2014世纪金榜课时提升作业(六十二) 第九章第六节

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2014版山东《复习方略》(人教A版数学理)课时提升作业第三章第六节简单的三角恒等变换

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课时提升作业(二十二)一、选择题1.2sin(1802)cos 1cos 2cos(90)︒+αα⋅+α︒+α等于 ( ) (A)-sin α (B)-cos α (C)sin α (D)cos α2.函数是 ( )(A)周期为2π的奇函数(B)周期为2π的偶函数(C)周期为4π的奇函数(D)周期为4π的偶函数3.(2013·淄博模拟)已知cos(α-4π)=4,则sin2α= ( )(C)34(D)-344.(2013·济南模拟)若函数f(x)=asin x-bcos x 在x=3π处有最小值-2，则常数a,b 的值分别是( )5.(2013·太原模拟)若函数f(x)=(sinx+cosx)2-2cos 2x-m 在[0,2π]上有零点,则实数m 的取值范围为( )(B)[-1,1]6.已知y=f(x)是奇函数,且图象关于x=3对称,f(1)=1,cosx-sinx=5, 则f(15sin 2xcos(x )4π+)= ( ) (A)-1 (B)0 (C)1 (D)2 二、填空题7.(能力挑战题)已知tan2θπ<2θ<2π,化简22cos sin 12)4θ-θ-πθ+= .8.(2013·温州模拟)函数y=(acosx+bsinx)cosx 有最大值2,最小值-1,则实数(ab)2的值为 . 9.函数y=cos x1sin x-的单调递增区间为 . 三、解答题10.(2013·潍坊模拟)已知函数()2f x sin (x)cos 2x 42π=+-. (1)求函数f(x)的最小正周期和单调递增区间.(2)函数f(x)的图象经过怎样的变换可以得到y=sin 2x 的图象？ 11.(2013·临沂模拟)已知函数f(x)=2sin(13x-6π),x ∈R.(1)求f(54π)的值. (2)设α,β∈[0,2π],f(3α+2π)=1013,f(3β+2π)=65,求cos(α+β)的值.12.(能力挑战题)已知函数f(x)=sin ωx ·sin(2π-φ)-sin(2π+ωx)sin(π+φ)是R 上的偶函数.其中ω>0,0≤φ≤π,其图象关于点M(34π,0)对称,且在区间[0,2π]上是单调函数,求φ和ω的值.答案解析1.【解析】选D.原式=2sin 2cos 1cos 2sin -αα⋅+α-α222sin cos cos 2cos sin -ααα=⋅α-α=cos α2.【思路点拨】利用倍角公式化简成y=Asin ωx 的形式,即可得其相应性质.【解析】选∴最小正周期为2.42ππ= ∵f(-x)=-f(x),∴函数是奇函数.3.【解析】选D.方法一:由cos(α-4π,得2cos α+2sin α=4,即sin α+cos α=12,平方得1+2sin αcos α=14, 故sin2α=-34.方法二:由cos(α-4π)=cos(4π-α), 所以cos(2π-2α)=2cos 2(4π-α)-1=2〃(4)2-1=-34.∵cos(2π-2α)=sin2α,∴sin2α=-34.4.【解析】选D.∵f(x)=asin x-bcos x )=-ϕ,∴2,a b1.1b22⎧=-⇒==-=-5.【解析】选A.f(x)=(sinx+cosx)2-2cos2x-m =1+sin 2x-2cos2x-m=1+sin 2x-1-cos 2x-m4π)-m.∵0≤x≤2π,∴0≤2x≤π,∴-4π≤2x-4π≤34π, ∴-1≤4π)故当-1≤m,f(x)在[0,2π]上有零点. 6.【解析】选A.∵∴1-sin2x=1825.∴sin2x=725,4π∴cos(x+4π)=3.571515sin 2x257.3cos(x)45⨯∴==π+f(7)=f(-1)=-f(1)=-1.7.【解析】原式=cos sin1tan.cos sin1tanθ-θ-θ=θ+θ+θ∵2θ∈(π,2π),∴θ∈(2π,π).而tan2θ=22tan1tanθ-θ2θ-tanθ即θ+1)(tanθ故tanθ=-2或tanθ舍去).∴11tan 1tan +-θ=+θ答案:8.【解析】y=acos 2x+bsinxcosx=1cos 2x ba 22+⋅+sin 2xφ)+a 2, a 2,2a 1,2=∴⎨⎪=-⎪⎩ ∴a=1,b 2=8,∴(ab)2=8. 答案:8【方法技巧】三角恒等变换的特点(1)三角恒等变换就是利用两角和与差的正弦、余弦、正切公式、倍角公式、半角公式等进行简单的恒等变换.三角恒等变换位于三角函数与数学变换的结合点上.(2)对于三角变换,由于不同的三角函数式不仅会有结构形式方面的差异,而且还会有所包含的角,以及这些角的三角函数种类方面的差异,因此三角恒等变换常常首先寻找式子所包含的各个角之间的联系,这是三角恒等变换的重要特点. 9.【思路点拨】利用倍角公式展开约分后化为正切再求解.【解析】222x xcos sin cos x 22y x x 1sin x (cos sin )22-==-- x x x cos sin 1tan222x x x cos sin 1tan222++==--=tan(x 2+4π).由k π-2π<x 2+4π<2π+k π,k ∈Z,知2k π-32π<x<2k π+2π,k ∈Z. 答案:(2k π-32π,2k π+2π),k ∈Z10.【解析】(1)f(x)=sin 2(4πcos 2x 1cos(2x)22π-+=11sin 2x 2221sin(2x ).23=+-π=+- 最小正周期T=π，单调递增区间为［5k ,k 1212ππ-π+π］,k ∈Z. (2)向左平移6π个单位，再向下平移12个单位.11.【解析】(1)f(54π)=2sin(512π-6π)=2sin 4π(2)f(3α+2π)=2sin α=10,13∴sin α=5.13又α∈[0, 2π],∴cos α=12,13f(3β+2π)=2sin(β+2π)=2cos β=6,5∴cos β=3.5又β∈[0, 2π],∴sin β=4,5∴cos(α+β)=cos αcos β-sin αsin β=16.6512.【解析】由已知得f(x)=sin ωxcos φ+cos ωxsin φ =sin(ωx+φ),∵f(x)是偶函数,∴φ=k π+2π,k ∈Z.又∵0≤φ≤π,∴φ=2π. ∴f(x)=sin(ωx+2π)=cos ωx.又f(x)关于(34π,0)对称, 故34πω=k π+2π,k ∈Z.即ω=4k 2,33+k ∈Z. 又ω>0,故k=0,1,2,…当k=0时,ω=23,f(x)=cos 23x 在[0, 2π]上是减函数. 当k=1时,ω=2,f(x)=cos2x 在[0, 2π]上是减函数.当k=2时,ω=103,f(x)=cos 103x 在[0, 2π]上不是单调函数, 当k>2时,同理可得f(x)在[0, 2π]上不是单调函数,综上,ω=23或ω=2.关闭Word 文档返回原板块。

人教版八年级物理上册课时提升卷及答案解析课时提升作业 (9)

课时提升作业(二十三)密度(30分钟40分)一、选择题(本大题共4小题,每小题4分,共16分)1.下列图像中,能正确反映同种物质的质量和体积关系的是()2.如表所示给出了在常温常压下一些物质的密度,阅读后请判断下面一些结论,其中正确的是( )物质密度/(kg·m-3) 物质密度/(kg·m-3)纯水 1.0×103冰0.9×1030.5×103煤油0.8×103干松木酒精0.8×103铜8.9×103水银13.6×103铅11.3×103A.固体的密度都比液体的大B.不同的物质,密度一定不同C.同种物质在不同状态下,其密度不同D.质量相等的实心铜块和实心铅块,铜块的体积比铅块小【解析】选C。

本题考查从密度表中获取信息的能力和对密度的理解。

从表中可以看出,冰、干松木、铜、铅的密度均小于水银的密度,故“固体的密度都比液体的大”说法错误;从表中可以看出,煤油和酒精的密度相等,故“不同的物质,密度一定不同”说法错误;从表中可以看出,水这种物质在液态和固态下,密度不相等,故“同种物质在不同状态下,其密度不同”说法正确;由密度公式变形V=可知,质量相等的实心铜块和实心铅块,密度大的,体积小,从表中可知铅的密度大于铜的密度,所以铅块的体积比铜块小,故“质量相等的实心铜块和实心铅块,铜块的体积比铅块小”说法错误。

故C选项正确,故选C。

3.(2013·天津中考)学完密度知识后,一位普通中学生对自己的身体体积进行了估算。

下列估算值最接近实际的是( ) A.30dm3 B.60dm3 C.100dm3 D.120dm3【解析】选B。

本题考查对密度公式的应用。

中学生的质量约为m=60kg,密度约为ρ=1.0×103kg/m3,中学生的体积V===0.06m3=60dm3。

故选B。

4.如图是在探究甲、乙两种物质质量跟体积关系时作出的图像,以下分析正确的是( )A.若V甲=V乙,则m甲<m乙B.若m甲=m乙,则V甲>V乙C.不同物质的质量跟体积的比值是不同的D.甲物质的质量跟体积的比值比乙物质小【解析】选C。

