创新设计文科作业本第3篇 步骤规范练1
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第三单元正确使用词语(包括熟语)板块一实词训练1.依次填入下列各句横线处的词语,最恰当的一组是()(1)大中城市居民消费调查结果显示,年收入5万元左右的三口之家,要将一年的生活费用控制在3万元以内,才有可能________2万元的子女教育费。
(2)1955年,13名优秀队员准备开赴西沙群岛进行资源________和敌情侦察,为保卫收复和开发建设西沙群岛做好战备。
(3)李小鹏从小受父母自立自强思想的________,勤奋学习,刻苦钻研,在生命科学方面取得了很大的成就。
A.节余勘察熏陶B.结余查看熏染C.结余查看熏陶D.节余勘察熏染解析结余:作动词,指结算后余下;作名词,指结算后余下的钱。
节余:作动词,指因节约而剩下;作名词,指节余的钱或东西。
结合语境应选“节余”。
查看:检查、观察(事物的情况)。
勘察:进行实地调查或查看(多用于采矿或工程施工前)。
结合语境应选“勘察”。
熏陶:长期接触的人或事物对人的生活习惯、思想行为、品行学问等逐渐产生某种影响(多指好的)。
熏染:长期接触的人或事物对人的生活习惯逐渐产生某种影响(多指坏的)。
结合语境应选“熏陶”。
答案 A2.依次填入下列各句横线处的词语,最恰当的一组是()(1)中国对南沙群岛拥有无可争辩的主权,这一点是不容________的,这是有充分的法理依据的。
(2)对腐败分子必须严惩不贷,及时摘除毒瘤,以保证社会________的健康。
(3)随着转基因食品的逐渐增加,我国政府已明确规定,在国内市场销售的转基因食品,必须进行________。
(4)这次消防安全工作会议十分重要,可他因要处理私事,没有和任何人打招呼,没等会议结束,就________离开,因此受到有关领导批评。
A.质疑机体标志径自B.质疑肌体标示竟自C.置疑肌体标示径自D.置疑机体标志竟自解析质疑:提出疑问。
臵疑:怀疑(多用于否定式)。
故应选“臵疑”。
机体:具有生命的个体的统称。
肌体:指身体,也用来比喻组织机构。
双基达标 (限时20分钟)thrift of excellent traditional and style lack deep of understanding, one-sided think in now of good situation Xia, economic smooth more fast development, and reform made major progress, and people living significantly improve, and social the career made new progress, no necessary always hard has, no awareness to China also has many poverty area, and also has many poverty population of exists, more living improve has, more need keep hard of style. Three is not strictly honest. Total thought units is a water sector, clean self-discipline away from himself too far, no real processing good living of improve and hard, and thrift excellent traditional of relationship, no effective do comply with Constitution and party of regulations must from I do up, no right mercy good bitter Le, and wealth of relationship, no right awareness to in comply with Constitution and party of regulations aspects everyone has accountability. Four, future direction and improvement measures 1, belief and faith,strengthen party spirit. One is to firmly establish the noble ideals of struggle for the ideals of communism, communist struggle for life, ready to sacrifice everything for the party and the people. Second is to continuously improve the quality and standard of political, conscientiously study Deng Xiaoping theory and "three represents" important thought and the scientific Outlook on development, implement the party's basic program for the primary stage of socialism, adhere to the correct political orientation, maintain highly consistent with the CPC Central Committee, is good at theories of socialism with Chinese characteristics to analyze and solve problems. Third, strengthening the party spirit and the world transformation, perseverance of the party Constitution, relive Party vows to establish correct world Outlook, Outlook on life and the world, practically embodies ideals and beliefs into action, combine lofty ideals and practical activities, transforming the objective world, activelyparticipate in the great practice of building socialism with Chinese characteristics. 2, keep the purpose in concept, changing the style of work. First, keep in mind that the purpose of serving, adhere to assuming power for the people, keep in mind the people and care about people, and the joys and sorrows of the masses, ... According to city discipline, and municipal organization Department requirements, today we held "implement implementation independent Commission against corruption guidelines effective strengthening led cadres style construction" topic democratic life, main task is close contact thought, and work actual, control check district Standing Committee team and the personal in implementation independent Commission against corruption guidelines Shang exists of problem, in-depth carried out criticism and self-critical, further clear rectification direction and measures, effective enhanced clean politics of consciousness and initiative, firm set good of ruling image. Before the meeting, the way we take surveys solicited a views, combed, related to team buildingEND如果输入x1=2,x2=3,那么执行此程序后,输出的结果是( ).A.7 B.10 C.5 D.8解析 ∵x1=2,x2=3,∴x1≠x2,∴y=x1+x2=2+3=5.答案 C4.给出下列程序:INPUT a,b,cIF a>b THEN a=bEND IFIF a>c THEN a=cEND IFPRINT aEND如果输入-10,-26,8,那么输出的是________.解析 由题知,输出的将是最小的数.答案 -265.已知程序如下:INPUT aIF a>=0 THEN PRINT aELSE PRINT -aEND IFEND若输入9,其运行结果是________.解析 因为9≥0,所以输出9.答案 96.函数y=Error!写出求函数的函数值的程序.