专题06 高三
专题06 功和功率 动能定理(解析版)
专题06 功和功率 动能定理目录题型一 功和功率的理解和计算 ..................................................................................................... 1 题型二 机车启动问题 ..................................................................................................................... 4 题型三 动能定理及其应用 ........................................................................................................... 12 题型四 功能中的图像问题 .. (22)题型一 功和功率的理解和计算【题型解码】1.要注意区分是恒力做功,还是变力做功,求恒力的功常用定义式.2.变力的功根据特点可将变力的功转化为恒力的功(如大小不变、方向变化的阻力),或用图象法、平均值法(如弹簧弹力的功),或用W =Pt 求解(如功率恒定的力),或用动能定理等求解.【典例分析1】(2023上·福建三明·高三校联考期中)如图所示,同一高度处有4个质量相同且可视为质点的小球,现使小球A 做自由落体运动,小球B 做平抛运动,小球C 做竖直上抛运动,小球D 做竖直下抛运动,且小球B 、C 、D 抛出时的初速度大小相同,不计空气阻力。
小球从释放或抛出到落地的过程中( )A .重力对4个小球做的功相同B .重力对4个小球做功的平均功率相等C .落地前瞬间,重力对4个小球的瞬时功率大小关系为A B CD P P P P =<= D .重力对4个小球做功的平均功率大小关系为A B C D P P P P =>= 【答案】AC【详解】A .4个质量相同的小球从同一高度抛出到落地的过程中,重力做功为G W mgh =故重力对4个小球做的功相同,故A 正确;BD .小球A 做自由落体运动,小球B 做平抛运动,小球C 做竖直上抛运动,小球D 做竖直下抛运动,小球从同一高度抛出到落地,运动时间关系为D A B C t t t t <=<重力对4个小球做功的平均功率为GW P t=可得重力对4个小球做功的平均功率大小关系为D A B C P P P P >=>故BD 错误;C .落地前瞬间,4个小球竖直方向有2A 2v gh =,2B 2v gh = 22C 02v v gh -=,22D 02v v gh -=4个小球竖直方向的速度关系为A B C D v v v v =<=落地前瞬间,重力对4个小球的瞬时功率y P mgv =落地前瞬间,重力对4个小球的瞬时功率大小关系为A B C D P P P P =<=故C 正确。
专题06 传送带模型--2024届新课标高中物理模型与方法(原卷版)
版新课标高中物理模型与方法)判断共速以后一定与传送带保持相对静止作匀速运动吗?(1)可能滑块一直加速;(2)可能滑块先加速后匀速;(1)v0<v时,可能一直加速,也可能先加速再匀速;(2)v0>v时,可能一直减速,传送带较短时,滑块一直减速达到左端.传送带较长时,滑块还要被传送带【模型演练1】(2023秋·安徽蚌埠·高三统考期末)如图甲为机场和火车站的安全检查仪,其传送装置可简化为如图乙模型,紧绷的传送带以1m/s的恒定速率运行。
旅客把行李无初速度地放在A处,设行李与传送带之间的动摩擦因数为0.1,AB间的距离为2m,g取10m/s。
行李从A到B的过程中()A.行李一直受到摩擦力作用,方向先水平向左,再水平向右B.行李到达B处时速率为1m/sC.行李到达B处所需的时间为2.5sD.行李与传送带间的相对位移为2m【模型演练1】.(2023·安徽宿州·统考一模)如图所示,水平传送带以8m/s的恒定速率逆时针运转,它与两侧的水平轨道分别相切于A、B两点,物块(视为质点)以初速度0v从B点滑上传送带,与轨道左端的竖直AB=,固定挡板P碰撞(无机械能损失)返回到B点。
已知物块与传送带、轨道间的动摩擦因数均为0.2,且6m AP=,取25mg。
物块的初速度0v可能是()10m/s=A.6m/s B.7m/s C.8m/s D.9m/s【模型演练3】.(2023春·四川成都·高三成都七中校考阶段练习)如图甲所示,一条绷紧的水平传送带AB 以恒定速率v1做匀速直线运动,传送带右端的光滑水平台面与传送带上表面等高,二者间的空隙极小不会影响滑块的运动。
滑块以速率v2向左从A点滑上传送带,在传送带上运动时动能随路程变化的E k-x图像如图乙所示,已知滑块质量m=2kg,可视为质点,重力加速度g=10m/s2。
则下列说法中正确的是()A .传送带长L 为24mB .若10v =,全程快递箱在传送带上留下的痕迹长为C .若121v v =,则全程快递箱的路程与传送带的路程之比为A .快件所受摩擦力的方向与其运动方向始终相反B .快件先受滑动摩擦力作用,后受静摩擦力作用C .快件与传送带的相对位移为0.5mD .快件由A 到B 的时间为5.5s【模型演练7】(2023·安徽·校联考模拟预测)如图为某自动控制装置的示意图,平台左右等高,在两平台中间有一个顺时针匀速转动的水平传送带,传送带的速度大小v 和长度L 都可以根据需要由自动驱动系统调节。
排列组合题型全归纳 专题06 染色问题(原卷版)