2014版英语复习方略(外研版通用) 课时提升作业(六) 必修1 Module 6 The Internet and Telecommunications

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课时提升作业(六)Ⅰ. 单项填空1. Computer lessons must be to students of all grades, so that they can master this modern communication and learning tool.A. avoidableB. accessibleC. valuableD. acceptable2. (2013·成都模拟)I find hard to concentrate on my studies with some people having small talk around me.A. themB. himC. thatD. it3. —How is Dennis getting along with his work?—Well, he could always new ideas for increasing sales.A. come up withB. come aboutC. get away fromD. get up4. (2013·哈尔滨模拟)The magician picked some persons from the audienceand asked them to help him with the performance.A. by accidentB. in vainC. at randomD. on average5. It is said that some software on learning subjects for pupils in a very practical way.A. designedB. is designedC. designsD. is designing6. (原创)Terribly, a Nepal’s plane crashed down near their capital, killing 19,7 British tourists.A. containingB. includingC. coveringD. counting7. (2013·福州模拟)Drunk driving used to occur, but now it is under control.A. frequentlyB. generallyC. practicallyD. gradually8. There were so many people talking in the main hal l that I couldn’t the words of the lecturer.A. concentrate onB. fix uponC. centre onD. devote to9. They were taken to the police station as they had entered the area without.A. permitB. permittedC. permissionD. a permission10. “Hope for the best and prepare for the worst” is a proverb, meaning that life is beautiful and full of frustration.A. stillB. as wellC. eitherD. in case11. Our car had a on the way to the airport last week.A. breakoutB. breakdownC. break-inD. break-up12. Some advertisements pictures or words of experts to show people howgood the products are.A. are consisted ofB. are made ofC. consist ofD. are made up13. Grandma’s very and does all her own shopping and cooking.A. financialB. individualC. friendlyD. independent14. (原创)Last year, China sold a total of 18. 5 million vehicles, 12. 8 million vehicles sold in the US.A. comparing withB. being compared withC. to compare withD. compared with15. Anyone not familiar with the Internet is at a serious when applying for a job in such an age of information.A. disadvantageB. advantageC. disagreementD. agreementⅡ. 阅读理解(2013·皖南八校模拟) Since the Internet has come into homes, the daily life has never been the same again. But the thing that worries most of us is that we can get viruses from the Internet. But can we catch viruses on our cell phones? A new study in the journal Science says yes, but the spread of such mobile software that can bring harm to our cell phones won’t reach dangerous levels until more cell phones are on the sameoperating system.Computers are easily attacked by viruses because they share data, especially over the Internet. Of course, nowadays, more people are using their cell phones more and more frequently. They use them for emailing, text messaging and downloading troublesome ring tones, etc. so it is obvious that cell phone viruses are a threat, as well.Scientists used nameless call data from more than six million cell phone users to help model a possible outbreak. And they concluded that viruses that spread from phone to phone by Bluetooth are not much concerned, because users have to be in close physical relation for their phones to“see”one another. However, viruses that spread through multimedia messaging services can move much faster, because they can come in disguise, such as a cool tune sent by a friend. The good news is that to be effective, these viruses need their victims to all use the same operating system, which not enough of us do. Because there is no Microsoft operating system for mobile phones, yet. Thank goodness. (272W)1. What’s the passage mainly about?A. The operating system of cell phones.B. The threat of cell phone viruses.C. The wide use of cell phones.D. Computer viruses.2. When, cell phone viruses can be dangerous.A. all the cell phones work on the same operating systemB. Microsoft operating system for cell phones is createdC. users can see each other on the phoneD. Bluetooth is widely used3. What can we infer about Bluetooth according to the scientists?A. It can increase the chances of cell phone viruses greatly.B. It can help us to stop the spreading of cell phone viruses.C. It won’t possibly cause the outbreak of cell phone viruses.D. It can make users have close physical relation with one another.4. We can learn from the passage that.A. it is impossible to catch viruses on our cell phonesB. cell phones are not well connected with the InternetC. it is dangerous for people to download ring tones to cell phonesD. more than one operating system is available for cell phone users nowⅢ. 阅读第二节(2013·延边模拟) 根据短文内容, 从短文后的选项中选出能填入空白处的最佳选项。

2014版英语复习方略课时提升作业(二) 必修1 Unit 2 English around the world(人教版通用)

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课时提升作业(二)必修1 Unit 2Ⅰ. 单项填空1. (2013·贵阳模拟)Today, rapid development of transportation and communication, the whole world seems to be smaller.A. regardless ofB. in spite ofC. instead ofD. because of2. (2013·南宁模拟)As soon as the project at the meeting, it attracted many people’s attention.A. comes upB. was come upC. came upD. had been come up3. (2013·太原模拟)With our knowledge practice, we can make great contributions to our country.A. based onB. basing onC. base onD. to base on4. (2013·兰州模拟)Every minute has been made full of by the girls ______ knowledge about etiquette(礼仪).A. using; learningB. use; to learnC. to use; to learnD. use; learning5. My English teacher advised me keep my mouth shut to improve my spokenEnglish.A. toB. not toC. notD. mustn’t6. (2013·唐山模拟)—I’m happy to take you there in my car.—You are so nice. Thank you.A. rather thanB. other thanC. more thanD. less than7. (2013·郑州模拟)He that the meeting room should be cleaned thoroughly.A. declaredB. announcedC. expressedD. commanded8. (2013·南阳模拟)More high-speed trains are to slow down to improve safety.A. requestedB. recognizedC. suggestedD. forbidden9. The film star wore dark glasses so that no one would him.A. knowB. reachC. recognizeD. realize10. Seeing the road with snow, we had to spend the holiday at home, watching TV.A. blockingB. to blockC. blockedD. to be blocked11. (2013·衡阳模拟)Many of them turned a deaf ear to his advice, they knew it to be valuable.A. as ifB. now thatC. even thoughD. so that12. —Have you seen a red pen on the desk?—Sorry. There is pen here.A. not suchB. such noC. no such aD. no such13. (2013·宁波模拟)According to the present economical, it is important to safeguard the regional growth and stability.A. expressionB. situationC. conditionD. translation14. She will tell us why she feels so strong that each of us has a role in making the earth a better place to live on.A. to have playedB. to playC. to be playedD. to be playing15. As is known to all language learners, the newly learnt words will soon be forgotten unless used in everyday communication.A. graduallyB. accuratelyC. familiarlyD. frequentlyⅡ. 阅读理解It started out as a noble intention to help travelers who lose their way in the mountains. Wang Gaoming never expected it to turn into a fulfilling career.“Initially, some travelers did not believe that I had better forest travel knowledge than them, ” says Wang from Hetaoping village in the Qinling Mountains in Taibaicounty, Shaanxi Province.While most of his fellow villagers had left for greener pastures(牧场), the 37-year-old stayed on. Living with his 80-year-old mother, Wang remains single because he “has been too poor to get married” .“Many villagers of my age have left for bigger cities but I like to stay to help develop the mountain tourism in our county, ” Wang says.He knows the mountains inside out because he used to follow his father to pick wild herbs(药草). Those trips also enhanced his knowledge of the various species of plants available and their characteristics.“It was hard work to pick herbs deep in the mountains and we often stayed for more than 20 days in the forest, ” Wang says. “Put me anywhere in the mountains, and I can easily tell the direction and find the way out. ”Taibai Mountain, a part of the Qinling Mountain range, is popular among hikers from across the country because of its unique and beautiful natural scenery.But there have been incidents of travelers losing their way. Some had died.“In June 2002, I learned that a college student died because he couldn’t find his way out of the mountains. I felt really sad that a young life ended that way and decided to do my part to help other tourists, ” he says.Very quickly, Wang earned a reputation as a forest guide, earning some 50 yuan ($8) a day. (288W)1. How does he get better forest travel knowledge?A. By picking wild herbs.B. By living in the village.C. By helping other tourists.D. By enjoying natural scenery.2. Which of the following can best describe Wang Gaoming?A. Reliable and generous.B. Tough and devoted.C. Careful and caring.D. Proud and wise.3. Why does he want to be a forest guide?A. Because he has lots of forest travel knowledge.B. Because he wants to help tourists.C. Because he wants to earn money.D. Because he used to pick herbs.4. What is the passage mainly about?A. The way to pick herbs.B. Measures to earn money.C. A forest guide.D. An advertisement of forest travel.Ⅲ. 阅读第二节(2013·敦煌模拟) 根据短文内容, 从短文后的选项中选出能填入空白处的最佳选项。

2014届高考生物总复习课时提升作业(七)第3章第3节物质的跨膜运输苏教版必修1

（金榜题库）2014届高考生物总复习课时提升作业（七）第3章第3节物质的跨膜运输苏教版必修1(45分钟 100分)一、选择题(包括12小题，每小题5分，共60分)1.RNA聚合酶参与细胞核内DNA的转录，其进入方式是( )A.简单扩散B.易化扩散C.主动运输D.非跨膜运输2.在培养玉米的营养液中加入某种负离子，结果发现玉米根细胞在吸收该种负离子的同时，对氯离子的吸收减少了，但对钾离子的吸收没有受影响，最可能的原因是( )A.该种负离子妨碍了ATP的形成B.该种负离子抑制了主动运输C.该种负离子抑制了细胞呼吸D.该种负离子和氯离子的载体蛋白相同3.下图表示物质跨膜运输的一种方式。

据图分析正确的是( )A.这种运输方式可逆浓度梯度进行B.水分子是以这种方式进入细胞的C.细胞产生的能量增加会提高物质的运输速率D.载体蛋白在物质运输过程中形状会发生改变4.用酶解法除去新鲜紫色洋葱鳞片叶的外表皮细胞的细胞壁得到原生质体，将其置于0.3 g/mL的蔗糖溶液中，一段时间后，原生质体一定不发生( )A.质壁分离B.渗透失水C.颜色加深D.吸水力增大5.(2013·某某模拟)在植物细胞质壁分离复原过程中，能正确表达细胞吸水速率变化过程的是( )6.如图所示，把体积与质量浓度相同的葡萄糖溶液与蔗糖溶液用半透膜(允许溶剂和葡萄糖通过，不允许蔗糖通过)隔开，开始时和一段时间后液面情况是( )A.甲高于乙B.乙高于甲C.甲先高于乙，乙后高于甲D.乙先高于甲，甲后高于乙7.(2013·某某模拟)下图表示小肠绒毛上皮细胞中的细胞膜对不同物质的运输(每种运输的方向由箭头表明，黑点的数量代表每种物质的浓度)，下列叙述正确的是( )A.a物质可能是氧气，b物质可能是葡萄糖B.a物质可能是水，b物质可能是甘油C.a物质可能是胆固醇，b物质可能是氧气D.a物质可能是葡萄糖，b物质可能是氨基酸8.(2013·某某联考)物质进出细胞的方式有跨膜运输(被动运输和主动运输)和非跨膜运输(内吞和外排)。