解 程序:project work, cadres, cadres selection and other aspects of a total of 19, further defined the direction of rectification, enhancing the pertinence and effectiveness of democratic life. Good for this meeting, we requested the Standing Committee of the municipal party Committee, municipal Secretary General Tang Wenjin, the City Commission for discipline inspection Standing Committee Yang Xu, Jiang Jixue, Deputy Director of the personnel office, and other leading comrades present guidance allows us to warm applause for their welcome and thanks! Below, I first District Committee member of the 20XX team annual democratic meeting implementation of the corrective measures, as well as member of the team and of my own report guidelines for the implementation of the independent Commission against corruption statement invited leaders and comrades criticism. First, annual party on the 20XX team, Member of democratic life action plan implementation year 20XX, districtdemocratic meeting a member of the team on "study and practice the scientific Outlook on development" theme, find a project supporting strong enough, the industry well enough, urban construction management level needs to be improved, livelihood improvement needs to be strengthened, district team itself, such as the five aspects. Year of 20XX, we find out the problems mentioned above, based on actual Shunqing construction and development, conscientiously sum up improvements, achieving significant development results and effectiveness of their work. (A) highlight the project handle, maintain high growth of investment. Adhere to the strategy as a Foundation, carrying out projects to promote increased stamina important. One is to increase the intensity of projects towards. Actively seek national and provincial policy support, superior special transfer payments, a variety of additional income will top 700 million Yuan, a number of key infrastructure projects, livelihoodprojects, industrial projects into the cage. Second, increasing the project attracts. This year a total of 23 signed a major project, of which 14 billion project, China Metallurgical group, vanke group, Guangdong hengda group, Taiwan Ruentex group, on the RI group, Shenzhen Kaisa, jifeng agricultural machinery, Mercedes-Benz, Audi 4S store and other large enterprises and projects have settled in Shunqing. Third, strengthen project construction. Annual implementation focus project 39 a, fixed assets investment total will breakthrough 12 billion yuan, created district yilai history high, South door dam, and five in shop old city transformation, and Ying China industrial concentrated district construction, and city main road construction, and Northeast steam Trade Center, and modern farmers industry Park, and vocational education city, and low rental, and also room construction, and features blocks build, and farmers market transformation, and "rural version" Street transformation, focusengineering speed up implementation, effectiveness obviously. Expected to be achieved this year, area-wide GDP 17 billion7.阅读下列程序,则该程序运行后,变量y的值为( ).A.4 B.16 C.6 D.8解析 因x=4满足“x>3”的条件,所以执行的是THEN后面的y=4×4=16.答案 B8.阅读下列程序:如果输入x=-2,则输出结果为.A.2解析 输入x=-2,则答案 D.阅读下面的程序:INPUT “x=”;xIF x<0 THENy=x+3ELSE IF x>0 THENy=x+5thrift of excellent traditional and style lack deep of understanding, one-sided think in now of good situation Xia, economic smooth more fast development, and reform made major progress, and people living significantly improve, and social the career made new progress, no necessary always hard has, no awareness to China also has many poverty area, and also has many poverty population of exists, more living improve has, more need keep hard of style. Three is not strictly honest. Total thought units is a water sector, clean self-discipline away from himself too far, no real processing good living of improve and hard, and thrift excellent traditional of relationship, no effective do comply with Constitution and party of regulations must from I do up, no right mercy good bitter Le, and wealth of relationship, no right awareness to in comply with Constitution and party of regulations aspects everyone has accountability. Four, future direction and improvement measures 1, belief and faith, strengthen party spirit. One is to firmly establish the noble ideals of struggle for the ideals of communism, communist struggle for life, ready to sacrifice everything for the party and the people. Second is to continuously improve the quality and standard of political, conscientiously study Deng Xiaoping theory and "three represents" important thought and the scientific Outlook on development, implement the party's basic program for the primary stage of socialism, adhere to the correct political orientation, maintain highly consistent with the CPC Central Committee, is good at theories of socialism with Chinese characteristics to analyze and solve problems. Third, strengthening the party spirit and the world transformation, perseverance of the party Constitution, relive Party vows to establish correct world Outlook, Outlook on life and the world, practically embodies ideals and beliefs into action, combine lofty ideals and practical activities, transforming the objective world, activelyparticipate in the great practice of building socialism with Chinese characteristics. 