专题06染色问题【方法技巧与总结】涂色问题常用方法:(1)根据分步计数原理,对各个区域分步涂色,这是处理区域染色问题的基本方法;(2)根据共用了多少种颜色讨论,分别计算出各种情形的种数,再用分类计数原理求出不同的涂色方法种数;(3)根据某两个不相邻区域是否同色分类讨论.从某两个不相邻区域同色与不同色入手,分别计算出两种情形的种数,再用分类计数原理求出不同涂色方法总数.k 种颜色圆周染色问题如图,把一个圆分成(2)n n ≥个扇形,每个扇形用k 种颜色之一染色,要求相邻扇形不同色,有(1)(1)(1)n n n a k k =-+-⨯-种方法.正常着色定理如图,用k (k 为正整数)种颜色给图的n 个顶点着色,则正常着色的方法为:,(1)(1)(1),2n n n k F k k n =-+--≥,1,k F k =.【典型例题】例1.(2023·全国·高三专题练习)如图是某届国际数学家大会的会标,现在有4种颜色给其中5个小区域涂色,规定每个区域只涂一种颜色,相邻区域颜色不相同,则不同的涂色方案种数为()A.72B.48C.36D.24例2.(2023·全国·高三专题练习)如图,湖北省分别与湖南、安徽、陕西、江西四省交界,且湘、皖、陕互不交界,在地图上分别给各省地域涂色,要求相邻省涂不同色,现有5种不同颜色可供选用,则不同的涂色方案数为()A.480B.600C.720D.840例3.(2023·全国·高三专题练习)给图中A,B,C,D,E,F六个区域进行染色,每个区域只染一种颜色,且相邻的区域不同色.若有4种颜色可供选择,则共有()种不同的染色方案.A.96B.144C.240D.360例4.(2023·全国·高三专题练习)有4种不同颜色的涂料,给图中的6个区域涂色,要求相邻区域的颜色不相同,则不同的涂色方法共有()A.1512种B.1346种C.912种D.756种例5.(2023·全国·高三专题练习)在一个正六边形的六个区域涂色(如图),要求同一区域同一种颜色,相邻的两块区域(有公共边)涂不同的颜色,现有5种不同的颜色可供选择,则不同涂色方案有()A.720种B.2160种C.4100种D.4400种例6.(2023·全国·高三专题练习)用红、黄、蓝、绿、橙五种不同颜色给如图所示的5块区域A、B、C、D、E涂色,要求同一区域用同一种颜色,有共公边的区域使用不同颜色,则共有涂色方法()A.120种B.720种C.840种D.960种例7.(2023秋·重庆沙坪坝·高三重庆八中校考开学考试)用黑白两种颜色随机地染如图所示表格中5个格子,每个格子染一种颜色,并且从左到右数,不管数到哪个格子,总有黑色格子不少于白色格子的染色方法种数为()A .6B .10C .16D .20例8.(2023·全国·高三专题练习)在如图所示的5个区域内种植花卉,每个区域种植1种花卉,且相邻区域种植的花卉不同,若有6种不同的花卉可供选择,则不同的种植方法种数是()A .1440B .720C .1920D .960例9.(2023·全国·高三专题练习)如图,用五种不同的颜色给图中的O ,A ,B ,C ,D ,E 六个点涂色(五种颜色不一定用完),要求每个点涂一种颜色,且图中每条线段的两个端点涂不同的颜色,则不同的涂法种数是()A .480B .720C .1080D .1200例10.(2023·全国·高三专题练习)用五种不同颜色给三棱柱111ABC A B C -的六个顶点涂色,要求每个顶点涂一种颜色,且每条棱的两个顶点涂不同颜色,则不同的涂法有()A .840种B .1200种C .1800种D .1920种例11.(2023·全国·高三专题练习)正方体六个面上分别标有A 、B 、C 、D 、E 、F 六个字母,现用5种不同的颜色给此正方体六个面染色,要求有公共棱的面不能染同一种颜色,则不同的染色方案有()种.A .420B .600C .720D .780例12.(2023春·内蒙古赤峰·高二赤峰二中校考阶段练习)如图,某伞厂生产的太阳伞的伞篷是由太阳光的七种颜色组成,七种颜色分别涂在伞篷的八个区域内,且恰有一种颜色涂在相对区域内,则不同颜色图案的此类太阳伞最多有().A.40320种B.5040种C.20160种D.2520种例13.(2023·全国·高三专题练习)如图,用四种不同的颜色给图中的A,B,C,D,E,F,G七个点涂色,要求每个点涂一种颜色,且图中每条线段的两个端点涂不同颜色,则不同的涂色方法有()A.192B.336C.600D.以上答案均不对例14.(2023·全国·高三专题练习)如图所示,将一个四棱锥的每一个顶点染上一种颜色,并使同一条棱上的两端异色,如果只有5种颜色可供使用,则不同的染色方法种数是()A.420B.210C.70D.35例15.(2023·全国·高二专题练习)如图,给图中的A,B,C,D,E,F六个点涂色,要求每个点涂一种颜色,且图中每条线段的两个端点涂不同颜色,若有四种颜色可供选择,则不同的涂色方法共有______种.例16.(2023·全国·高三专题练习)如图,用四种不同颜色给图中的A,B,C,D,E,F,G,H八个点涂色,要求每个点涂一种颜色,且图中每条线段上的点颜色不同,则不同的涂色方法有___________种.例17.(2023·全国·高三专题练习)如图,一个地区分为5个行政区域,现给地图着色,要求相邻区域不得使用同一种颜色,共有5种颜色可供选择,则不同的着色方法共有________种(以数字作答).例18.