2014版高中数学复习方略课时提升作业：5.1数列(北师大版)(北师大版·数学理·通用版)

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课时提升作业(三十)一、选择题1.已知数列,,,…,,…,下面各数中是此数列中的项的是( )(A)(B)(C)(D)2.由a 1=1,a n+1=,给出的数列{a n}的第34项为( )(A)(B)100(C)(D)3.(2013·南昌模拟)已知数列{a n}的前n项和S n=2-2n+1,则a3= ( )(A)-1 (B)-2 (C)-4 (D)-84.已知数列{a n}的前n项和S n=2n2-3n+1,则a4+a5+a6+a7+a8+a9+a10的值为( )(A)150 (B)161 (C)160 (D)1715.(2013·西安模拟)在数列{a n}中,a1=1,a n a n-1=a n-1+(-1)n(n≥2,n∈N+),则的值是( )(A)(B)(C)(D)6.在数列{a n}中,a1=2,a n+1=a n+ln(1+),则a n= ( )(A)2+lnn (B)2+(n-1)lnn(C)2+nlnn (D)1+n+lnn7.已知数列{a n}的前n项和S n=n2-9n,第k项满足5<a k<8,则k等于( )(A)9 (B)8 (C)7 (D)68.(能力挑战题)定义:F(x,y)=y x(x>0,y>0),已知数列{a n}满足:a n=(n∈N+),若对任意正整数n,都有a n≥a k(k∈N+)成立,则a k的值为( )(A)(B)2 (C)3 (D)4二、填空题9.数列-,,-,,…的一个通项公式可以是.10.数列{a n}的前n项和记为S n,a1=1,a n+1=2S n+1(n≥1,n∈N+),则数列{a n}的通项公式是.11.(2013·赣州模拟)已知数列{a满足a1=,a n-1-a n=(n≥2),则该数列的通项公式a n= .12.(能力挑战题)已知数列{a n}满足:a1=m(m为正整数),a n+1=若a6=1,则m所有可能的值为.三、解答题13.已知数列{a n}满足前n项和S n=n2+1,数列{b n}满足b n=,且前n项和为T n,设c n=T2n+1-T n.(1)求数列{b n}的通项公式.(2)判断数列{c n}的增减性.14.(能力挑战题)解答下列各题:(1)在数列{a n}中,a1=1,a n+1=ca n+c n+1(2n+1)(n∈N+),其中实数c≠0.求{a n}的通项公式.(2)数列{a n}满足:a1=1,a n+1=3a n+2n+1(n∈N+),求{a n}的通项公式.15.(2012·广东高考)设数列{a n}前n项和为S n,数列{S n}的前n项和为T n,满足T n=2S n-n2,n∈N+.(1)求a1的值.(2)求数列{a n}的通项公式.答案解析1.【解析】选B.∵42=6×7,故选B.2.【解析】选C.把递推式取倒数得=+3,所以=+3×(34-1)=100,所以a34=.3.【解析】选D.a3=S3-S2=-14-(-6)=-8.4.【解析】选B.S10-S3=(2×102-3×10+1)-(2×32-3×3+1)=161.5.【解析】选C.当n=2时,a2·a1=a1+(-1)2,∴a2=2.当n=3时,a3a2=a2+(-1)3,∴a3=.当n=4时,a4a3=a3+(-1)4,∴a4=3.当n=5时,a 5a4=a4+(-1)5,∴a5=,∴=.6.【思路点拨】根据递推式采用“叠加”方法求解.【解析】选A.∵a n+1=a n+ln(1+)=a n+ln=a n+ln(n+1)-lnn,∴a2=a1+ln2,a3=a2+ln3-ln2,…,a n=a n-1+lnn-ln(n-1),将上面n-1个式子左右两边分别相加得a n=a1+ln2+(ln3-ln2)+(ln4-ln3)+…+[lnn-ln(n-1)]=a1+lnn=2+lnn.7.【解析】选B.a n=即a n=∵n=1时也适合a n=2n-10,∴a n=2n-10.∵5<a k<8,∴5<2k-10<8,∴<k<9.又∵k∈N+,∴k=8.,==,2n2-(n+1)2=n2-2n-1,只有当n=1,28.【解析】选 A.a时,2n2<(n+1)2,当n≥3时,2n2>(n+1)2,即当n≥3时,a n+1>a n,故数列{a n}中的最小项是a1,a2,a3中的较小者,a1=2,a2=1,a3=,故a k的值为.9.【解析】正负相间使用(-1)n,观察可知第n项的分母是2n,分子比分母的值少1,故a n=(-1)n.答案：a n=(-1)n10.【思路点拨】根据a n和S n的关系转换a n+1=2S n+1(n≥1)为a n+1与a n的关系或者S n+1与S n的关系.【解析】方法一:由a n+1=2S n+1可得a n=2S n-1+1(n≥2),两式相减得a n+1-a n=2a n,a n+1=3a n(n≥2).又a2=2S1+1=3,∴a2=3a1,故{a n}是首项为1,公比为3的等比数列,∴a n=3n-1.方法二:由于a n+1=S n+1-S n,a n+1=2S n+1,所以S n+1-S n=2S n+1,S n+1=3S n+1,把这个关系化为S n+1+=3(S n+),即得数列{S n+}为首项是S1+=,公比是3的等比数列,故S n+=×3n-1=×3n,故S n=×3n-.所以,当n≥2时,a n=S n-S n-1=3n-1,由n=1时a1=1也适合这个公式,知所求的数列{a n}的通项公式是a n=3n-1.答案：a n=3n-1【方法技巧】a n和S n关系的应用技巧在根据数列的通项a n与前n项和的关系求解数列的通项公式时,要考虑两个方面,一个是根据S n+1-S n=a n+1把数列中的和转化为数列的通项之间的关系;一个是根据a n+1=S n+1-S n把数列中的通项转化为前n项和的关系,先求S n再求a n.11.【解析】由递推公式变形,得-==-,则-=1-,-=-,…,-=-,各式相加得-=1-,即=,∴a n=.答案：12.【解析】根据递推式以及a1=m(m为正整数)可知数列{a n}中的项都是正整数.a 6=1,若a6=,则a5=2,若a6=3a5+1,则a5=0,故只能是a5=2.若a 5=,则a4=4,若a5=3a4+1,则a4=,故只能是a4=4.若a 4=,则a3=8,若a4=3a3+1,则a3=1.(1)当a 3=8时,若a3=,则a2=16,若a3=3a2+1,则a2=,故只能是a2=16,若a2=,则a1=32,若a2=3a1+1,则a1=5.(2)当a 3=1时,若a3=,则a2=2,若a3=3a2+1,则a2=0,故只能是a2=2.若a 2=,则a1=4,若a2=3a1+1,则a1=,故只能是a1=4.综上所述:a1的值,即m的值只能是4或5或32.答案：4或5或32【变式备选】已知数列{a n}中,a1=,a n+1=1-(n≥2),则a16= .【解析】由题可知a2=1-=-1,a3=1-=2,a4=1-=,∴此数列为循环数列,a1=a4=a7=a10=a13=a16=.答案：13.【解析】(1)a1=2,a n=S n-S n-1=2n-1(n≥2).∴b n=(2)∵c n=b n+1+b n+2+…+b2n+1=++…+,∴c n+1-c n=+-=<0,∴{c n}是递减数列.14.【解析】(1)由原式得=+(2n+1).令b n=,则b1=,b n+1=b n+(2n+1),因此对n≥2有b n=(b n-b n-1)+(b n-1-b n-2)+…+(b2-b1)+b1=(2n-1)+(2n-3)+…+3+=n2-1+,因此a n=(n2-1)c n+c n-1,n≥2.又当n=1时上式成立.因此a n=(n2-1)c n+c n-1,n∈N+.(2)两端同除以2n+1得,=·+1,即+2=(+2),即数列{+2}是首项为+2=,公比为的等比数列,故+2=×()n-1,即a n=5×3n-1-2n+1.15.【解析】(1)当n=1时,T1=2S1-1.因为T1=S1=a1,所以a1=2a1-1,求得a1=1.(2)当n≥2时,S n=T n-T n-1=2S n-n2-[2S n-1-(n-1)2]=2S n-2S n-1-2n+1,所以S n=2S n-1+2n-1 ①,所以S n+1=2S n+2n+1 ②,②-①得a n+1=2a n+2,所以a n+1+2=2(a n+2),即=2(n≥2),求得a1+2=3,a2+2=6,则=2.所以{a n+2}是以3为首项,2为公比的等比数列,所以a n+2=3·2n-1,所以a n=3·2n-1-2,n∈N+.关闭Word文档返回原板块。

课时提升作业(九) 必修1 5.1&5.2

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课时提升作业(九)降低化学反应活化能的酶细胞的能量“通货”——ATP(45分钟100分)一、单项选择题(包括7题,每题4分,共28分。

每题只有一个选项符合题意。

)1.(2014·云浮模拟)下列关于酶的叙述中,正确的是( )A.人体中酶只能在人体的内环境中起作用B.酶的形成都要经过核糖体的合成、内质网和高尔基体的加工等几个阶段C.基因控制生物的性状有些是通过控制酶的合成来控制相应代谢过程实现的D.酶均是由内分泌腺或外分泌腺的细胞合成的,具有高效性、专一性2.(2014·漳州模拟)如图表示不同pH及温度对某反应产物生成量的影响,下列相关叙述正确的是( )A.随着pH的升高,酶的活性先降低后增大B.该酶的最适温度是35℃C.酶的最适pH是一定的,不随着温度的升高而升高D.随着温度的升高,酶的活性逐渐降低3.(2012·海南高考)下列操作中,不可能导致淀粉酶活性发生变化的是( )A.淀粉酶溶液中加入强酸B.淀粉酶溶液中加入蛋白酶C.淀粉酶溶液中加入淀粉溶液D.淀粉酶经高温烘干制成粉剂4.(2014·延吉模拟)关于探究淀粉酶最适用量的实验,叙述不正确的是( )A.各组需加入等量不同浓度的淀粉酶溶液B.要保证各组适宜并相同的pH和温度等条件C.需要检测不同反应时间条件下的生成物量D.几组实验之间可形成相互对照,不需单设空白对照5.下列有关细胞的能量“通货”——ATP变化的叙述错误的是( )A.ATP与ADP相互转化的能量供应机制是生物界的共性B.人体在紧张或愤怒状态下,细胞内产生ATP的速率大大超过产生ADP的速率C.ATP中的能量可以来源于光能或化学能D.人体在剧烈运动时,通过机体的神经调节和体液调节,细胞产生ATP的速率迅速增大6.(2014·南京模拟)下列曲线中能正确表示人体消化酶作用规律的是( )A.Ⅰ和ⅢB.Ⅱ和ⅢC.Ⅰ和ⅣD.Ⅱ和Ⅳ7.(2014·成都模拟)ATP是细胞内直接的能源物质,可通过多种途径产生,如图所示。