2, keep the purpose in concept, changing the style of work. First, keep in mind that the purpose of serving, adhere to assuming power for the people, keep in mind the people and care about people, and the joys and sorrows of the masses, ... According to city discipline, and municipal organization Department requirements, today we held "implement implementation independent Commission against corruption guidelines effective strengthening led cadres style construction" topic democratic life, main task is close contact thought, and work actual, control check district Standing Committee team and the personal in implementation independent Commission against corruption guidelines Shang exists of problem, in-depth carried out criticism and self-critical, further clear rectification direction and measures, effective enhanced clean politics of consciousness and initiative, firm set good of ruling image. Before the meeting, the way we take surveys solicited a views, combed, related to team buildingINPUT xIF x<0 THENy=(x+1)*(x+1)ELSEy=(x-1)*(x-1)END IFPRINT yEND解析 程序对应的函数是y=Error!由Error!或Error!得x=-6或x=6.答案 -6或611.已知函数y=Error!根据输入x的值,计算y的值,设计一个算法并写出相应程序.解 算法分析:第一步,输入x的值.第二步,判断x的范围:若x>2.5,则用y=x2-1求函数值.若x≤2.5,则用y=x2+1求函数值.第三步,输出y的值.程序如下:INPUT“x=”;xIF x>2.5 THEN y=x^2-1ELSE y=x^2+1END IFPRINT“y=”;yEND12.(创新拓展)读下面的程序,并回答问题.project work, cadres, cadres selection and other aspects of a total of 19, further defined the direction of rectification, enhancing the pertinence and effectiveness of democratic life. Good for this meeting, we requested the Standing Committee of the municipal party Committee, municipal Secretary General Tang Wenjin, the City Commission for discipline inspection Standing Committee Yang Xu, Jiang Jixue, Deputy Director of the personnel office, and other leading comrades present guidance allows us to warm applause for their welcome and thanks! Below, I first District Committee member of the 20XX team annual democratic meeting implementation of the corrective measures, as well as member of the team and of my own report guidelines for the implementation of the independent Commission against corruption statement invited leaders and comrades criticism. First, annual party on the 20XX team, Member of democratic life action plan implementation year 20XX, district democratic meeting a member of the team on "study and practice the scientific Outlook on development" theme, find a project supporting strong enough, the industry well enough, urban construction management level needs to be improved, livelihood improvement needs to be strengthened, district team itself, such as the five aspects. Year of 20XX, we find out the problems mentioned above, based on actual Shunqing construction and development, conscientiously sum up improvements, achieving significant development results and effectiveness of their work. (A) highlight the project handle, maintain high growth of investment. Adhere to the strategy as a Foundation, carrying out projects to promote increased stamina important. One is to increase the intensity of projects towards. Actively seek national and provincial policy support, superior special transfer payments, a variety of additional income will top 700 million Yuan, a number of key infrastructure projects, livelihood projects, industrial projects into the cage. Second, increasing the project attracts. This year a total of 23 signed a major project, of which 14 billion project, China Metallurgical group, vanke group, Guangdong hengda group, Taiwan Ruentex group, on the RI group, Shenzhen Kaisa, jifeng agricultural machinery, Mercedes-Benz, Audi 4S store and other large enterprises and projects have settled in Shunqing. Third, strengthen project construction. Annual implementation focus project 39 a, fixed assets investment total will breakthrough 12 billion yuan, created district yilai history high, South door dam, and five in shop old city transformation, and Ying China industrial concentrated district construction, and city main road construction, and Northeast steam Trade Center, and modern farmers industry Park, and vocational education city, and low rental, and also room construction, and features blocks build, and farmers market transformation, and "rural version" Street transformation, focus engineering speed up implementation, effectiveness obviously. Expected to be achieved this year, area-wide GDP 17 billion该程序的作用是输入x的值,输出(1)画出该程序对应的程序框图;(2)若要使输入的x值与输出的解 (1)程序对应的程序框图如图所示.或x=1.此时均满足x≤2;若2x-3=x,则x5.综上可知,满足题设条件的x值有3个.=3.。