(2023·全国·高三专题练习)如图,用4种不同的颜色给图中的8个区域涂色,每种颜色至少使用一次,每个区域仅涂一种颜色,且相邻区域所涂颜色互不相同,则区域A,B,C,D和1A,1B,1C,1D分别各涂2种不同颜色的涂色方法共有_________种;区域A,B,C,D和1A,1B,1C,1D分别各涂4种不同颜色的涂色方法共有_________种.例19.(2023·陕西宝鸡·统考一模)七巧板是古代劳动人民智慧的结晶.如图是某同学用木板制作的七巧板,它包括5个等腰直角三角形、一个正方形和一个平行四边形.若用四种颜色给各板块涂色,要求正方形板块单独一色,其余板块两块一种颜色,而且有公共边的板块不同色,则不同的涂色方案有______种.例20.(2023秋·甘肃张掖·高三高台县第一中学校考阶段练习)如图,节日花坛中有5个区域,现有4种不同颜色的花卉可供选择,要求相同颜色的花不能相邻栽种,则符合条件的种植方案有_____________种.例21.(2023·高二课时练习)如图,用5种不同的颜色给图中的A、B、C、D、E、F6个不同的点涂色,要求每个点涂1种颜色,且图中每条线段的两个端点涂不同的颜色,则不同的涂色方法共有______种.例22.(2023·全国·高三专题练习)在如图所示的十一面体ABCDEFGHI中,用3种不同颜色给这个几何体各个顶点染色,每个顶点染一种颜色,要求每条棱的两端点异色,则不同的染色方案种数为__________.例23.(2023·全国·高二专题练习)埃及胡夫金字塔是古代世界建筑奇迹之一,它的形状可视为一个正四棱锥,如图所示.将一个正四棱锥的每一个顶点染上一种颜色,并使同一条棱的两端异色,如果只有5种颜色可供使用,求不同的染色方法种数.例24.(2023·全国·高三专题练习)如图,一个正方形花圃被分成5份.(1)若给这5个部分种植花,要求相邻两部分种植不同颜色的花,已知现有5种颜色不同的花,求有多少种不同的种植方法?(2)若向这5个部分放入7个不同的盆栽,要求每个部分都有盆栽,问有多少种不同的放法?例25.(2023·全国·高三专题练习)用()3,n n n N *≥∈种不同的颜色给如图所示的A 、B 、C 、D 四个区域涂色.(1)若相邻区域能用同一种颜色,则图①有多少种不同的涂色方案?(2)若相邻区域不能用同一种颜色,当6n =时,图①、图②各有多少种不同的涂色方案?(3)若相邻区域不能用同一种颜色,图③有180种不同的涂色方案,求n 的值.例26.(2023·全国·高二专题练习)如图所示的A ,B ,C ,D 按照下列要求涂色.(1)用3种不同颜色填涂图中A,B,C,D四个区域,且使相邻区域不同色,若按从左到右依次涂色,有多少种不同的涂色方案?(2)若恰好用3种不同颜色给A,B,C,D四个区域涂色,且相邻区域不同色,共有多少种不同的涂色方案?(3)若有3种不同颜色,恰好用2种不同颜色涂完四个区域,且相邻区域不同色,共有多少种不同的涂色方案?例27.(2023·全国·高二专题练习)(1)从5种颜色种选出3种颜色,涂在一个四棱锥的五个顶点上,每一个顶点涂一种颜色,并使同一条棱上的两个顶点异色,则不同的涂色方法有______种;(2)从5种颜色种选出4种颜色,涂在一个四棱锥的五个顶点上,每一个顶点涂一种颜色,并使同一条棱上的两个顶点异色,则不同的涂色方法有______种.。
2024年高考物理试题分项解析专题06功和功率第01期
专题6 功和功率一.选择题1.(2024江苏泰州12月联考)中国已成为世界上高铁系统技术最全、集成实力最强、运营里程最长、运行速度最高、在建规模最大的国家。
报道称新一代高速列车牵引功率达9000kW,持续运行速度为350km/h,则新一代高速列车从北京开到杭州全长约为1300km,则列车在动力上耗电约为()A.3.3×103kW·hB.3.3×104kW·hC.3.3×105kW·hD.3.3×106kW·h【参考答案】B2.【济宁模拟】一汽车在水平平直路面上,从静止起先以恒定功率P运动,运动过程中所受阻力大小不变,汽车最终做匀速运动。
汽车运动速度的倒数1v与加速度a的关系如图所示。
下列说法正确的是( )A .汽车运动的最大速度为v 0B .阻力大小为02PvC .汽车的质量为002Pa v D .汽车的质量为00Pa v【参考答案】AD3.【郑州2025届质量检测】如图所示,不行伸长的轻绳通过定滑轮将物块甲、乙(均可视为质点)连接,物块甲套在固定的竖直光滑杆上,用外力使两物块静止,轻绳与竖直方向夹角θ=37°,然后撤去外力,甲、乙两物块从静上起先无初速释放,物块甲能上升到最高点Q ,己知Q 点与滑轮上缘O 在同一水平线上,甲、乙两物块质量分别为m 、M ,sin 37°=0.6,cos 37°=0.8,重力加速度为g ,不计空气阻力,不计滑轮的大小和摩擦。
设物块甲上升到最高点Q 时加速度为a ,则下列说法正确的是( )A .M =3mB .M =2mC .a =0D .a =g 【参考答案】BD【名师解析】当甲上升到最高点时,甲和乙的速度均为零,此时设甲上升的高度为h ,则乙下降的高度为,由能量关系可知,则M=2m,选项B正确,A错误;甲在最高点时,竖直方向只受重力作用,则a=g,选项C错误,D正确。
专题06 金属及其化合物(讲)(学生版)
8.Na-K合金常温下呈液态,是原子反应堆的导热剂。
9.铝是活泼金属,但铝抗腐蚀性相当强,因为铝表面生成一层致密的氧化物薄膜。由于Al2O3的熔点高于Al的熔点,故在酒精灯上加热铝箔直至熔化,发现熔化的铝并不滴落。
10.铝热反应不仅仅是单质铝与Fe2O3反应,还包含制取其他难熔金属的反应,由于铝热剂是混合物,故铝热反应不能用于工业上冶炼铁。注意铝热反应是中学化学中唯一一类金属单质与金属氧化物在高温条件下的置换反应。