2014高考英语全程复习方略课时提升作业九：Unit3 Back to the past(牛津译林版必修3)

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课时提升作业(九)必修3Unit 3Ⅰ. 单项填空1. The “Titanic” sank slowly to the bottom of the ocean with 1, 500 passengers ______.A. on the boardB. on boardC. getting on boardD. across the board2. Shopping online is ______ accepted. More and more people are buying what they need on the Internet.A. fortunatelyB. graduallyC. eventuallyD. faithfully3. At the beginning of class, the noise of desks ______ could be heard outside the classroom.A. opened and closedB. to be opened and closedC. being opened and closedD. to open and close4. (2013·台州模拟)A survey of China’s netizens found that some smokers are unwilling to kick the habit because they aren’t fully ______ of the harm it does to health.A. fondB. curiousC. afraidD. aware5. It ______ have been Tom that parked the car here, as he is the only one with a car.A. mayB. canC. mustD. should6. —Miss Brown is very busy, isn’t she?—So she is. She has ______ greater responsibilities since she was promoted.A. taken offB. taken afterC. taken inD. taken on7. Ladies and gentlemen，please remain ______ until the plane has come to a complete stop.A. seatedB. seatingC. to seatD. seat8. Mr Brown is willing to listen to his students’ problems but won’t give them any direct ______.A. solutionB. demandC. measureD. function9. At the conference the Chinese foreign minister ______ his opinion that China was strongly against the terrorism.A. declaredB. statedC. announcedD. showed10. ______ China’s Youth’s Day, a c elebration evening took place on our campus. It was a success.A. In place ofB. In hopes ofC. Instead ofD. In memory of11. (2013·南京模拟)Mozart left more than 600 works, but who knows how many more pieces by the master lie ______ to be discovered?A. waitingB. waitedC. to waitD. wait12. At this rate, the forest will be completely ______ within the next 30 years.A. damagedB. ruinedC. spoiledD. destroyed13. The topic we are going to talk about today is actually of great importance in living a low-carbon life, though ______ a small one.A. thought of asB. thought asC. thinking of asD. thinking as14. (2013·无锡模拟) She is in a poor ______ of health, which worries her mother a lot.A. positionB. situationC. stateD. condition15. If ______ in wet sand, the vegetables can remain fresh for a longtime.A. being buriedB. having buriedC. buriedD. buryingⅡ. 阅读理解(A)(2013·盐城模拟) The largest campaign of killing rats in history is set to poison millions of rats on the sub-Antarctic island of South Georgia. Scientists say the ca mpaign planned for 2013 and 2014 will restore beautiful South Georgia to the position it once held as the world’s most important nesting site for seabirds.It was sailors in the late 18th century who unintentionally introduced rats to what had been a fresh environment. “If we can destroy the rats, at least 100 million birds will return to their home on South Georgia. ” says Tony Martin, a biology professor at the University of Dundee who was invited to lead the project.South Georgia is by far the largest island to get rid of animals that destroy native wildlife after being introduced deliberately or accidentally by people. Though rats and mice have done the most damage, cats, foxes, pigs, goats, deer, rabbits and other species have been targeted in the campaigns around the world.South Georgia is seven times the size of New Zealand’s CampbellIsland, currently the largest area ever killing rats. The successful war against Campbell Island rats was carded out in 2001 with 132 tons of poison dropped from five helicopters.“New Zealand pioneered the techniques for ridding islands of rats and in fact our operation on South Georgia is based on New Zealand’s technology. ” says Martin. “Some New Zealanders will be helping our campaign, including our chief pilot, Peter Garden, who was also chief pilot for the projects at Campbell Island and Rat Island, in the Aleutian chain of the north Pacific. ”The second and third stages in 2013 and 2014 will involve dropping as much as 300 tons of poison from the air onto every part of the island where rats might live. It is a huge operation, carried out during the stormy southern autumn when the rats are hungry and the risks of poisoning native wildlife are less than in the spring and summer months. “Ideally we’d do it in winter but the weather makes that too risky, ” Martin says.The ecological payback will be priceless. But Martin says. “The full benefits will take decades to arrive，because some of these birds are very slow to hatch. ”(370W)1. According to the passage, how did the rats appear on the sub-Antarctic island of South Geor gia?A. They were attracted there by wildlife.B. They escaped there from Campbell Island.C. They were introduced there by sailors accidentally.D. They were brought in by people deliberately.2. Which of the following is TRUE about Peter Garden?A. He is in charge of the campaign on the sub-Antarctic island.B. He will be the only pilot for the project on the sub-Antarctic island.C. He will benefit a lot from the campaign on the sub-Antarctic island.D. He made great contributions to the projects at Campbell Island and Rat Island.3. The operation of ridding South Georgia of rats is to be carried out in autumn because ______.A. the war against Campbell Island rats failed in all seasons except autumnB. only then do the New Zealanders to help the operation have spare timeC. rats then need more food and the operation does less harm to native wildlifeD. the poison kills the rats more effectively than it does in any other season4. What can we infer from the passage?A. The campaign of killing rats will benefit the native wildlife in a short time.B. Rats aren’t the only species to be blamed for the disappearance of wildlife.C. The first stage of killing rats on the sub-Antarctic island didn’t make great achievements.D. The campaign in South Georgia will fully follow in the footsteps of that on Campbell Island.(B)As China comes down from a travel rush duringthe combined eight-day holiday for Mid-AutumnFestival and National Day of 2012, experts calledfor a return of the Labor Day Golden Week holidayas a way to ease the travel peak.By Friday, 79 million passengers were estimated to have used the railways up about 8percent over last year. Waterways had shipped around 2 million, an increase of about 17 percent on last year, Cntv. cn, the website of China Central Television, said on Saturday.As large numbers of tourists swarmed scenic spots around China, it caused huge crowds and numerous complaints.The Forbidden City in Beijing has long been a big draw for travelers. On Tuesday alone, the museum reported, it received more than 180, 000 visitors, about six times higher than a regular day. According to China Central Television, restless visitors demanded their money back from the tourism committee, and police were sent to help deal with the crisis.The Beijing-HongKong-Macao expressway, the Shanghai-Kunmingexpressway and the route from Beijing to Kunming saw large increases in traffic on Saturday, according to Cntv. cn.As there is only one long holiday in the country and paid leave is not well carried out by employers, people have limited chances for travel, Dai Bin, the director of China Tourism Academy, said in a report by Beijing Times on Saturday.The travel rush during the “Golden Week” holidays happens because people do not take long journeys during shorter holidays, said Liu Simin, a researcher with the China Academy of Social Sciences, in a report by Beijing Times on Saturday.Liu said that now the most important task is to release the pressure from the huge numbers of tourists. When the Labor Day Golden Week holiday was abolished in 2007, a chance to travel was reduced.In 2008, the government shortened the Labor Day Golden Week holiday, usually occurring from May 1 to May 7, to three days and added three other short vacations to the list. (333W)5. Why does the travel rush during the “Golden Week” holidays happen?A. Because people only have one long holiday to take long journeys.B. Because the weather is fine during the “Golden Week” holidays.C. Because the scenic spots are free during the “Golden Week” holidays.D. Because the sc en ic spots are more beautiful during the “Golden Week” holidays.6. The Labor Day Golden Week holiday was shortened by ______.A. one dayB. two daysC. three daysD. four days7. The underlined word “swarmed” in the third paragraph proba bly means “______”.A. crowded intoB. leftC. escaped fromD. planned to go to8. Which of the following statements is TRUE according to the passage?A. About 8 percent passengers were estimated to have used the railways.B. Three expressways were mentioned in the passage.C. Experts called for a return of the Labor Day Golden Week holiday.D. Employers had nothing to do with limited chances for travel.9. What is the passage mainly about？A. Golden Week sees huge crowds on popular tourist spots.B. The problem during the Eight-day Golden Week Holiday.C. Travel complaints.D. The pressure from the huge numbers of tourists.【语篇随练】把阅读理解中的段落译成汉语In 2008, the government shortened the Labor Day Golden Week holiday, usually occurring from May 1 to May 7, to three days and added three other short vacations to the list.【译】___________________________________________________________ ______________________________________________________________ _________答案解析Ⅰ.1.【解析】选B。

课时提升作业(二) 必修1 2.1&2.5

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课时提升作业(二)细胞中的元素和化合物细胞中的无机物(45分钟100分)一、单项选择题(包括7题,每题4分,共28分。

每题只有一个选项符合题意。

)1.(2014·济宁模拟)对火星进行探测表明,火星过去曾有丰富的水资源,由此推测火星上可能存在过生命。

这一推测的理由是( )A.水是细胞中含量最多的化合物B.结合水是细胞和生物体的成分C.一切生命活动都离不开水D.代谢旺盛的细胞中水的含量高2.(2014·南昌模拟)下列有关细胞中元素和化合物的叙述,正确的是( )A.C是构成细胞的基本元素,在人体活细胞中含量最多B.脂肪分子中含H比糖类多,是主要的能源物质C.氨基酸脱水缩合产生水,水中的氧来自氨基酸的羧基D.RNA和DNA主要组成元素的种类不同,碱基种类不完全相同3.如图表示人体细胞中四种主要元素占细胞鲜重的百分比,其中表示碳元素的是( )A.甲B.乙C.丙D.丁4.(2014·潍坊模拟)下列关于无机盐的叙述,错误的是( )A.铁是合成血红蛋白不可缺少的成分B.所有蛋白质中都含有硫元素C.植物缺镁会影响光合作用D.食盐中加碘可预防地方性甲状腺肿5.农科院研究员在指导农民生产的过程中发现一位农民种植的某块农田小麦产量总是比邻近地块的低。