单元写作学案训练目标文章的思路与结构。
文章的思路与结构指的是文章内容的组织和排列形式,主要包括开头和结尾、段落和层次、过渡和照应等。
这些部分在文章结构的总体设计中各自承担着不同的任务。
高考语文《考试说明》中作文部分对结构的两个基本要求是“结构完整”与“层次分明”,主要体现在以下三个方面:(1)开头与结尾的构思,(2)段落与层次的安排,(3)过渡与照应的设置。
一、文本借鉴所谓“思路”,就是作者在思索并行文时思想发展、前进的路线。
理清思路是作者思想认识系统化、明晰化的过程,也是作者对文章的整体构思,如怎样开头,如何承接、过渡,怎样围绕中心把选择的材料组织起来,哪些地方需要为文章下一步发展作好铺垫,哪些地方要与文章的前一部分有所照应,怎样结尾等,予以梳理,使之条理化、文字化的过程。
这一系列的思索、谋划,即理清思路的工作。
理顺思路非常重要,思路清楚了,才能把文章写得流畅。
而所谓“结构”,就是文章的组织构造、布局安排。
它是作者思路的外化体现。
思路与结构紧密相关,思路决定结构,结构体现思路;思路是文章的内在脉络,结构是思路的外在呈现。
这正像盖房子,对房屋的整体构想相当于“思路”,盖好的房子所体现的构造、样式即为“结构”,而盖房子的过程就是“写作”。
文章的结构并没有固定的格式,就像人们可以设计、建造出各式各样的房子一样。
但文章结构的确立,又都会受到两个条件的制约:一个是生活本身固有的规律,如人物、事件的发展变化总有一个过程,环境、景物的构成总有一定的层次,阐发道理也总会有一定的顺序、步骤;一个是文章主旨的制约作用,主旨是文章的“统帅”,文章的结构形式当然也必须为表现主旨服务。
文章的结构形式尽管可以多种多样,但都不能有悖于以上两条。
所以文章虽无“定体”,但其结构形式还是有规律可循的,可以总结出一个“大体”,即常见的文章“结构模式”。
常见的文章结构模式有“纵向式”“横向式”和“纵横结合式”。
纵向结构即按照时间顺序或事物发展的过程结构文章。
第4讲 函数y =A sin(ωx +φ)的图象及应用基础巩固题组 (建议用时:40分钟)一、选择题1.将函数y =sin x 的图象向左平移φ(0≤φ<2π)个单位后,得到函数y =sin ⎝ ⎛⎭⎪⎫x -π6的图象,则φ等于( ).A .π6 B .5π6 C .7π6D .11π6解析 将函数y =sin x 向左平移φ(0≤φ<2π)个单位得到函数y =sin(x +φ). 只有φ=116π时有y =sin ⎝ ⎛⎭⎪⎫x +116π=sin ⎝ ⎛⎭⎪⎫x -π6.答案 D2.(2014·深圳二模)如果函数f (x )=sin(πx +θ)(0<θ<2π)的最小正周期为T ,且当x =2时,f (x )取得最大值,那么( ).A .T =2,θ=π2 B .T =1,θ=π C .T =2,θ=πD .T =1,θ=π2解析 T =2ππ=2,当x =2时,由π×2+θ=π2+2k π(k ∈Z ),得θ=-3π2+2k π(k ∈Z ),又0<θ<2π,∴θ=π2. 答案 A3.已知函数f (x )=2sin(ωx +φ)(ω>0)的图象关于直线x =π3对称,且f ⎝ ⎛⎭⎪⎫π12=0,则ω的最小值为( ).A .2B .4C .6D .8解析 由f ⎝ ⎛⎭⎪⎫π12=0知⎝ ⎛⎭⎪⎫π12,0是f (x )图象的一个对称中心,又x =π3是一条对称轴,所以应有⎩⎪⎨⎪⎧ω>0,2πω≤4⎝ ⎛⎭⎪⎫π3-π12,解得ω≥2,即ω的最小值为2,故选A.答案 A4.(2014·长春模拟)函数f (x )=sin(2x +φ)⎝ ⎛⎭⎪⎫|φ|<π2向左平移π6个单位后是奇函数,则函数f (x )在⎣⎢⎡⎦⎥⎤0,π2上的最小值为( ).A .-32 B .-12 C .12D .32解析 函数f (x )=sin(2x +φ)⎝ ⎛⎭⎪⎫|φ|<π2向左平移π6个单位后得到函数为f ⎝ ⎛⎭⎪⎫x +π6=sin ⎣⎢⎡⎦⎥⎤2⎝ ⎛⎭⎪⎫x +π6+φ=sin ⎝ ⎛⎭⎪⎫2x +π3+φ,因为此时函数为奇函数,所以π3+φ=k π(k ∈Z ),所以φ=-π3+k π(k ∈Z ).因为|φ|<π2,所以当k =0时,φ=-π3,所以f (x )=sin ⎝ ⎛⎭⎪⎫2x -π3.当0≤x ≤π2时,-π3≤2x -π3≤2π3,即当2x -π3=-π3时,函数f (x )=sin ⎝ ⎛⎭⎪⎫2x -π3有最小值为sin ⎝ ⎛⎭⎪⎫-π3=-32. 答案 A5.若函数f (x )=sin ωx (ω>0)在区间⎣⎢⎡⎦⎥⎤0,π3上单调递增,在区间 ⎣⎢⎡⎦⎥⎤π3,π2上单调递减,则ω=( ).A .23 B .32 C .2D .3解析 由于函数f (x )=sin ωx (ω>0)的图象经过坐标原点,由题意知f (x )的一条对称轴为直线x =π3,和它相邻的一个对称中心为原点,则f (x )的周期T =4π3,从而ω=32.答案 B 二、填空题6.函数y =A sin(ωx +φ)(A ,ω,φ为常数,A >0,ω>0)在闭区间[-π,0]上的图象如图所示,则ω=________.解析 由图象可以看出32T =π,∴T =23π=2πω,因此ω=3. 答案 37.(2014·山东省实验中学诊断)已知函数y =g (x )的图象由f (x )=sin 2x 的图象向右平移φ(0<φ<π)个单位得到,这两个函数的部分图象如图所示,则φ=________.解析 函数f (x )=sin 2x 的图象在y 轴右侧的第一个对称轴为2x =π2,所以x =π4,π8关于x =π4对称的直线为x =3π8,由图象可知,通过向右平移之后,横坐标为x =3π8的点平移到x =17π24,所以φ=17π24-3π8=π3. 答案 π38.(2014·武汉模拟)设y =f (t )是某港口水的深度y (米)关于时间t (时)的函数(其中0≤t ≤24).下表是该港口某一天从0时至24时记录的时间t 与水深y 的关系:象.最能近似表示表中数据间对应关系的函数是________.解析 由数据可知函数的周期T =12,又T =12=2πω,所以ω=π6;函数的最大值为7.5,最小值为2.5,即h +A =7.5,h -A =2.5,解得h =5.0,A =2.5,所以函数为y =f (x )=5.0+2.5sin ⎝ ⎛⎭⎪⎫π6t +φ,又y =f (3)=5.0+2.5sin ⎝ ⎛⎭⎪⎫π6×3+φ=7.5,所以sin ⎝ ⎛⎭⎪⎫π2+φ=cos φ=1,即φ=2k π(k ∈Z ),所以最能近似表示表中数据间对应关系的函数是y =5.0+2.5sin π6t . 答案 y =5.0+2.5sin π6t 三、解答题9.(2014·苏州调研)已知函数f (x )=A sin(ωx +φ)⎝ ⎛⎭⎪⎫其中A >0,ω>0,0<φ<π2的周期为π,且图象上有一个最低点为M ⎝ ⎛⎭⎪⎫2π3,-3.(1)求f (x )的解析式;(2)求使f (x )<32成立的x 的取值集合. 解 (1)由题意知:A =3,ω=2, 由3sin ⎝ ⎛⎭⎪⎫4π3+φ=-3,得φ+4π3=-π2+2k π,k ∈Z , 即φ=-11π6+2k π,k ∈Z . 而0<φ<π2,所以k =1,φ=π6. 故f (x )=3sin ⎝ ⎛⎭⎪⎫2x +π6.(2)f (x )<32等价于3sin ⎝ ⎛⎭⎪⎫2x +π6<32,即sin ⎝ ⎛⎭⎪⎫2x +π6<12,于是2k π-7π6<2x +π6<2k π+π6(k ∈Z ), 解得k π-2π3<x <k π(k ∈Z ),故使f (x )<32成立的x 的取值集合为⎩⎨⎧⎭⎬⎫x |k π-2π3<x <k π,k ∈Z . 10.(2013·济宁测试)已知函数f (x )=23sin x cos x +2sin 2 x -1,x ∈R . (1)求函数f (x )的最小正周期和单调递增区间;(2)将函数y =f (x )的图象上各点的纵坐标保持不变,横坐标缩短到原来的12,再把所得到的图象向左平移π6个单位长度,得到函数y =g (x )的图象,求函数y =g (x )在区间⎣⎢⎡⎦⎥⎤-π6,π12上的值域.解 (1)因为f (x )=23sin x cos x +2sin 2x -1 =3sin 2x -cos 2x =2sin ⎝ ⎛⎭⎪⎫2x -π6,∴函数f (x )的最小正周期为T =π, 由-π2+2k π≤2x -π6≤π2+2k π,k ∈Z , ∴-π6+k π≤x ≤π3+k π,k ∈Z ,∴f (x )的单调递增区间为⎣⎢⎡⎦⎥⎤-π6+k π,π3+k π,k ∈Z . (2)函数y =f (x )的图象上各点的纵坐标保持不变,横坐标缩短到原来的12,得到y =2sin ⎝ ⎛⎭⎪⎫4x -π6;再把所得到的图象向左平移π6个单位长度,得到g (x )=2sin ⎣⎢⎡⎦⎥⎤4⎝⎛⎭⎪⎫x +π6-π6=2sin ⎝ ⎛⎭⎪⎫4x +π2=2cos 4x ,当x ∈⎣⎢⎡⎦⎥⎤-π6,π12时,4x ∈⎣⎢⎡⎦⎥⎤-2π3,π3,所以当x =0时,g (x )max =2,当x =-π6时,g (x )min =-1.∴y =g (x )在区间⎣⎢⎡⎦⎥⎤-π6,π12上的值域为[-1,2].能力提升题组 (建议用时:25分钟)一、选择题1.当x =π4时,函数f (x )=A sin(x +φ)(A >0,-π<φ<0)取得最小值,则函数y =f ⎝ ⎛⎭⎪⎫3π4-x 是( ).A .奇函数且图象关于点⎝ ⎛⎭⎪⎫π2,0对称B .偶函数且图象关于点(π,0)对称C .奇函数且图象关于直线x =π2对称 D .偶函数且图象关于点⎝ ⎛⎭⎪⎫π2,0对称解析 当x =π4时,函数f (x )=A sin(x +φ)(A >0)取得最小值,即π4+φ=-π2+2k π(k ∈Z ),即φ=-3π4+2k π(k ∈Z ),又-π<φ<0,所以φ=-3π4,所以f (x )=A sin ⎝ ⎛⎭⎪⎫x -3π4(A >0),所以y =f ⎝ ⎛⎭⎪⎫3π4-x =A sin ⎝ ⎛⎭⎪⎫3π4-x -3π4=-A sin x ,所以函数f (x )为奇函数且图象关于直线x =π2对称,选C. 