11.并不是Al与所有金属氧化物均能组成铝热剂,该金属氧化物对应的金属活泼性应比铝弱。
12.Al2O3、Al(OH)3与NaOH溶液的反应常用于物质的分离提纯。Al(OH)3不溶于氨水,所以实验室常用铝盐和氨水来制备Al(OH)3。
核心素养
宏观辨识与微观探析科学态度与社会责任
高频考点一金属单质的性质
1.常见金属的主要性质
(1)钠及其化合物
①等物质的量的金属钠被氧化成Na2O和Na2O2时转移的电子数相同。
②钠与酸反应时,先与酸反应,酸不足再与水反应。
③钠与盐的溶液反应:钠不能置换出溶液中的金属,钠直接与水反应,反应后的碱再与溶液中的其他物质反应。
和水的反应
与冷水剧
烈反应
与沸水缓
慢反应
高温与水
蒸气反应
—
不跟水反应
生成碱和氢气
生成氧化
物和氢气
和酸的
反应
剧烈反应
反应逐渐减缓
—
不能置换稀酸中的氢
与非氧化性酸反应放出H2;与浓硫酸、浓硝酸及稀硝酸反应,不放出H2,一般产物为:盐+水+成酸元素的低价化合物
专题06 不等式 真题专项训练(全国竞赛+强基计划专用)原卷版
【高中数学竞赛真题•强基计划真题考前适应性训练】专题06不等式真题专项训练(全国竞赛+强基计划专用)一、单选题1.(2020·北京·高三强基计划)若正实数x ,y ,z ,w 满足x y w ≥≥和2()x y z w +≤+,则w z x y+的最小值等于()A .34B .78C .1D .前三个答案都不对2.(2021·北京·高三强基计划)已知,,a b c +∈R ,且111()3a b c a b c ⎛⎫+-+-= ⎪⎝⎭,则()444444111ab c a b c ⎛⎫++++ ⎪⎝⎭的最小值是()A.417+B.417-C .417D .以上答案都不对3.(2021·北京·高三强基计划)若a ,b ,c 为非负实数,且22225a b c ab bc ca ++---=,则a b c ++的最小值为()A .3B .5C .7D .以上答案都不对二、填空题4.(2021·北京·高三强基计划)在锐角ABC 中,tan tan 2tan tan 3tan tan A B B C C A ++的最小值是_________.5.(2021·全国·高三竞赛)已知正实数122020,,,a a a 满足1220201a a a +++= ,则222202012122320201a a a a a a a a a ++++++ 的最小值为________.6.(2022·浙江·高二竞赛)设a ,b ,c ,d +∈R ,1abcd =,则21914a a+∑∑的最小值为______.7.(2021·全国·高三竞赛)设正实数122020,,,a a a 满足202011i i a ==∑,则120201min1i ii kk a a ≤≤=+∑最大值为_________.8.(2021秋·天津河北·高三天津外国语大学附属外国语学校校考阶段练习)设0,0,25y x y x >>+=,则当=x _______时,12y y x +取到最大值.三、解答题9.(2023·全国·高三专题练习)设0()R[]nii i f x a x x ==∈∑,满足00,1,2,,.i a a i n ≤≤= 又设()0,1,,2i b i n = 满足22[()]ni i i f x b x ==∑,证明:()2111.2n b f +⎡⎤≤⎣⎦10.(2023·全国·高三专题练习)设0()nii i f x a x ==∑,1()n ii i g x c x +==∑是两个实系数非零多项式,且存在实数r 使得()()().g x x r f x =-记{}{}001max ,max i i i n i n a a c c ≤≤≤≤+==,证明:()1.a n c ≤+11.(2021·全国·高三竞赛)已知:a ,b ,0,2c a b c ≥++=,求证:11()1()1()bc ca ababc a b abc b c abc c a ++≤++++++.12.(2021·全国·高三竞赛)求所有的正实数a ,使得存在实数x 满足22sin cos22x x a a +≥.13.(2022·新疆·高二竞赛)(1)若实数x ,y ,z 满足2221++=x y z,证明:||||||-+-+-≤x y y z z x ;(2)若2023个实数122023,,, x x x 满足2221220231+++= x x x ,求12232022202320231-+-++-+- x x x x x x x x 的最大值.14.(2021·全国·高三竞赛)设m 为正整数,且21n m =+,求所有的实数组12,,,n x x x ,使得22221221i i nmx x x x x =++++ ,对所有1,2,,i n = 成立.15.(2021·全国·高三竞赛)求最大的正实数λ,使得对任意正整数n 及正实数01,,,n x x x ,均有010111.nnk k k kx x x x λ==≥+++∑∑ .16.(2021·全国·高三竞赛)已知01({0,1,,10})i x i <<∈ 证明:存在,{0,1,2,,10}i j ∈ ,使得()1030i j j i x x x x <-<.17.(2021·全国·高三专题练习)已知:0a >,0b >,1a b +=.2<.18.(2021·全国·高三专题练习)已知a ,b 为正数,且a b ¹,证明2112a b a b+>>>+.19.