他怀疑该农田可能是缺少某种元素,为此将该块肥力均匀的农田分成面积相等的五小块,进行田间试验。

除施肥不同外,其他田间处理措施相同,试验结果如下表：从表中可判断,该农田最可能缺少的元素是( )A.KB.NC.PD.S6.(2014·三明模拟)科学工作者研究了钙和硼对某种植物花粉粒萌发和花粉管生长的影响,结果如图所示。

下列结论错误的是( )A.适宜浓度的钙有利于花粉管的生长,适宜浓度的硼有利于花粉的萌发B.钙或硼对花粉萌发和花粉管生长有同样的影响C.钙在一定浓度范围内几乎不影响花粉的萌发D.硼在一定浓度范围内几乎不影响花粉管的生长7.(2014·荆州模拟)人体红细胞呈圆饼状,具有运输氧气的功能。

高中数学必修一第一章集合与函数概念1-1集合课时提升作业及解析

综上可知 k=0 或 1. 【误区警示】解答本题时易不考虑二次项系数 k 是否为 0 而直接利用根与系数的关系求解致错. 6.某研究性学习小组共有 8 位同学,记他们的学号分别为 1,2,3,„,8.现指导老师决定派某些同学去市图书馆查询有关数据,分派的原则为:若 x 号同学去,则 8-x 号同学也去.请你根据老师的要求回答下列问题: (1)若只有一个名额,请问应该派谁去? (2)若有两个名额,则有多少种分派方法? 【解析】本题实质是考查集合中元素的特性,只有一个名额等价于 x=8-x,有两个名额则为 x 和 8-x. 分派去图书馆查数据的所有同学组成一个集合,记作 M,则有 x∈M,8-x∈M. (1)若只有一个名额,即 M 中只有一个元素,必须满足 x=8-x,故 x=4,所以应该派学号为 4 的同学去. (2)若有两个名额,即 M 中有且仅有两个不同的元素 x 和 8-x,从而全部含有两个元素的集合 M 含有元素的情况为:1,7 或 2,6 或 3,5,也就是有两个名额的分派方法有 3 种.
高中数学必修一 1-1 集合课时提升作业(一) 集合
的含义
(25 分钟 60 分) 一、选择题(每小题 5 分,共 25 分) 1.下列指定的对象,不能组成集合的是 ( ) A.一年中有 31 天的月份 B.平面上到点 O 距离是 1 的点 C.满足方程 x2-2x-3=0 的 x D.某校高一(1)班性格开朗的女生【解析】选 D.因为 A,B,C 所给的对象都是确定的,从而可以组成集合,而 D 中所给的对象没有具体的标准来衡量一名女生怎样才能算性格开朗,故不能组成集合. 【补偿训练】(2015·昆明高一检测)下列对象能组成集合的是 ( ) A.中国大的城市 B.方程 x2-9=0 在实数范围内的解 C.直角坐标平面内第一象限的一些点 D. 的近似值的全体【解析】选 B.A 中的城市大到什么程度不明确,所以不能组成集合;B 能组成集合;C 中“一些点”无明确的标准,对于某个点是否在“一些点”中无法确定,因此“直角坐标平面内第一象限的一些点”不能组成集合;D 中“ 的近似值”不明确精确到什么程度,因此很难判断一个数如“2”是不是它的近似值,所以不能组成集合. 2.(2015·黄山高一检测)若 a 是 R 中的元素,但不是 Q 中的元素,则 a 可以

人教部编版六年级语文上册课时练第12课《桥》(含答案)

部编版六年级上册语文12.桥课时练一、选择题1．下列句子中，加点成语使用不当的一项是（）A．山洪咆哮着，像一群受惊的野马，从山谷里狂奔而来，势不可当．．．．。

B．成长是跌跌撞撞．．．．还依然向前的勇气。

C．老师对成天不专心读书、不求上进的同学指手画脚．．．．。

D．当救援者从废墟中救出一名女孩时，在场的许多父母潸然泪下．．．．。

2．下面句子没有使用排比修辞方法的一项是（）A．我是亲友之间交往的礼品，我是婚礼的冠冕，我是生者赠予死者最后的祭献。

B．起初是全场肃静，只听见炮声和乐曲声，只听见国旗和其他许多旗帜飘拂的声音。

C．花开了，就像睡醒了似的。

鸟飞了，就像在天上逛似的。

虫子叫了，就像在说话似的。

D．漓江的水真静啊，静得让你感觉不到它在流动；漓江的水真清啊，清得可以看见江底的沙石；漓江的水真绿啊，绿得仿佛那是一块无瑕的翡翠。

3．下列三个句子中，画横线的词语，使用恰当的选项是（）A．垂直绿化工程成了北京新奇观。

B．瀑布、山川、河流构成了十分壮观的景色。

C．海上日出是伟大的景观。

4．下面四句话中不是拟人句的是（）。

A．近一米高的洪水已经在路面上跳舞了。

B．木桥开始发抖，开始痛苦地呻吟。

C．死亡在洪水的狞笑声中逼近。

D．山洪咆哮着，像一群受惊的野马。

5．下列作品不属于《汤姆·索亚历险记》的作者写的是( )A．《哈克贝利.费恩历险记》B．《竞选州长》C．《假如给我三天光明》二、填空题6．写出下列句子的修辞方法。

（1）他像一座山。

( )（2）水渐渐窜上来，放肆地舔着人们的腰。

( )（3）山洪咆哮着，像一群受惊的野马。

( )（4）木桥开始发抖，开始痛苦地呻吟。

( )7．她来祭奠两个人。

她丈夫和她儿子。

①“她丈夫”指的是__________，“她儿子”指的是_________。

②读到此处，你有什么感受？课文这样安排有什么好处？_______________________________________8．补充词语。

课时提升作业(四) 气体摩尔体积

课时提升作业(四)气体摩尔体积(30分钟50分)一、选择题(本题包括8小题,每小题3分,共24分)1.下列说法正确的是( )A.同温同压下,相同质量的气体都占有相同的体积B.同温同压下,相同体积的气体都含有相同数目的分子C.在标准状况下,1 mol酒精的体积约是22.4 LD.22.4 L N2含1.204×1024个氮原子【解析】选B。

同温同压下,质量相同的不同气体物质的量不一定相同,所以不一定占有相同的体积,故A项错误;B项为阿伏加德罗定律的内容,故正确;由于酒精在标准状况下不是气体,故C项错误;D项应注明N2为标准状况,故D项错误。

2.下列有关气体体积的叙述中,正确的是( )A.一定温度和压强下,各种气体物质体积的大小由构成气体的分子大小决定B.一定温度和压强下,各种气态物质体积的大小,由构成气体的原子数决定C.无论外界条件是否相同,不同的气体,若物质的量不同,则它们所含的分子数也不同D.相同状况下,相同微粒数的Fe、H2O、H2的体积相同【解析】选C。

一定温度和压强下,各种气体物质体积的大小由构成气体的分子数目决定;无论外界条件是否相同,只要物质的量相同,气体所含的分子数都相同;相同状态下,Fe、H2O、H2的状态不同,体积不同。

3.(2015·惠州高一检测)用N A表示阿伏加德罗常数的值,下列叙述正确的是( )A.0.012 g12C中含有N A个12C原子B.24 g Mg含有的镁原子的个数是2N AC.含有N A个氦原子的氦气在标准状况下的体积约为22.4 LD.标准状况下,22.4 L酒精含有的分子数为N A【解析】选C。

0.012 kg12C中含有N A个12C原子,A不正确;24 g Mg是1 mol,含有N A个镁原子,B不正确;由于He是单原子分子,所以含有N A个氦原子的氦气在标准状况下的体积约为22.4 L,C正确;酒精在标准状况下是液体,D错误。

2014版高中数学复习方略课时提升作业：单元评估检测(二)(北师大版)