答案 C2.(2014·长沙一模)定义⎪⎪⎪⎪⎪⎪a 1 a 2a 3 a 4=a 1a 4-a 2a 3,若函数f (x )=⎪⎪⎪⎪⎪⎪sin 2x cos 2x 1 3,则将f (x )的图象向右平移π3个单位所得曲线的一条对称轴的方程是 ( ).A .x =π6 B .x =π4 C .x =π2D .x =π解析 由定义可知,f (x )=3sin 2x -cos 2x =2sin ⎝ ⎛⎭⎪⎫2x -π6,将f (x )的图象向右平移π3个单位得到y =2sin ⎣⎢⎡⎦⎥⎤2⎝⎛⎭⎪⎫x -π3-π6=2sin ⎝ ⎛⎭⎪⎫2x -5π6,由2x -5π6=π2+k π(k ∈Z ),得对称轴为x =2π3+k π2(k ∈Z ),当k =-1时,对称轴为x =2π3-π2=π6. 答案 A 二、填空题3.已知函数f (x )=sin(ωx +φ)⎝ ⎛⎭⎪⎫ω>0,-π2≤φ≤π2的图象上的两个相邻的最高点和最低点的距离为22,且过点⎝ ⎛⎭⎪⎫2,-12,则函数解析式f (x )=________.解析 据已知两个相邻最高和最低点距离为22,可得⎝ ⎛⎭⎪⎫T 22+(1+1)2=22,解得T =4,故ω=2πT =π2,即f (x )=sin ⎝ ⎛⎭⎪⎫πx 2+φ,又函数图象过点⎝ ⎛⎭⎪⎫2,-12,故f (2)=sin ⎝ ⎛⎭⎪⎫π2×2+φ=-sin φ=-12,又-π2≤φ≤π2,解得φ=π6,故f (x )=sin ⎝ ⎛⎭⎪⎫πx 2+π6. 答案 sin ⎝ ⎛⎭⎪⎫πx 2+π6三、解答题4.(2013·淄博二模)已知函数f (x )=3sin ωx ·cos ωx +cos 2ωx -12(ω>0),其最小正周期为π2.(1)求f (x )的表达式;(2)将函数f (x )的图象向右平移π8个单位,再将图象上各点的横坐标伸长到原来的2倍(纵坐标不变),得到函数y =g (x )的图象,若关于x 的方程g (x )+k =0在区间⎣⎢⎡⎦⎥⎤0,π2上有且只有一个实数解,求实数k 的取值范围.解 (1)f (x )=3sin ωx ·cos ωx +cos 2ωx -12 =32sin 2ωx +cos 2ωx +12-12=sin⎝ ⎛⎭⎪⎫2ωx +π6,由题意知f (x )的最小正周期T =π2,T =2π2ω=πω=π2, 所以ω=2,所以f (x )=sin ⎝ ⎛⎭⎪⎫4x +π6.(2)将f (x )的图象向右平移π8个单位后,得到y =sin ⎝ ⎛⎭⎪⎫4x -π3的图象;再将所得图象所有点的横坐标伸长到原来的2倍(纵坐标不变),得到y =sin ⎝ ⎛⎭⎪⎫2x -π3的图象,所以g (x )=sin ⎝ ⎛⎭⎪⎫2x -π3, 因为0≤x ≤π2,所以-π3≤2x -π3≤2π3,所以g (x )∈⎣⎢⎡⎦⎥⎤-32,1又g (x )+k =0在区间⎣⎢⎡⎦⎥⎤0,π2上有且只有一个实数解,即函数y =g (x )与y =-k在区间⎣⎢⎡⎦⎥⎤0,π2上有且只有一个交点,由正弦函数的图象可知-32≤-k <32或-k =1,解得-32<k ≤32或k =-1,所以实数k 的取值范围是⎝ ⎛⎦⎥⎤-32,32∪{-1}.。
Evaluation Only. Created with Aspose.Words. Copyright 2003-2016 Aspose Pty Ltd.第三节培育创新意识眼前有景道不得“诗仙”李白漫游湖北武昌,打算到出名的黄鹤楼登高赋诗。
黄鹤楼建筑在一个险峻的山头上,面临浩浩荡荡的长江,格外壮丽,加之传奇有位仙人乘黄鹤在这里停留过,更加令人神往。
李白登上此楼,历史的沧桑感,怀才不遇的惆怅,以及天际浪子的思乡之情,一齐涌上心头。
李白心中已有了一首诗,刚要提笔,突然抬头观察了墙壁上崔颢的诗歌,连忙呆住了:自己想好的诗句,无论是文字还是意境,都没有超出这首诗。
他写诗的态度是格外严峻的,没有新意宁肯不写,也不愿步人后尘。
于是他投笔叹息道:“眼前有景道不得,崔颢题诗在上头。
”正是由于坚持没有新意宁肯不写,所以李白不写则已,一写就下笔有神,感天地,泣鬼神,为后世留下了很多脍炙人口的诗章。
“有创新”属于高考作文的进展等级。
具体要求是“见解新颖,材料新颖,构思新巧,推抱负象有独到之处,有共性颜色”。
见解新颖,是就作者的主见、看法、生疏而言的。
“新颖”指对客观事物有独到的见解,这就要对材料进行客观地、多角度地分析,与众不同而又顺理成章,合理透彻而又自圆其说。
材料新颖,是就写作的素材而言的,要力避陈旧、俗气的材料,紧密与当前的生活、社会、现实结合起来,多用给人以新颖感、现实感、时代感的材料。
这就要求多看书报,了解当前国际国内的社会和生活状况。
构思精致,是就文章结构而言的,即故事构思引人入胜,以巧取胜,不落俗套;谈论能选准角度,提炼精确而又独到的观点。
“有创新”包括两个方面:一是内容上有开拓,二是形式上与众不同。
1.立意创新文以载道,立意是文章的灵魂。
立意创新就是要有自己的独特感受和发觉,它来源于深刻感受生活并细心提炼,要学会多角度、多侧面去思维、感悟,或正面确定,挖掘闪光点;或逆向思维,反弹琵琶;或多向思维,纵横比较;或辩证剖析,推论因果等。
第1讲导数的概念及运算基础巩固题组(建议用时:40分钟)一、选择题1.(2015·深圳中学模拟)曲线y=x3在原点处的切线() A.不存在B.有1条,其方程为y=0C.有1条,其方程为x=0D.有2条,它们的方程分别为y=0,x=0=0,∴曲线y=x3在原点处的切线方程为y=解析∵y′=3x2,∴k=y′|x=00.答案 B2.若曲线y=x4的一条切线l与直线x+4y-8=0垂直,则l的方程为() A.4x-y-3=0 B.x+4y-5=0C.4x-y+3=0 D.x+4y+3=0解析切线l的斜率k=4,设y=x4的切点的坐标为(x0,y0),则k=4x30=4,∴x0=1,∴切点为(1,1),即y-1=4(x-1),整理得l的方程为4x-y-3=0.答案 A3.(2014·长春模拟)曲线y=x e x+2x-1在点(0,-1)处的切线方程为() A.y=3x-1 B.y=-3x-1C.y=3x+1 D.y=-2x-1解析根据导数运算法则可得y′=e x+x e x+2=(x+1)e x+2,则曲线y=x e x+2x-1在点(0,-1)处的切线斜率为y′|x=0=1+2=3.故曲线y=x e x+2x-1在点(0,-1)处的切线方程为y+1=3x,即y=3x-1.答案 A4.已知f1(x)=sin x+cos x,f n+1(x)是f n(x)的导函数,即f2(x)=f1′(x),f3(x)=f′2(x),…,f n+1(x)=f n′(x),n∈N*,则f2 015(x)等于()A.-sin x-cos x B.sin x-cos xC.-sin x+cos x D.sin x+cos x解析∵f1(x)=sin x+cos x,∴f2(x)=f1′(x)=cos x-sin x,∴f3(x)=f2′(x)=-sin x-cos x,∴f4(x)=f3′(x)=-cos x+sin x,∴f5(x)=f4′(x)=sin x+cos x,∴f n(x)是以4为周期的函数,∴f2 015(x)=f3(x)=-sin x-cos x,故选A.答案 A5.(2014·陕西卷)如图,修建一条公路需要一段环湖弯曲路段与两条直道平滑连接(相切).已知环湖弯曲路段为某三次函数图象的一部分,则该函数的解析式为()A.y=12x3-12x2-x B.y=12x3+12x2-3xC.y=14x3-x D.y=14x3+12x2-2x解析设三次函数的解析式为y=ax3+bx2+cx+d(a≠0),则y′=3ax2+2bx +c.由已知得y=-x是函数y=ax3+bx2+cx+d在点(0,0)处的切线,则y′|x=0=-1⇒c=-1,排除B、D.又∵y=3x-6是该函数在点(2,0)处的切线,则y′|x =2=3⇒12a+4b+c=3⇒12a+4b-1=3⇒3a+b=1.只有A项的函数符合,故选A.答案 A二、填空题6.(2015·珠海一模)若曲线y=ax2-ln x在点(1,a)处的切线平行于x轴,则a=________.解析y′=2ax-1x,∴y′|x=1=2a-1=0,∴a=12.答案 127.(2014·广东卷)曲线y =-5e x +3在点(0,-2)处的切线方程为__________________.解析 由y =-5e x +3得,y ′=-5e x ,所以切线的斜率k =y ′|x =0=-5,所以切线方程为y +2=-5(x -0),即5x +y +2=0. 答案 5x +y +2=08.(2014·江苏卷)在平面直角坐标系xOy 中,若曲线y =ax 2+b x (a ,b 为常数)过点P (2,-5),且该曲线在点P 处的切线与直线7x +2y +3=0平行,则a +b 的值是______.解析 y =ax 2+b x 的导数为y ′=2ax -b x 2,直线7x +2y +3=0的斜率为-72.由题意得⎩⎪⎨⎪⎧4a +b2=-5,4a -b 4=-72,解得⎩⎨⎧a =-1,b =-2,则a +b =-3.答案 -3 三、解答题9.已知曲线y =13x 3+43.(1)求曲线在点P (2,4)处的切线方程; (2)求曲线过点P (2,4)的切线方程.解 (1)∵P (2,4)在曲线y =13x 3+43上,且y ′=x 2, ∴在点P (2,4)处的切线的斜率为y ′|x =2=4.∴曲线在点P (2,4)处的切线方程为y -4=4(x -2),即4x -y -4=0.(2)设曲线y =13x 3+43与过点P (2,4)的切线相切于点A ⎝⎛⎭⎪⎫x 0,13x 30+43,则切线的斜率为y ′|x =x 0=x 20.∴切线方程为y -⎝ ⎛⎭⎪⎫13x 30+43=x 20(x -x 0),即y =x 20·x -23x 30+43.∵点P (2,4)在切线上,∴4=2x 20-23x 30+43,即x 30-3x 20+4=0,∴x 30+x 20-4x 20+4=0,∴x 20(x 0+1)-4(x 0+1)(x 0-1)=0,∴(x 0+1)(x 0-2)2=0,解得x 0=-1,或x 0=2,故所求的切线方程为x -y +2=0,或4x -y -4=0.10.设函数f (x )=ax -bx ,曲线y =f (x )在点(2,f (2))处的切线方程为7x -4y -12=0. (1)求f (x )的解析式;(2)曲线f (x )上任一点处的切线与直线x =0和直线y =x 所围成的三角形面积为定值,并求此定值.