(2022·湖北武汉·高三统考强基计划)设()1,,2n x x n ⋅⋅⋅≥皆为正数,且满足1211112022202220222022n x x x ++⋅⋅⋅+=+++2022≥20.(2023·全国·高三专题练习)实数,,a b c 和正数λ使得()32f x x ax bx c =+++有三个实数根123,,x x x .且满足:(1)21x x λ-=;(2)()31212x x x >+,求332279a c abλ+-的最大值.21.(2021·全国·高三竞赛)设,1,2,,i a i n +∈=R ,记:121kk k ni i i kD C aa a =+++∑ ,其中求和是对1,2,…,n 的所有kn C 个k 元组合12,,,k i i i 进行的,求证:1.(1,2,,1)k k D D k n +≥=- .22.(2021·全国·高三竞赛)已知12,,,n a a a R ∈L ,且满足222121n a a a +++= ,求122311n n n a a a a a a a a --+-++-+-L 的最大值.23.(2021·全国·高三竞赛)已知正实数12,,,(2)n a a a n > 满足121n a a a +++= .证明:23131212121222(1)n n n n a a a a a a a a a a n a n a n n -+++≤+-+-+-- .24.(2021·浙江金华·高三浙江金华第一中学校考竞赛)数列{}n a 定义为11a =,()11111nn k k a a n n +==+≥∑.证明,存在正整数n ,使得2020n a >.25.(2021·全国·高三竞赛)给定正整数3n ≥.求最大的实数M .使得211nk k k k a M a a =+⎛⎫≥ ⎪+⎝⎭∑对任意正实数12,,,n a a a 恒成立,其中11n a a +=.26.(2019·河南·高二校联考竞赛)锐角三角形ABC 中,求证:cos()cos()cos()8cos cos cos B C C A A B A B C --- .27.(2022·贵州·高二统考竞赛)正数a ,b 满足+=1a b ,求证:2332211318a b a b ⎛⎫⎛⎫⎛⎫-- ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭.28.(2022·江苏南京·高三强基计划)已知整数1n >,证明:11!32nnn n n ++⎛⎫⎛⎫<< ⎪ ⎪⎝⎭⎝⎭.29.(2022·浙江杭州·高三学军中学校考竞赛)设实数12,,,n a a a 满足11(1)(1)n ni i i i a a ==+=-∏∏,求1ni i a =∑的最小值.30.(2021·浙江·高二竞赛)设x ,y ,0z >1=,证明4224224225552221()()()x y z y z x z y x x y z y z x z y x +++++≥+++.。
新高三英语提分培优通关练:一轮语法 专题06 现在分词(高考真题+名校模拟+写作升格)原卷版
新高三英语提分培优通关练(高考真题+名校模拟)第02辑(一轮语法专辑)专题06 现在分词(高考真题+名校模拟+写作升格)原卷版目录高考真题专区1名校模拟专区2写作升格专区4高考真题专区:练真题,明方向;练技巧,提能力;练速度,提分数!考点一:考查现在分词短语作伴随、时间状语(与句子谓语动词构成主动关系)1.(2024新课标II卷) (recall)watching a Chinese opera version of Shakespeare’s play Richard III in Shanghai and meeting Chinese actors who came to Stratford a few years ago to perform parts of The Peony Pavilion, Edmondson said, “It was very exciting to hear the Chinese language and see how Tang’s play was being performed.”2. (2024北京卷)Just then, some kids ran at him, (knock) his books out of his arms. His glasses went flying and landed in the grass.3.(2023全国甲卷)“There was once a town in the heart of America, where all life seemed to enjoy peaceful existence with is surroundings,” her fable begins, (borrow) some familiar words from many age-old fables.4.(2023全国乙卷)(visit) several times over the last 10 years, I was amazed by the co-existence of old and new, and how a city was able to keep such a rich heritage (遗产) while constantly growing.5.(2023天津3月卷)________(date) back to the 18th century, Peking Opera has over two hundred years of history.6.