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单元评估检测(二)第二章(120分钟150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.函数y=的定义域为( )(A)(0,8] (B)(-2,8] (C)(2,8] (D)[8,+∞)2.(2013·咸阳模拟)下列函数中,既是偶函数又在(0,+∞)上是增加的函数是( ) (A)y=x3(B)y=|x|+1(C)y=-x2+1 (D)y=2-|x|3.已知实数a=log45,b=()0,c=log30.4,则a,b,c的大小关系为( )(A)b<c<a (B)b<a<c(C)c<a<b (D)c<b<a4.若已知函数f(x)=则f(f(1))+f(log3)的值是( )(A)7 (B)2 (C)5 (D)35.(2013·合肥模拟)已知偶函数f(x)在区间[0,+∞)上是增加的,则满足f(2x-1)<f()的x的取值范围是( )(A)(,) (B)[,) (C)(,) (D)[,)6.(2013·芜湖模拟)函数f(x)=xcosx-sinx在下面哪个区间内是增加的( )(A)(,) (B)(π,2π)(C)(,) (D)(2π,3π)7.已知函数f(x)=是(-∞,+∞)上的减函数,则a的取值范围是( )(A)(0,3) (B)(0,3] (C)(0,2) (D)(0,2]8.(2013·抚州模拟)已知f(x)是定义在R上的偶函数,对任意x∈R,都有f(x+4)=f(x)+2f(2),且f(-1)=2,则f(2013)等于( )(A)1 (B)2 (C)3 (D)49.(2013·大连模拟)函数f(x)=ln(1-x2)的图象只可能是( )10.(2013·长春模拟)若y=f(x)在x>0上可导,且满足:xf′(x)-f(x)>0恒成立,又常数a,b满足a>b>0,则下列不等式一定成立的是( )(A)bf(a)>af(b) (B)af(a)>bf(b)(C)bf(a)<af(b) (D)af(a)<bf(b)二、填空题(本大题共5小题,每小题5分,共25分.请把正确答案填在题中横线上)11.f(x)=3x+sinx+1(x∈R),若f(t)=2,则f(-t)的值为.12.已知函数f(x)=lnx+2x,若f(x2+2)<f(3x),则实数x的取值范围是.13.(2013·宝鸡模拟)已知函数y=f(x)=x3+3ax2+3bx+c在x=2处有极值,其图像在x=1处的切线平行于直线6x+2y+5=0,则f(x)的极大值与极小值之差为.14.方程2x3+7=6x2在(0,2)内的实根个数为.15.(2013·上饶模拟)对于函数f(x),若存在区间M=[a,b](a<b),使得{y|y=f(x),x∈M}=M,则称区间M为函数f(x)的一个“稳定区间”.给出下列4个函数:①f(x)=e x;②f(x)=x3;③f(x)=cos x;④f(x)=lnx+1.其中存在“稳定区间”的函数有(填上所有符合要求的序号).三、解答题(本大题共6小题,共75分.解答时应写出必要的文字说明、证明过程或演算步骤)16.(12分)函数f(x)=log2(4x)·log2(2x),≤x≤4.(1)若t=log2x,求t的取值范围.(2)求f(x)的最值,并给出取最值时对应的x的值.17.(12分)(2013·太原模拟)若g(x)=x+(x>0),g(x)=m有零点,求m的取值范围.18.(12分)已知函数f(x)=log2(-1≤x≤1)为奇函数,其中a为不等于1的常数.(1)求a的值.(2)若对任意的x∈[-1,1],f(x)>m恒成立,求m的取值范围.19.(12分)(2013·黄山模拟)已知某公司生产某品牌服装的年固定成本为10万元,每生产一千件,需要另投入2.7万元,设该公司年内共生产该品牌服装x千件并全部销售完,每千件的销售收入为R(x)万元,且R(x)=(1)写出年利润W(万元)关于年产量x(千件)的函数关系式.(2)年生产量为多少千件时,该公司在这一品牌服装的生产中所获年利润最大?20.(13分)(2013·榆林模拟)已知函数f(x)=ax2-(2a+1)x+2lnx,求f(x)的单调区间.21.(14分)(2012·湖北高考)设函数f(x)=ax n(1-x)+b(x>0),n为整数,a,b为常数.曲线y=f(x)在(1,f(1))处的切线方程为x+y=1.(1)求a,b的值.(2)求函数f(x)的最大值.(3)证明:f(x)<.答案解析1.【解析】选B.由⇒⇒-2<x≤8.2.【解析】选B.对于A:y=x3是奇函数,不合题意;对于C,D:y=-x2+1和y=2-|x|在(0, +≦)上是减少的,不合题意;对于B:y=|x|+1的图像如图所示,知y=|x|+1符合题意,故选B.3.【解析】选D.由题知,a=log45>1,b=()0=1,c=log30.4<0,故c<b<a.4.【解析】选 A.f(1)=log21=0,所以f(f(1))=f(0)=2.因为log3<0,所以f(log3)=+1=+1=+1=+1=4+1=5,所以f(f(1))+f(log3)=2+5=7,故选A.5.【解析】选A.f(x)是偶函数,其图像关于y轴对称,又f(x)在[0,+≦)上是增加的,≨f(2x-1)<f()⇔f(|2x-1|)<f(),则|2x-1|<,解得<x<.6.【解析】选B.f′(x)=(xcosx-sinx)′=cosx-xsinx-cosx=-xsinx,由函数是增加的,则f′(x)≥0,又各选项均为正实数区间,所以sinx≤0,故选B.7.【解析】选D.≧f(x)为(-≦,+≦)上的减函数,≨解得0<a≤2.8.【解析】选B.在f(x+4)=f(x)+2f(2)中,令x=-2得f(2)=f(-2)+2f(2),即f(2)=f(2)+2f(2),故f(2)=0.因此f(x+4)=f(x),即f(x)是以4为周期的函数.又2013=4〓503+1,所以f(2013)=f(1)=f(-1)=2.9.【解析】选A.函数f(x)=ln(1-x2)的定义域为(-1,1),且f(x)为偶函数,当x ∈(0,1)时,函数f(x)=ln(1-x2)为减少的;当x∈(-1,0)时,函数f(x)为增加的,且函数值都小于零,所以其图象为A.10.【思路点拨】令g(x)=,根据g(x)的单调性比较大小.【解析】选A.令g(x)=,则g′(x)=,由已知得,当x>0时,g′(x)>0. 故函数g(x)在(0,+≦)上是增加的,又a>b>0,故g(a)>g(b),即bf(a)>af(b). 11.【解析】由f(t)=3t+sint+1=2得3t+sint=1,所以f(-t)=-3t-sint+1=-1+1=0. 答案:012.【解析】由f(x)=lnx+2x⇒f′(x)=+2x ln 2>0(x∈(0,+≦)),所以f(x)在(0, +≦)上是增加的,又f(x2+2)<f(3x)⇒0<x2+2<3x⇒x∈(1,2).答案:(1,2)13.【解析】因为f′(x)=3x2+6ax+3b,又所以解得因此f′(x)=3x2-6x=3x(x-2),由f′(x)=0得x=0或x=2.所以f(x)极大值-f(x)极小值=f(0)-f(2)=4.答案:414.【解析】设f(x)=2x3-6x2+7,则f′(x)=6x2-12x=6x(x-2),因为x∈(0,2),所以有f′(x)<0,所以f(x)在(0,2)内是减少的,又f(0)=7>0,f(2)=-1<0,所以在(0,2)内存在唯一的x0,使f(x0)=0,因此,方程2x3+7=6x2在(0,2)内的实根个数为1.答案:115.【思路点拨】由“稳定区间”的定义可知存在“稳定区间”的函数即为定义域和值域相同的函数.【解析】①若存在稳定区间[a,b],因为f(x)=e x在R上是增函数,则即方程e x=x有两个不等实根,即函数y=e x-x的图像与x轴有两个不同的交点,y′=e x-1, x∈(-≦,0),y′<0;x∈(0,+≦),y′>0,且x=0时,y=1,所以y≥1,即函数y=e x-x 的图像与x轴没有交点,所以假设不成立,即不存在稳定区间;②显然存在稳定区间[0,1]或[-1,0]或[-1,1];③显然存在稳定区间[0,1];④因为y=lnx+1-x的导函数y′=-1=,在(0,1)上,y′>0;在(1,+≦)上,y′<0,且x=1时,y=0,所以y=lnx+1-x≤0在(0,+≦)上恒成立,即函数y=lnx+1,y=x只有1个交点,所以不存在稳定区间,故存在稳定区间的是②③.答案:②③16.【解析】(1)≧t=log2x,≤x≤4,≨log2≤t≤log24即-2≤t≤2.(2)f(x)=(log2x)2+3log2x+2,≨令t=log2x,则y=t2+3t+2=(t+)2-,当t=-,即log2x=-,x=时,f(x)min=-.当t=2,即x=4时,f(x)max=12.17.【解析】方法一:≧g(x)=x+≥2=2e,等号成立的条件是x=e,故g(x)的值域是[2e,+≦),因而只需m≥2e,则g(x)=m就有零点.方法二:作出g(x)=x+(x>0)的大致图象.如图,可知若使g(x)=m有零点,则只需m≥2e.方法三:由g(x)=m得x2-mx+e2=0.此方程有大于零的根且e2>0,故根据根与系数的关系得m>0,故等价于故m≥2e.18.【解析】(1)≧f(x)=log2(-1≤x≤1)为奇函数,≨f(-x)=-f(x)⇒log2=-log2,⇒=对x∈[-1,1]恒成立,所以(5+ax)(5-ax)=(5+x)(5-x)⇒a=〒1,因为a为不等于1的常数,所以a=-1.(2)≧f(x)=log2(-1≤x≤1),设t=(-1≤x≤1),≨f(t)=log2t,因为t==-1+在[-1,1]上是减少的,所以≤t≤,又因为f(t)=log2t在[,]上是增加的,所以f(t)min=log2.因为对任意的x∈[-1,1],f(x)>m恒成立,所以f(x)min>m,所以m<log2.19.【解析】(1)当0<x≤10时,W=xR(x)-(10+2.7x)=8.1x--10; 当x>10时,W=xR(x)-(10+2.7x)=98--2.7x.≨年利润W(万元)关于年产量x(千件)的函数关系式为W=(2)当0<x≤10时,由W′=8.1->0⇒0<x<9,即年利润W在(0,9)上增加,在(9,10)上减少,≨当x=9时,W取得最大值,且W max=38.6(万元).时取“=”,综上可知,当当x>10时,W=98-(+2.7x)≤98-2=38,仅当x=1009年产量为9千件时,该公司这一品牌服装的生产中所获年利润最大,最大值为38.6万元.【变式备选】(2013·宿州模拟)据环保部门测定,某处的污染指数与附近污染源的强度成正比,与到污染源距离的平方成反比,比例常数为k(k>0).现已知相距18km的A,B两家化工厂(污染源)的污染强度分别为a,b,若线段AB上任意一点C 处的污染指数y等于两化工厂对该处的污染指数之和.设AC=xkm.(1)试将y表示为x的函数.(2)若a=1,x=6时,y取得最小值,试求b的值.【解析】(1)由题意知点C受A污染源污染指数为,点C受B污染源污染指数为,其中k为比例系数,且k>0.从而点C处的污染指数y=+(0<x<18).(2)因为a=1,所以y=+,y′=k[+],令y′=0,得x=,又此时x=6,解得b=8,经验证符合题意.所以,污染源B的污染强度b的值为8.20.【思路点拨】求导后转化为二次不等式问题,结合二次项系数的符号,相应二次方程根的大小,以及两根是否大于0进行分类讨论.【解析】f′(x)==(x>0).≨①当a≤0时,x>0,ax-1<0,在区间(0,2)上,f′(x)>0;在区间(2,+≦)上,f′(x)<0,故f(x)的递增区间是(0,2),递减区间是(2,+≦).②当0<a<时,>2,在区间(0,2)和(,+≦)上,f′(x)>0;在区间(2,)上,f′(x)<0,故f(x)的递增区间是(0,2)和(,+≦),递减区间是(2,).③当a=时,f′(x)=≥0,故f(x)的递增区间是(0,+≦),④当a>时,0<<2,在区间(0,)和(2,+≦)上,f′(x)>0;在区间(,2)上,f′(x)<0,故f(x)的递增区间是(0,)和(2,+≦),递减区间是(,2).【方法技巧】分类讨论思想分类讨论是基本逻辑方法之一,也是一种数学思想,在近几年的高考中,都把分类讨论列为重要的思想方法来考查,当我们面临的数学问题不能用统一形式解决或因为一种形式无法进行概括,不分类就不能再进行下去,这时,就要使用分类讨论思想了,分类时要遵循不重不漏的分类原则,对于每一类情况都要给出问题的解答.分类讨论的一般步骤:(1)确定标准.(2)恰当分类.(3)逐类讨论. (4)归纳结论.21.【思路点拨】本题(1)易解,(2)问中直接求导,根据零点讨论单调性求解;(3)要构造函数利用函数的单调性证明.【解析】(1)因为f(1)=b,由点(1,b)在x+y=1上,可得1+b=1,即b=0. 因为f′(x)=anx n-1-a(n+1)x n,所以f′(1)=-a,又因为切线x+y=1的斜率为-1,所以-a=-1,即a=1.故a=1,b=0.(2)由(1)知,f(x)=x n(1-x)=x n-x n+1,f′(x)=(n+1)x n-1(-x).令f′(x)=0,解得x=,即f′(x)在(0,+≦)上有唯一零点x0=. 在(0,)上,f′(x)>0,f(x)是增加的;而在(,+≦)上,f′(x)<0,f(x)是减少的.故f(x)在(0,+≦)上的最大值为f()=()n(1-)=.(3)令φ(t)=lnt-1+(t>0),则φ′(t)=-=(t>0).在(0,1)上,φ′(t)<0,φ(t)是减少的;在(1,+≦)上,φ′(t)>0,φ(t)是增加的.故φ(t)在(0,+≦)上的最小值为φ(1)=0,所以φ(t)>0(t>1),即lnt>1-(t>1).令t=1+,得ln>,即ln()n+1>ln e,所以()n+1>e,即<.由(2)知,f(x)≤<,故所证不等式成立.【变式备选】已知函数f(x)=e x-1-x.(1)求y=f(x)在点(1,f(1))处的切线方程.(2)若存在x∈[-1,ln],使a-e x+1+x<0成立,求a的取值范围.(3)当x≥0时,f(x)≥tx2恒成立,求t的取值范围.【解析】(1)f′(x)=e x-1,f(1)=e-2,f′(1)=e-1.≨f(x)在(1,f(1))处的切线方程为y-e+2=(e-1)(x-1),即y=(e-1)x-1. (2)a<e x-1-x,即a<f(x).令f′(x)=e x-1=0,x=0.≧x>0时,f′(x)>0,x<0时,f′(x)<0,≨f(x)在(-≦,0)上是减少的,在(0,+≦)上是增加的.又x∈[-1,ln],≨f(x)的最大值在区间端点处取到.f(-1)=e-1-1+1=,f(ln)=-1-ln,f(-1)-f(ln)=-+1+ln=-+ln>0,≨f(-1)>f(ln),≨f(x)在[-1,ln]上的最大值为,故a的取值范围是a<.(3)由已知得x≥0时,e x-x-1-tx2≥0恒成立,设g(x)=e x-x-1-tx2,≨g′(x)=e x-1-2tx.由(2)知e x≥1+x,当且仅当x=0时等号成立,故g′(x)≥x-2tx=(1-2t)x,从而当1-2t≥0,即t≤时,g′(x)≥0(x≥0),≨g(x)是增加的,又g(0)=0,于是当x≥0时,g(x)≥0,即f(x)≥tx2,≨t≤时符合题意.由e x>1+x(x≠0)可得e-x>1-x(x≠0),从而当t>时,g′(x)<e x-1+2t(e-x-1)=e-x(e x-1)(e x-2t),故当x∈(0,ln 2t)时,g′(x)<0,≨g(x)是减少的,又g(0)=0,于是当x∈(0,ln 2t)时,g(x)<0,即f(x)≤tx2,故t>,不符合题意.综上可得t的取值范围为(-≦,].关闭Word文档返回原板块。