解 (1)方程7x -4y -12=0可化为y =74x -3, 当x =2时,y =12.又f ′(x )=a +bx 2,于是⎩⎪⎨⎪⎧2a -b 2=12,a +b 4=74,解得⎩⎨⎧a =1,b =3.故f (x )=x -3x .(2)设P (x 0,y 0)为曲线上任一点,由y ′=1+3x 2知曲线在点P (x 0,y 0)处的切线方程为y -y 0=(1+3x 20)(x -x 0),即y-(x 0-3x 0)=(1+3x 20)(x -x 0).令x =0,得y =-6x 0,从而得切线与直线x =0交点坐标为⎝ ⎛⎭⎪⎫0,-6x 0.令y =x ,得y =x =2x 0,从而得切线与直线y =x 的交点坐标为(2x 0,2x 0).所以点P (x 0,y 0)处的切线与直线x =0,y =x 所围成的三角形的面积为S =12⎪⎪⎪⎪⎪⎪-6x 0|2x 0|=6.故曲线y =f (x )上任一点处的切线与直线x =0,y =x 所围成的三角形面积为定值,且此定值为6.能力提升题组 (建议用时:25分钟)11.已知曲线y =1e x +1,则曲线的切线斜率取得最大值时的直线方程为 ( ) A .x +4y -2=0 B .x -4y +2=0 C .4x +2y -1=0D .4x -2y -1=0解析 y ′=-e x(e x +1)2=-1e x+1e x +2,因为e x >0,所以e x+1e x ≥2e x ×1e x =2(当且仅当e x =1e x ,即x =0时取等号),则e x +1e x +2≥4,故y ′=-1e x+1e x +2≤-14(当x =0时取等号).当x =0时,曲线的切线斜率取得最大值,此时切点的坐标为⎝ ⎛⎭⎪⎫0,12,切线的方程为y -12=-14(x -0),即x +4y -2=0.故选A .答案 A12.(2014·开封二模)过点A (2,1)作曲线f (x )=x 3-3x 的切线最多有 ( )A .3条B .2条C .1条D .0条解析 由题意得,f ′(x )=3x 2-3,设切点为(x 0,x 30-3x 0),那么切线的斜率为k =3x 20-3,利用点斜式方程可知切线方程为y -(x 30-3x 0)=(3x 20-3)(x -x 0),将点A (2,1)代入可得关于x 0的一元三次方程2x 30-6x 20+7=0.令y =2x 30-6x 20+7,则y ′=6x 20-12x 0.由y ′=0得x 0=0或x 0=2.当x 0=0时,y =7>0;x 0=2时,y =-1<0.结合函数y =2x 30-6x 20+7的单调性可得方程2x 30-6x 20+7=0有3个解.故过点A (2,1)作曲线f (x )=x 3-3x 的切线最多有3条,故选A . 答案 A13.(2014·武汉中学月考)已知曲线f (x )=x n +1(n ∈N *)与直线x =1交于点P ,设曲线y =f (x )在点P 处的切线与x 轴交点的横坐标为x n ,则log 2 016x 1+log 2 016x 2+…+log 2 016x 2 015的值为________.解析 f ′(x )=(n +1)x n ,k =f ′(1)=n +1, 点P (1,1)处的切线方程为y -1=(n +1)(x -1), 令y =0,得x =1-1n +1=n n +1,即x n =nn +1, ∴x 1·x 2·…·x 2 015=12×23×34×…×2 0142 015×2 0152 016=12 016,则log 2 016x 1+log 2 016x 2+…+log 2 016x 2 015=log 2 016(x 1x 2…x 2 015)=-1. 答案 -114.设抛物线C: y =-x 2+92x -4,过原点O 作C 的切线y =kx ,使切点P 在第一象限. (1)求k 的值;(2)过点P 作切线的垂线,求它与抛物线的另一个交点Q 的坐标. 解 (1)设点P 的坐标为(x 1,y 1),则y 1=kx 1,① y 1=-x 21+92x 1-4,②①代入②得x 21+⎝ ⎛⎭⎪⎫k -92x 1+4=0. ∵P 为切点,∴Δ=⎝ ⎛⎭⎪⎫k -922-16=0得k =172或k =12.当k =172时,x 1=-2,y 1=-17.当k =12时,x 1=2,y 1=1. ∵P 在第一象限,∴所求的斜率k =12.(2)过P 点作切线的垂线,其方程为y =-2x +5.③ 将③代入抛物线方程得x 2-132x +9=0. 设Q 点的坐标为(x 2,y 2),即2x 2=9, ∴x 2=92,y 2=-4.∴Q 点的坐标为⎝ ⎛⎭⎪⎫92,-4.。
第2讲 导数与函数的单调性、极值、最值基础巩固题组 (建议用时:40分钟)一、选择题1.(2015·威海模拟)函数f (x )=(x -3)e x 的单调递增区间是( ) A .(-∞,2) B .(0,3) C .(1,4)D .(2,+∞)解析 函数f (x )=(x -3)e x 的导数为f ′(x )=[(x -3)e x ]′=e x +(x -3)e x =(x -2)e x .由函数导数与函数单调性的关系,得当f ′(x )>0时,函数f (x )单调递增,此时由不等式f ′(x )=(x -2)e x >0,解得x >2. 答案 D2.函数y =x e x 的最小值是( ) A .-1 B .-e C .-1eD .不存在解析 y ′=e x +x e x =(1+x )e x ,令y ′=0,则x =-1,因为x <-1时,y ′<0,x >-1时,y ′>0,所以x =-1时,y min =-1e . 答案 C3.(2013·浙江卷)已知函数y =f (x )的图象是下列四个图象之一,且其导函数y =f ′(x )的图象如图所示,则该函数的图象是( )解析 由y =f ′(x )的图象知,y =f (x )的图象为增函数,且在区间(-1,0)上增长速度越来越快,而在区间(0,1)上增长速度越来越慢. 答案 B4.设a ∈R ,若函数y =e x +ax ,x ∈R 有大于零的极值点,则( ) A .a <-1B .a >-1C .a >-1eD .a <-1e解析 ∵y =e x +ax ,∴y ′=e x +a .∵函数y =e x +ax 有大于零的极值点, 则方程y ′=e x +a =0有大于零的解, ∵x >0时,-e x <-1,∴a =-e x <-1. 答案 A5.(2014·青岛模拟)已知函数f (x )=x 3+ax 2+(a +6)x +1有极大值和极小值,则实数a 的取值范围是( ) A .(-1,2) B .(-∞,-3)∪(6,+∞) C .(-3,6)D .(-∞,-1)∪(2,+∞)解析 ∵f ′(x )=3x 2+2ax +(a +6), 由已知可得f ′(x )=0有两个不相等的实根, ∴Δ=4a 2-4×3(a +6)>0,即a 2-3a -18>0. ∴a >6或a <-3. 答案 B 二、填空题6.已知函数f (x )=x 3-12x +8在区间[-3,3]上的最大值与最小值分别为M ,m ,则M -m =________.解析 由题意,得f ′(x )=3x 2-12,令f ′(x )=0,得x =±2,又f (-3)=17, f (-2)=24,f (2)=-8,f (3)=-1,所以M =24,m =-8,M -m =32. 答案 327.(2015·广州模拟)已知f (x )=x 3+3ax 2+bx +a 2在x =-1时有极值0,则a -b =________.解析 由题意得f ′(x )=3x 2+6ax +b ,则⎩⎨⎧ a 2+3a -b -1=0,b -6a +3=0,解得⎩⎨⎧ a =1,b =3或⎩⎨⎧a =2,b =9,经检验当a =1,b =3时,函数f (x )在x =-1处无法取得极值,而a =2,b =9满足题意,故a -b =-7. 答案 -78.若函数f (x )=-13x 3+12x 2+2ax 在⎣⎢⎡⎭⎪⎫23,+∞上存在单调递增区间,则a 的取值范围是________.解析 对f (x )求导,得f ′(x )=-x 2+x +2a =-⎝ ⎛⎭⎪⎫x -122+14+2a .当x ∈⎣⎢⎡⎭⎪⎫23,+∞时,f ′(x )的最大值为f ′⎝ ⎛⎭⎪⎫23=29+2a .令29+2a >0,解得a >-19. 所以a 的取值范围是⎝ ⎛⎭⎪⎫-19,+∞.答案 ⎝ ⎛⎭⎪⎫-19,+∞三、解答题9.设f (x )=a (x -5)2+6ln x ,其中a ∈R ,曲线y =f (x )在点(1,f (1))处的切线与y 轴相交于点(0,6). (1)确定a 的值;(2)求函数f (x )的单调区间与极值. 解 (1)因为f (x )=a (x -5)2+6ln x , 故f ′(x )=2a (x -5)+6x .令x =1,得f (1)=16a ,f ′(1)=6-8a , 所以曲线y =f (x )在点(1,f (1))处的切线方程为 y -16a =(6-8a )(x -1),由点(0,6)在切线上,可得6-16a =8a -6,解得a =12. (2)由(1)知,f (x )=12(x -5)2+6ln x (x >0), f ′(x )=x -5+6x =(x -2)(x -3)x .令f ′(x )=0,解得x 1=2,x 2=3. 当0<x <2或x >3时,f ′(x )>0,故f (x )的递增区间是(0,2),(3,+∞);当2<x <3时,f ′(x )<0,故f (x )的递减区间是(2,3).由此可知f (x )在x =2处取得极大值f (2)=92+6ln 2,在x =3处取得极小值f (3)=2+6ln 3.10.(2014·湘潭检测)已知函数f (x )=-x 3+ax 2+bx +c 在点P (1,f (1))处的切线方程为y =-3x +1.(1)若函数f (x )在x =-2时有极值,求f (x )的解析式; (2)函数f (x )在区间[-2,0]上单调递增,求实数b 的取值范围.解 f ′(x )=-3x 2+2ax +b ,函数f (x )在x =1处的切线斜率为-3,所以f ′(1)=-3+2a +b =-3,即2a +b =0 ①,又f (1)=-1+a +b +c =-2得a +b +c =-1 ②.(1)函数f (x )在x =-2时有极值,所以f ′(-2)=-12-4a +b =0③,由①②③解得a =-2,b =4,c =-3, 所以f (x )=-x 3-2x 2+4x -3.(2)因为函数f (x )在区间[-2,0]上单调递增,所以导函数f ′(x )=-3x 2-bx +b 在区间[-2,0]上的值恒大于或等于零,则⎩⎨⎧f ′(-2)=-12+2b +b ≥0,f ′(0)=b ≥0,得b ≥4,所以实数b 的取值范围是[4,+∞).