) is not allowed in our school. 答案hairstyle 6．She enjoys playing the piano and hopes to be a ________ (钢琴家) when she grows up. 答案pianist 7．Mr. Smith has ________ (剃) all of his moustache and looks more handsome. 答案shaved 8．Plants get their supplies of food from the ground through their ________ (根)．

答案roots 9．The business card is your ________ (身份), you must have one at all times. 答案identity 10．We can't judge a person only by his or her ________ (外貌)．答案appearance Ⅱ.同义句转换 1．The little girl surprised the audience by performing a short play in English. →The little girl ________ the audience ________ by performing a short play in English. 答案made; surprised 2．He got too nervous to express himself clearly. →He got ________ nervous ________ he could not express himself clearly. 答案so; that 3．The bus arrived very late because the weather was bad. →The bus arrived very late ________ ________ the bad weather. 答案because of 4．Because of the terrible flood, farmers had a poor harvest that year. →________ ________ ________ ________ the terrible flood, farmers had a poor harvest that year. 答案As a result of 5．When he was only ten years old, the boy began to earn his living by selling newspapers.