2014版高中数学复习方略课时提升作业：4.5数系的扩充与复数的引入(北师大版)(北师大版·数学理·通用版)

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课时提升作业(二十九)一、选择题1.(2013·蚌埠模拟)复数z=的实部是( )(A)4 (B)1 (C)-1 (D)-42.(2013·景德镇模拟)复数(m2-3m)+mi(m∈R)是纯虚数,则实数m的值是 ( )(A)3 (B)0(C)0或3 (D)0或1或33.复数z=对应的点在复平面位于( )(A)第一象限(B)第二象限(C)第三象限(D)第四象限4.已知复数z=1+i,则等于( )(A)2i (B)-2i (C)2 (D)-25.若+(1+i)2=a+bi(a,b∈R),则a-b= ( )(A)2 (B)-2(C)2+2 (D)2-26.若i为虚数单位,图中复平面内点Z表示复数z,则表示复数的点是( )(A)E (B)F (C)G (D)H7.设0<θ<,a∈R,(a+i)(1-i)=cosθ+i,则θ的值为( )(A)π(B)π(C)(D)8.复数z=(m∈R,i为虚数单位)在复平面上对应的点不可能位于( )(A)第一象限 (B)第二象限(C)第三象限 (D)第四象限9.已知m(1+i)=2-ni(m,n∈R),其中i是虚数单位,则()3等于( )(A)1 (B)-1 (C)i (D)-i10.(能力挑战题)若sin2θ-1+i(cosθ+1)是纯虚数,则θ的值为( )(A)2kπ-,k∈Z (B)2kπ+,k∈Z(C)2kπ±,k∈Z (D)π+,k∈Z二、填空题11.(2013·芜湖模拟)若(1+ai)2=-1+bi(a,b∈R,i是虚数单位),则|a+bi|= .12.定义一种运算如下:=x1y2-x2y1,则复数z=(i是虚数单位)的共轭复数是.13.(能力挑战题)已知复数z1=cosθ-i,z2=sinθ+i,则z1·z2的实部的最大值为,虚部的最大值为.14.若复数z=cosθ+isinθ且z2+=1,则sin2θ= .三、解答题15.已知关于x的方程:x2-(6+i)x+9+ai=0(a∈R)有实数根b.(1)求实数a,b的值.(2)若复数满足|-a-bi|-2|z|=0,求z为何值时,|z|有最小值,并求出|z|的最小值.答案解析1.【解析】选C.∵z====-1-2i,∴z的实部是-1.2.【解析】选A.∵(m2-3m)+mi是纯虚数,∴m2-3m=0且m≠0,∴m=3.3.【思路点拨】先计算所给的复数,根据实部、虚部确定对应点所在的象限. 【解析】选D.z===,故对应的点在第四象限.4.【解析】选A.===2i.【变式备选】已知x,y∈R,i为虚数单位,且(x-2)i-y=-1+i,则(1+i)x+y的值为( ) (A)4 (B)4+4i (C)-4 (D)2i【解析】选C.由(x-2)i-y=-1+i,得x=3,y=1,∴(1+i)4=[(1+i)2]2=(2i)2=-4.5.【思路点拨】先化简等号左边的复数,再根据复数相等解题.【解析】选B.+(1+i)2=1-i-2+2i=-1+(2-1)i=a+bi,则a=-1,b=2-1,故a-b=-2.6.【解析】选D.依题意得z=3+i,====2-i,该复数对应的点的坐标是(2,-1),选D.7.【解析】选D.由条件得a++(-a)i=cosθ+i,∴解得cosθ=.又0<θ<,∴θ=.8.【思路点拨】先把z化成a+bi(a,b∈R)的形式,再进行判断.【解析】选A.z===+i,显然>0与->0不可能同时成立,则z=对应的点不可能位于第一象限.【一题多解】选 A.z==+i,设x=,y=,则2x+y+2=0.又直线2x+y+2=0不过第一象限,则z=对应的点不可能位于第一象限.【方法技巧】复数问题的解题技巧(1)根据复数的代数形式,通过其实部和虚部可判断一个复数是实数,还是虚数.(2)复数z=a+bi,a∈R,b∈R与复平面上的点Z(a,b)是一一对应的,通过复数z的实部和虚部可判断出其对应点在复平面上的位置.9.【解析】选C.由m(1+i)=2-ni,得m+mi=2-ni,故m=2,m=-n,故m=2,n=-2,故()3=()3=i.10.【解析】选B.由题意,得解得∴θ=2kπ+,k∈Z.11.【解析】∵(1+ai)2=-1+bi,∴1-a2+2ai=-1+bi,∴解得或∴|a+bi|===.答案:12.【解析】由定义知,z=(+i)i-(-i)×(-1)=-1+(-1)i,故=-1-(-1)i.答案:-1-(-1)i13.【解析】z1〃z2=(cosθsinθ+1)+i(cosθ-sinθ).实部为cosθsinθ+1=1+sin 2θ≤,所以实部的最大值为.虚部为cosθ-sinθ=sin(-θ)≤,所以虚部的最大值为.答案:14.【解析】z2+2z=(cosθ+isinθ)2+(cosθ-isinθ)2=2cos 2θ=1⇒cos 2θ=,所以sin2θ==.答案:15.【思路点拨】(1)把b代入方程,根据复数的实部、虚部等于0解题即可.(2)设z=s+ti(s,t∈R),根据所给条件可得s,t间的关系,进而得到复数z对应的轨迹,根据轨迹解决|z|的最值问题.【解析】(1)∵b是方程x2-(6+i)x+9+ai=0(a∈R)的实根,∴(b2-6b+9)+(a-b)i=0,∴解得a=b=3.(2)设z=s+ti(s,t∈R),其对应点为Z(s,t),由|-3-3i|=2|z|,得(s-3)2+(t+3)2=4(s2+t2),即(s+1)2+(t-1)2=8,∴Z点的轨迹是以O 1(-1,1)为圆心,2为半径的圆,如图所示,当Z点在OO1的连线上时,|z|有最大值或最小值.∵|OO 1|=,半径r=2,∴当z=1-i时,|z|有最小值且|z|min=.【变式备选】若虚数z同时满足下列两个条件:①z+是实数;②z+3的实部与虚部互为相反数.这样的虚数是否存在?若存在,求出z;若不存在,请说明理由. 【解析】设z=a+bi(a,b∈R,b≠0),则z+=a+bi+=a(1+)+b(1-)i.又z+3=a+3+bi,z+是实数,根据题意有∵b≠0,∴解得或∴z=-1-2i或z=-2-i.关闭Word文档返回原板块。