能力提升题组 (建议用时:25分钟)11.函数f (x )=x 3-3x -1,若对于区间[-3,2]上的任意x 1,x 2,都有|f (x 1)-f (x 2)|≤t ,则实数t 的最小值是( ) A .20 B .18 C .3D .0解析 因为f ′(x )=3x 2-3=3(x -1)(x +1)令f ′(x )=0得x =±1,可知-1,1为函数的极值点.又f (-3)=-19,f (-1)=1,f (1)=-3,f (2)=1,所以在区间[-3,2]上f (x )max =1,f (x )min =-19.由题设知在区间[-3,2]上f (x )max -f (x )min ≤t ,从而t ≥20,所以t 的最小值是20. 答案 A12.(2015·福州质量检测)若函数f (x )=x 33-a 2x 2+x +1在区间⎝ ⎛⎭⎪⎫12,3上有极值点,则实数a 的取值范围是( ) A .⎝ ⎛⎭⎪⎫2,52 B .⎣⎢⎡⎭⎪⎫2,52 C .⎝ ⎛⎭⎪⎫2,103D .⎣⎢⎡⎭⎪⎫2,103解析 若函数f (x )在区间⎝ ⎛⎭⎪⎫12,3上无极值,则当x ∈⎝ ⎛⎭⎪⎫12,3时,f ′(x )=x 2-ax +1≥0恒成立或当x ∈⎝ ⎛⎭⎪⎫12,3时,f ′(x )=x 2-ax +1≤0恒成立.当x ∈⎝ ⎛⎭⎪⎫12,3时,y =x +1x 的值域是⎣⎢⎡⎭⎪⎫2,103;当x ∈⎝ ⎛⎭⎪⎫12,3时,f ′(x )=x 2-ax +1≥0,即a ≤x +1x 恒成立,a ≤2;当x ∈⎝ ⎛⎭⎪⎫12,3,f ′(x )=x 2-ax +1≤0,即a ≥x +1x 恒成立,a ≥103.因此要使函数f (x )在⎝ ⎛⎭⎪⎫12,3上有极值点,实数 a 的取值范围是⎝ ⎛⎭⎪⎫2,103,故选C .答案 C13.已知函数f (x )=-12x 2+4x -3ln x 在区间[t ,t +1]上不单调,则t 的取值范围是________.解析 由题意知f ′(x )=-x +4-3x =-(x -1)(x -3)x ,由f ′(x )=0得函数f (x )的两个极值点为1和3,则只要这两个极值点有一个在区间(t ,t +1)内,函数f (x )在区间[t ,t +1]上就不单调,由t <1<t +1或t <3<t +1,得0<t <1或2<t <3. 答案 (0,1)∪(2,3)14.(2014·安徽卷)设函数f (x )=1+(1+a )x -x 2-x 3,其中a >0. (1)讨论f (x )在其定义域上的单调性;(2)当x ∈[0,1]时,求f (x )取得最大值和最小值时的x 的值. 解 (1)f (x )的定义域为(-∞,+∞), f ′(x )=1+a -2x -3x 2. 令f ′(x )=0,得x 1=-1-4+3a 3,x 2=-1+4+3a3,x 1<x 2, 所以f ′(x )=-3(x -x 1)(x -x 2).当x <x 1或x >x 2时,f ′(x )<0;当x 1<x <x 2时,f ′(x )>0.故f (x )在(-∞,x 1)和(x 2,+∞)内单调递减,在(x 1,x 2)内单调递增.(2)因为a>0,所以x1<0,x2>0.①当a≥4时,x2≥1,由(1)知,f(x)在[0,1]上单调递增,所以f(x)在x=0和x=1处分别取得最小值和最大值.②当0<a<4时,x2<1,由(1)知,f(x)在[0,x2]上单调递增,在[x2,1]上单调递减,所以f(x)在x=x2=-1+4+3a3处取得最大值.又f(0)=1,f(1)=a,所以当0<a<1时,f(x)在x=1处取得最小值;当a=1时,f(x)在x=0处和x=1处同时取得最小值;当1<a<4时,f(x)在x=0处取得最小值.。
创新基础作业策划书3篇篇一《创新基础作业策划书》一、作业背景随着社会的发展和科技的进步,创新能力在各个领域都变得至关重要。
本次基础作业旨在培养学生的创新思维和实践能力,通过具体的任务和活动,激发学生的创造力和解决问题的能力。
二、作业目标1. 提升学生的创新意识和思维能力。
2. 培养学生独立思考和团队协作的能力。
3. 鼓励学生提出新颖的想法并付诸实践。
三、作业内容及要求1. 个人创新挑战每位学生需选择一个日常生活中的问题或现象,运用创新方法提出至少三种独特的解决方案。
要求对每个方案进行详细描述和分析其可行性。
2. 团队创新项目将学生分成小组,每个小组共同确定一个创新主题,进行深入研究和探讨。
小组需制定详细的项目计划,包括目标、步骤、分工等,并在规定时间内完成项目成果展示。
3. 创新案例分析四、作业时间安排[具体时间段 1]:个人创新挑战完成。
[具体时间段 2]:团队创新项目进行。
[具体时间段 3]:创新案例分析完成及作业汇报。
五、作业评估标准1. 创新性:方案或项目的新颖程度和独特性。
2. 可行性:方案或项目在实际中的可操作性和实施可能性。
3. 完整性:作业内容的完整性和规范性。
4. 团队协作:小组作业中团队成员的协作情况和分工合理性。
六、作业成果展示组织作业成果汇报会,学生可以通过展示 PPT、实物模型等形式展示个人和团队的作业成果。
邀请教师和其他同学进行点评和交流。
七、注意事项1. 鼓励学生大胆创新,不要局限于传统思维。
2. 强调团队合作的重要性,促进学生之间的沟通与协作。
3. 教师应及时给予学生指导和反馈,帮助学生更好地完成作业。
通过本次创新基础作业,希望学生能够在创新实践中不断提升自己的能力和素质,为未来的学习和发展打下坚实的基础。
策划人:[姓名]日期:[具体日期]篇二《创新基础作业策划书》一、作业背景随着时代的发展和科技的进步,创新能力已成为个人和社会发展的核心竞争力。
本次创新基础作业旨在培养学生的创新思维和实践能力,通过具体的任务和项目,激发学生的创造力和想象力,提升其解决问题的能力。
第1篇一、教学目标1. 知识目标:- 让学生掌握……(具体知识点)- 理解……(概念、原理)- 能够运用……(技能、方法)2. 能力目标:- 培养学生……(观察、分析、归纳、总结等能力) - 提高学生……(实际操作、解决问题等能力)3. 情感目标:- 激发学生对……(学科、知识、问题)的兴趣- 培养学生……(严谨、创新、合作等)的价值观二、教学重难点1. 教学重点:- 重点讲解……(核心知识点)- 突出……(关键步骤、方法)2. 教学难点:- 难点在于……(学生理解困难、操作复杂的问题) - 需要引导学生……(逐步突破难点)三、教学过程1. 导入新课- 引入背景:介绍……(课题相关背景知识)- 提出问题:提出……(与课题相关的问题)- 引导思考:引导学生……(思考问题、激发兴趣)2. 新课讲授- 讲解新知:讲解……(新知识、新概念)- 举例说明:举例……(具体实例,帮助学生理解)- 互动讨论:组织学生……(讨论问题、分享心得)3. 练习巩固- 基础练习:布置……(基础题目,巩固知识点)- 能力提升:设计……(有一定难度的题目,提升学生能力) - 课堂小结:总结……(本节课所学内容)4. 课堂小结- 回顾重点:回顾……(本节课的重点内容)- 总结收获:总结……(学生本节课的收获)- 布置作业:布置……(课后作业,巩固所学知识)四、板书设计1. 板书标题:……(课题名称)2. 板书内容:- 知识点1:……(第一知识点)- 举例:……(举例说明)- 知识点2:……(第二知识点)- 举例:……(举例说明)- ……- 课堂小结:……(本节课所学内容总结)3. 板书风格:- 简洁明了:板书内容要简练,突出重点- 逻辑清晰:板书内容要有层次,便于学生理解- 色彩鲜明:适当使用颜色,提高视觉效果4. 板书布局:- 中心位置:放置板书标题- 横向布局:知识点横向排列,便于学生观察- 纵向布局:知识点纵向排列,便于学生记忆五、教学反思1. 教学效果:分析……(本节课的教学效果)2. 学生反馈:了解……(学生对本节课的反馈)3. 教学改进:针对……(存在的问题)提出……(改进措施)通过以上教案过程板书设计模板,教师可以根据实际教学需求进行调整和优化,以提高教学效果。
2024年中学各科作业布置与批改细则一、语文作业布置和批改(一)作业布置1.布置作业要目的明确,精心选题,有较强的针对性,使其起到巩固和反馈的作用。
2.作业的数量和难度应适当,符合学生实际,分层次要求,体现因材施教原则.注意作业形式的多样性,减少简单的机械抄写的作业。
3.作业要求:(1)作业包括听说读写四个方面,口答笔答两种形式。
(2)课后练习应在教师指导下课内完成,课外教师要精选少量作业布置给学生,以“课堂作业本”为主,以“课时特训”为辅。
(3)每单元字词听写(含古诗默写)不少于____次。
(4)每学期大作文不少于____篇,一般在课内完成;随笔或读书笔记等课外练笔一般一周一次。
(5)各年级每天书面课外作业量要适当:(以中等水平学生平均作业时间为衡量标准)七、八、九年级:____分钟以内。
(二)作业辅导1.指导学生按时、认真、独立完成作业,培养良好的作业习惯。
对学生作业的书写应有明确要求,字体端正,字迹清楚,书写格式、步骤符合规范要求。
对不认真完成作业或作业有困难的学生应及时了解情况,分别给以教育帮助和指导。
2.每班每周下班指导学生早读(含听写检查)不少于____次,以值日老师记录为主。
(三)作业批改1.教师批改的字迹要工整,为学生起示范作用,每次作业情况要及时、详细登记。
2.作业等级分四档:优秀、良好、合格、不合格,也可根据老师个人习惯折合成相应分数。
听写(写字)作业同样依次批改。
测验、考试先以百分制结分,期终评定学期成绩时再折合成等级。
优秀作业和写得好的字可加“★”。
3.学生出现错误,应要求及时订正,并写好“订正”两字(除练习册原地订正外),其余作业都要另外订正。
错别字要求改正后组成词抄写三遍,字数较多的可以句为单位订正。
(具体实行时,要注意从效果出发,反对简单重复劳动)。
4.作文批改要重视指导讲评,要多种方法综合运用。
原则上三分之一面批,三分之一精批,三分之一略批。
要划出学生作文中出现的错别字,并在右边划半方框要求学生订正,病句可划出并修改。