(2022新课标I卷)___________ (cover)an area about three times the size of Yellowstone National Park, the GPNP will be one of the first national parks in the country.7.(2022全国甲卷)Now, Cao has started the second part of his dream to walk along the Belt and Road route. He flew 4, 700 kilometers from Xi’an to Kashgar on seat 20, ___________ (plan) to hike back to Xi’an in five months.8.(2022全国乙卷)To strengthen the connection with young people, the event included a number of public promotional activities on social media, ___________ (invite) twenty-nine tea professionals from around the world to have thirty-six hours of uninterrupted live broadcasts.9.(2021新课标II卷)I was upset to learn that many sea animals eat plastic garbage, ___________ (think)it is food.10.(2021北京卷)From 2000 to 2019, there were 7,348 major natural disasters around the world, ___________ (result) in USD 2,970 billion in economic loss.11.(2019全国III卷)On the last day of our week-long stay, we were invited to attend a private concert on a beautiful farm on the North Shore under the stars, __________ (listen) to musicians and meeting interesting locals.考点二:考查现在分词短语作定语(表示一个主动的、正在进行的动作)12.(2023新课标II卷)They talk to the flood of international tourists and to 40 (visit) Chinese zookeepers who often come to check on the pandas, which are on loan from China.13.(2023天津6月卷)The Palace Museum has the most remarkable collection of fine clocks in the world, mainly _________(originate) from Europe and China.14.(2022新课标II卷)When he saw a young child hanging from a sixth-floor apartment balcony (阳台), Henry ran one hundred metres, jumped over a 1.2-metre fence, and held out his arms to catch the ___________ (fall) child.15.(2022浙江卷)Blind people recognize shapes with their ___________ (exist) senses, in a way similar to that of ___________ (sight )people, says Ella Striem-Amit, a Harvard scientist.16.(2021新课标I卷)Though it is the only unnatural thing on your way up the mountain, still it highlights the whole adventure and offers a place where you can sit down to rest your ___________ (ache) legs.17.(2021浙江1月卷)This may be due to some disadvantages for people ___________ (live) in the countryside, including lower levels of income and education, higher costs of healthy foods, and fewer sports facilities.18.(2019全国II卷)When we got a call ___________(say)she was short-listed,we thought it was a joke.