高中数学课时提升作业(一)第课时棱柱、棱锥、棱台的结构特征

__________,ABCD是它的__________.
(3)图(3)中的几何体叫做__________,它是由棱锥__________被平行于底面 ABCD
的平面__________截得的,AA′,BB′为其__________,BCC′B′,DAA′D′为其
__________.
【解析】根据棱柱,棱锥,棱台的结构特征,可依次得出答案.
【解析】选 C.图 C 不能围成正方体. 3.关于棱台,下列说法正确的是 ( ) A.两底面可以不相似 B.侧面都是全等的梯形 C.侧棱长一定相等
-1-
D.侧棱延长后交于一点【解析】选 D.只有 D 符合棱台的特征.选项 A,B,C均不正确. 4.下列命题中正确的是 ( ) A.由五个平面围成的多面体只能是四棱锥 B.棱锥的高线可能在几何体之外 C.仅有一组对面平行的六面体是棱台 D.有一个面是多边形,其余各面是三角形的几何体是棱锥【解析】选 B.三棱柱是由五个平面围成的多面体,故 A 不正确. 仅有一组对面平行的六面体可以是四棱柱,故 C 不正确. D 中,当这些三角形不共用一顶点时,不一定是棱锥,故 D 不正确,故 B 正确. 5.(2013·嘉兴高一检测)如图都是正方体的表面展开图,还原成正方体后,其中两个完全一样的是 ( )
MN= (23a)2 ― (2a)2
= 22a. 二、填空题(每小题 5 分,共 10分)
5.在如图①～④4 个平面图形中,哪几个是下面各侧棱都相等的四面体的展开图? 其序号是________.(把你认为正确的序号都填上)
-8-
【解析】将①②沿折痕折起,为三棱锥;③④沿折痕折起,构不成空间封闭图形. 答案:①② 【误区警示】本题易得答案①③,①正确,③不正确.错误的原因是思维想象能力较差,可动手制作几何体,观察其展开图,提高识图能力. 6.如图,这是一个正方体的表面展开图,若把它再折回成正方体后,有下列命题:

2014版广西《复习方略》(数学文)课时提升作业第二章第六节指数、指数函数

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课时提升作业(九)一、选择题1.(2013·玉林模拟)函数y=(a2-3a+3)a x是指数函数,则有( )(A)a=1或a=2 (B)a=1(C)a=2 (D)a>0且a≠12.化简[(-2)6-(-1)0的结果为( )(A)-9 (B)7 (C)-10 (D)93.已知函数f(x)=4+a x-1(a>0且a≠1)的图象恒过定点P,则点P的坐标是( )(A)(1,5) (B)(1,4)(C)(0,4) (D)(4,0)4.(2013·杭州模拟)函数y=a|x|(a>1)的图象是( )5.函数f(x)=a x(a>0且a≠1)对于任意的实数x,y,都有( )(A)f(xy)=f(x)f(y)(B)f(xy)=f(x)+f(y)(C)f(x+y)=f(x)f(y)(D)f(x+y)=f(x)+f(y)6.若102x =25,则10-x 等于( ) (A)- (B)(C) (D)7.(2013·河池模拟)函数y=21x 112（）的值域为( ) (A)(-∞,1) (B)(,1) (C)[,1) (D)[,+∞) 8.定义运算a ⊗b=则函数f(x)=1⊗2x 的图象是( )9.(2013·玉林模拟)已知f(x)=则f(8)等于( )(A)4 (B)0 (C) (D)210.(2013·钦州模拟)设y 1=40.9,y 2=80.48,y 3=()-1.5,则( ) (A)y 3>y 1>y 2 (B)y 2>y 1>y 3 (C)y 1>y 2>y 3 (D)y 1>y 3>y 211.(2013·柳州模拟)函数y=a x -(b+1)(a>0且a ≠1)的图象在第一、三、四象限,则必有( )(A)0<a<1,b>0 (B)0<a<1,b<0 (C)a>1,b<1 (D)a>1,b>0 12.下列各式正确的是( ) (A)=(B)=a (C)= (D)×=二、填空题13.(2013·南宁模拟)若a=4,b=2,则= .14.若直线y=2a与函数y=|a x-1|(a>0且a≠1)的图象有两个公共点,则a的取值范围是.15.(2012·山东高考)若函数f(x)=a x(a>0,a≠1)在[-1,2]上的最大值为4,最小值为m,且函数g(x)=(1-4m)在[0,+∞)上是增函数,则a= .16.(能力挑战题)函数y=a2x-2(a>0,a≠1)的图象恒过点A,若直线l:mx+ny-1=0经过点A,则坐标原点到直线l的距离的最大值为.三、解答题17.(能力挑战题)已知函数f(x)=(.(1)若a=-1,求f(x)的单调区间.(2)若f(x)有最大值3,求a的值.(3)若f(x)的值域是(0,+∞),求a的值.答案解析1.【解析】选C.≧y=(a2-3a+3)a x是指数函数,≨解得a=2.2.【解析】选B.原式=(26-1=23-1=7.3.【解析】选A.≧x=1时,f(x)=4+a1-1=5为定值,≨函数f(x)的图象恒过定点P(1,5).【变式备选】函数f(x)=+m(a>1)恒过点(1,10),则m= . 【解析】方法一:≧f(x)=+m 在x 2+2x-3=0时过定点(1,1+m)或(-3,1+m),≨1+m=10,解得m=9.方法二:由已知得x=1时,f(x)=10, 即+m=10,解得m=9.答案:94.【解析】选B.y=a |x|=当x ≥0时,与指数函数y=a x (a>1)的图象相同;当x<0时,y=a -x 与y=a x 的图象关于y 轴对称,由此判断B 正确. 5.【解析】选C.f(x)f(y)=a x ·a y =a x+y =f(x+y). 6.【解析】选D.由102x =25,得10x =5, ≨10-x =5-1=.7.【思路点拨】函数y=a f(x)的值域的求解,先确定f(x)的值域,再根据指数函数的单调性,确定y=a f(x)的值域. 【解析】选C.≧x ∈R,0<≤1,≨y=1x 112+（）≥()1=且y=1x 112+（）<()0=1,≨y ∈[,1). 8.【解析】选A.当x ≤0时,2x ≤1,f(x)=2x ;当x>0时,2x >1,f(x)=1,即f(x)=故选A.9.【解析】选C.f(8)=f(6)=f(4)=f(2)=f(0)=f(-2)=2-2=.10.【解析】选D.将三个幂都化成以2为底的幂,y 1=40.9=21.8,y 2=80.48= (23)0.48=21.44,y 3=()-1.5=21.5,因为函数y=2x 是增函数,且1.8>1.5>1.44,所以y 1>y 3>y 2.11.【解析】选D.借助指数函数图象则解得a>1,b>0.12.【解析】选C.≧==≠,≨选项A错.≧=|a|≠a,≨选项B错.≧=,≨选项C正确.≧,无意义,≨选项D错.13.【思路点拨】先化简,再代入求值.【解析】=====2.答案:214.【思路点拨】对a进行分类讨论,画出y=|a x-1|(a>0且a≠1)的图象,利用与动直线y=2a有两个公共点求出a的取值范围.【解析】y=|a x-1|(a>0且a≠1)的图象如图所示,y=2a与y=|a x-1|的图象有两个公共点,则0<2a<1,0<a<.答案:(0,)15.【思路点拨】本题关键是分a>1和0<a<1两种情况讨论,再代入到函数g(x)=(1-4m)内检验是否为增函数.【解析】当a>1时,有a2=4,a-1=m,此时a=2,m=,此时g(x)=-为减函数,不合题意.当0<a<1时,则a-1=4,a2=m,故a=,m=,经检验知符合题意.答案:16.【解析】由题意知点A(1,1),而A∈l,≨m+n-1=0,即m+n=1,由基本不等式得:m2+n2≥(m+n)2=.≨坐标原点到直线l的距离:d=≤=.≨坐标原点到直线l的距离的最大值为.答案:17.【思路点拨】(1)根据复合函数的单调性法则“同增异减”求得.(2)等价转化为幂指数ax2-4x+3有最小值-1求解.(3)考虑使得ax2-4x+3取到所有实数的a值.【解析】(1)当a=-1时,f(x)=(,令g(x)=-x2-4x+3,由于g(x)在(-≦,-2)上单调递增,在(-2,+≦)上单调递减,而y=()t在R上单调递减,所以f(x)在(-≦,-2)上单调递减,在(-2,+≦)上单调递增,即函数f(x)的单调递增区间是(-2,+≦),单调递减区间是(-≦,-2).(2)令h(x)=ax2-4x+3,f(x)=()h(x),由于f(x)有最大值3,所以h(x)应有最小值-1,因此必有解得a=1,即当f(x)有最大值3时,a的值等于1.(3)由指数函数的性质知,要使y=()h(x)的值域为(0,+≦).应使h(x)=ax2-4x+3的值域为R,因此只能a=0.(因为若a≠0,则h(x)为二次函数,其值域不可能为R).故a的值为0.关闭Word文档返回原板块。

2014世纪金榜课时提升作业(六十二) 第九章 第六节

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