步骤规范练——三角函数及三角函数的图象与性质(建议用时:90分钟)一、选择题 1.sin 600°的值为( ).A .32 B .-32 C .-12D .12解析 sin 600°=sin(720°-120°)=-sin 120°=-32. 答案 B2.若角α的终边经过点P (1,-2),则tan 2α的值为 ( ).A .-43 B .43 C .34D .-34解析 tan α=-21=-2, tan 2α=2tan α1-tan 2α=2×(-2)1-4=43.答案 B3.(2013·广州一模)下列函数中周期为π且为偶函数的是 ( ).A .y =sin ⎝ ⎛⎭⎪⎫2x -π2B .y =cos ⎝ ⎛⎭⎪⎫2x -π2C .y =sin ⎝ ⎛⎭⎪⎫x +π2D .y =cos ⎝ ⎛⎭⎪⎫x +π2解析 y =sin ⎝ ⎛⎭⎪⎫2x -π2=-cos 2x 为偶函数,且周期是π,故选A.答案 A4.(2014·郑州模拟)将函数y =cos x 的图象向右平移π2个单位长度,再向上平移1个单位长度,则所得的图象对应的解析式为 ( ).A .y =1-sin xB .y =1+sin xC .y =1-cos xD .y =1+cos x解析 函数y =cos x 的图象向右平移π2个单位长度,得到函数为y =cos ⎝ ⎛⎭⎪⎫x -π2,再向上平移1个单位长度,得到y =cos ⎝ ⎛⎭⎪⎫x -π2+1=1+sin x .答案 B5.(2013·温岭中学模拟)函数f (x )=sin x sin ⎝ ⎛⎭⎪⎫x +π2的最小正周期为( ).A .4πB .2πC .πD .π2解析 f (x )=sin x sin ⎝ ⎛⎭⎪⎫x +π2=sin x cos x =12sin 2x ,故最小正周期为T =2π2=π. 答案 C6.(2014·浙江五校联盟)要得到函数y =sin ⎝ ⎛⎭⎪⎫2x -π4的图象,只要将函数y =sin 2x的图象( ).A .向左平移π4单位B .向右平移π4单位C .向右平移π8单位D .向左平移π8单位解析 y =sin 2x ――→向右平移π8个单位y =sin 2⎝ ⎛⎭⎪⎫x -π8=sin ⎝ ⎛⎭⎪⎫2x -π4. 答案 C7.已知f (x )=2sin(ωx +φ)的部分图象如图所示,则f (x )的表达式为( ).A .f (x )=2sin ⎝ ⎛⎭⎪⎫32x +π4B .f (x )=2sin ⎝ ⎛⎭⎪⎫32x +5π4C .f (x )=2sin ⎝ ⎛⎭⎪⎫43x +2π9D .f (x )=2sin ⎝ ⎛⎭⎪⎫43x +2518π解析 由函数的部分图象可知34T =5π6-⎝ ⎛⎭⎪⎫-π6,则T =4π3,结合选项知ω>0,故ω=2πT =32,排除C ,D ;又因为函数图象过点⎝ ⎛⎭⎪⎫5π6,2,代入验证可知只有B 项满足条件. 答案 B8.(2013·昆明模拟)已知函数f (x )=2sin ⎝ ⎛⎭⎪⎫ωx -π6(ω>0)的最小正周期为π,则f (x )的单调递增区间为( ).A .⎣⎢⎡⎦⎥⎤k π+π3,k π+5π6(k ∈Z )B.⎣⎢⎡⎦⎥⎤2k π-π6,2k π+π3(k ∈Z ) C .⎣⎢⎡⎦⎥⎤k π-π3,k π+π6(k ∈Z )D.⎣⎢⎡⎦⎥⎤k π-π6,k π+π3(k ∈Z ) 解析 因为T =2πω=π,所以ω=2,所以函数为f (x )=2sin ⎝ ⎛⎭⎪⎫2x -π6,由-π2+2k π≤2x -π6≤π2+2k π,得-π6+k π≤x ≤π3+k π,即函数f (x )的单调递增区间是⎣⎢⎡⎦⎥⎤-π6+k π,π3+k π(k ∈Z ). 答案 D9.(2014·成都模拟)将函数f (x )=3sin ⎝ ⎛⎭⎪⎫4x +π6图象上所有点的横坐标伸长到原来的2倍,再向右平移π6个单位长度,得到函数y =g (x )的图象,则y =g (x )图象的一条对称轴是 ( ).A .x =π12 B .x =π6 C .x =π3D .x =2π3解析 将函数f (x )=3sin ⎝ ⎛⎭⎪⎫4x +π6图象上所有点的横坐标伸长到原来的2倍,得到函数y =3sin ⎝ ⎛⎭⎪⎫2x +π6,再向右平移π6个单位长度,得到y =3sin ⎣⎢⎡⎦⎥⎤2⎝ ⎛⎭⎪⎫x -π6+π6=3sin ⎝ ⎛⎭⎪⎫2x -π6,即g (x )=3sin ⎝ ⎛⎭⎪⎫2x -π6.当2x -π6=k π+π2时,解得x =k π+π3,又当k =0时,x =π3,所以x =π3是一条对称轴,故选C.10.(2013·长沙一模)若函数f (x )=sin ⎝ ⎛⎭⎪⎫ωx +π3的图象向右平移π3个单位后与原函数的图象关于x 轴对称,则ω的最小正值是 ( ).A .12 B .1 C .2D .3解析 若函数向右平移π3个单位后与原函数的图象关于x 轴对称,函数f (x )的周期的最大值满足T 2=π3,所以T =2π3,所以T =2π3=2πω,即ω=3,所以选D. 答案 D 二、填空题11.(2013·长沙模拟)已知角α的终边上一点的坐标为⎝ ⎛⎭⎪⎫sin 5π6,cos 5π6,则角α的最小正值为________.解析 因为tan α=cos 5π6sin 5π6=-3212=-3,且sin 5π6=12>0,cos 5π6=-32<0,所以α为第四象限角,所以α的最小正值为5π3. 答案 5π312.(2013·宁波十校测试)函数y =sin(x +10°)+cos(x +40°)(x ∈R )的最大值=________.解析 y =sin(x +10°)+cos(x +40°) =sin(x +10°)+cos[(x +10°)+30°]=sin(x +10°)+32cos(x +10°)-12sin(x +10°) =12sin(x +10°)+32cos(x +10°) =sin(x +10°+60°) =sin(x +70°),答案 113.如图所示的是函数y =A sin(ωx +φ)⎝ ⎛⎭⎪⎫A >0,ω>0,|φ|<π2图象的一部分,则其函数解析式是________.解析 由图象知A =1,T 4=π6-⎝ ⎛⎭⎪⎫-π3=π2,得T =2π,则ω=1,所以y =sin(x+φ).由图象过点⎝ ⎛⎭⎪⎫π6 ,1,可得φ=2k π+π3(k ∈Z ),又|φ|<π2,所以φ=π3,所以所求函数解析式是y =sin ⎝ ⎛⎭⎪⎫x +π3.答案 y =sin ⎝ ⎛⎭⎪⎫x +π314.已知函数f (x )=A sin(ωx +φ)(A >0,ω>0,|φ|<π)的图象与直线y =b (0<b <A )的三个相邻交点的横坐标分别是2,4,8,则f (x )的单调递增区间是________.解析 根据分析可得函数的周期为6,即2πω=6,得ω=π3,由三角函数的对称性可知,函数在x =3处取得最大值,即A sin ⎝ ⎛⎭⎪⎫π3×3+φ=A ,即sin φ=-1,所以φ=2k π-π2(k ∈Z ).又|φ|<π,所以φ=-π2,故函数的解析式为f (x )=A sin ⎝ ⎛⎭⎪⎫π3x -π2,令2k π-π2≤π3x -π2≤2k π+π2(k ∈Z ),得6k ≤x ≤6k +3(k ∈Z ).故函数f (x )的单调递增区间是[6k,6k +3](k ∈Z ). 答案 [6k,6k +3](k ∈Z )三、解答题15.函数f (x )=A sin ⎝ ⎛⎭⎪⎫ωx -π6+1(A >0,ω>0)的最大值为3,其图象相邻两条对称轴之间的距离为π2. (1)求函数f (x )的解析式;(2)设α∈⎝ ⎛⎭⎪⎫0,π2,f ⎝ ⎛⎭⎪⎫α2=2,求α的值. 解 (1)∵函数f (x )的最大值为3, ∴A +1=3,即A =2,∵函数图象的相邻两条对称轴之间的距离为π2, ∴最小正周期T =π,∴ω=2,故函数f (x )的解析式为y =2sin ⎝ ⎛⎭⎪⎫2x -π6+1.(2)f ⎝ ⎛⎭⎪⎫α2=2sin ⎝ ⎛⎭⎪⎫α-π6+1=2, 即sin ⎝ ⎛⎭⎪⎫α-π6=12,∵0<α<π2,∴-π6<α-π6<π3, ∴α-π6=π6,故α=π3.16.(2014·烟台期末考试)已知角α的顶点在原点,始边与x 轴的非负半轴重合,终边经过点P (3,-1). (1)求sin 2α-tan α的值;(2)若函数f (x )=sin 2x ·cos α+cos 2x ·sin α,求f (x )在⎣⎢⎡⎦⎥⎤0,2π3上的单调递增区间.解 (1)∵角α的终边经过点P (3,-1), ∴sin α=-12,cos α=32,tan α=-33, ∴sin 2α-tan α=2sin αcos α-tan α=-36. (2)f (x )=sin 2x ·cos α+cos 2x ·sin α=32sin 2x -12cos 2x =sin ⎝ ⎛⎭⎪⎫2x -π6.∵0≤x ≤2π3,∴0≤2x ≤4π3,∴-π6≤2x -π6≤7π6.当-π6≤2x -π6≤π2时,即0≤x ≤π3时,函数f (x )单调递增.所以函数f (x )单调递增区间是⎣⎢⎡⎦⎥⎤0,π3. 17.(2014·衡水模拟)已知函数f (x )=1+sin x cos x . (1)求函数f (x )的最小正周期和单调递减区间; (2)若tan x =2,求f (x )的值.解 (1)已知函数可化为f (x )=1+12sin 2x , 所以T =2π2=π,令π2+2k π≤2x ≤3π2+2k π(k ∈Z ), 则π4+k π≤x ≤3π4+k π(k ∈Z ),即函数f (x )的单调递减区间是⎣⎢⎡⎦⎥⎤π4+k π,3π4+k π(k ∈Z ).(2)由已知f (x )=sin 2 x +sin x cos x +cos 2xsin 2 x +cos 2x=tan 2 x +tan x +1tan 2 x +1,∴当tan x =2时,f (x )=22+2+122+1=75.18.(2014·江苏省七校联考)已知m =(a sin x ,cos x ),n =(sin x ,b sin x ),其中a ,b ,x ∈R .若f (x )=m ·n 满足f ⎝ ⎛⎭⎪⎫π6=2,且f (x )的导函数f ′(x )的图象关于直线x=π12对称. (1)求a ,b 的值;(2)若关于x 的方程f (x )+log 2k =0在区间⎣⎢⎡⎦⎥⎤0,π2上总有实数解,求实数k 的取值范围.解 (1)f (x )=m ·n =a sin 2x +b sin x cos x . 由f ⎝ ⎛⎭⎪⎫π6=2,得a +3b =8.①∵f ′(x )=a sin 2x +b cos 2x ,且f ′(x )的图象关于直线x =π12对称,∴f ′(0)=f ′⎝ ⎛⎭⎪⎫π6, ∴b =32a +12b ,即b =3a .② 由①②得,a =2,b =2 3.(2)由(1)得f (x )=1-cos 2x +3sin 2x =2sin ⎝ ⎛⎭⎪⎫2x -π6+1.∵x ∈⎣⎢⎡⎦⎥⎤0,π2,∴-π6≤2x -π6≤5π6,∴-12≤sin ⎝ ⎛⎭⎪⎫2x -π6≤1,∴0≤2sin ⎣⎢⎡⎦⎥⎤2x -π6+1≤3,即f (x )∈[0,3].又f (x )+log 2k =0在⎣⎢⎡⎦⎥⎤0,π2上有解,即f (x )=-log 2k 在⎣⎢⎡⎦⎥⎤0,π2上有解,∴-3≤log 2k ≤0,解得18≤k ≤1,即k ∈⎣⎢⎡⎦⎥⎤18,1.。