考点三:考查现在分词短语作宾语补足语(与句子宾语之间构成主动关系)19.(2023新课标I卷)No matter where I buy them, one steamer is rarely enough, yet two seems greedy, so I am always left ___________ (want) more next time.20.(2020全国III卷)And when he saw the mists rising from the river and the soft clouds, ___________ (surround) the mountain tops, he was reduced to tears.名校模拟专区:做好题才有好成绩!练技能,补漏洞,提分数,强信心!1. You may already know of UNESCO’s famous list of World Heritage Sites, _____________ (consist) of places selected for special protection because of their value to the world.2. Misunderstanding _____________ (arise) from the lack of communication, unless handled properly, may lead to serious problems.3. _____________ (live) abroad for many years, the old man finally returned to his hometown.4. With so many people _____________ (focus) their eyes on him, he felt very nervous.5. Weather _____________ (permit), a field trip will be organized for the whole class.6. The room _____________ (measure) five meters across is comfortable to live in.7. While _____________ (wait) for the bus, she didn’t forget to listen to VOA.8. Mr Wang and his students spend their time restoring clocks _____________ (date) back hundreds of years.9. The study further strengthens the evidence _____________ (link) smoking with early death.10. When she looked around, she noticed a man _____________ (put) his hand into a passenger’s pocket. She rushed to stop it in time.11. Her husband died ten years ago, _____________ (leave) her with three children to look after.12. Jim was listening attentively to the lecture, (fix) all his attention on it.13. _____________ (receive) the letter from her best friend, Lucy ran towards home excitedly.14. The old man sat in front of the television every evening, happily _____________ (watch) anything that happened to be on.15. I found Janet _____________ (sit) on a bench in the backyard alone, watching the bright moon.16. _____________ (live) in Canada for three years, he still can`t speak English well.17. _____________ (judge) from his cheerful manner, he must have enjoyed his meal.18. They will get a studio flat _____________ (measure) 23 square meters with a private kitchen, bathroom and a balcony.19. The money will go to the Glide Foundation, a charity _____________ (provide) food, health care, housing and job training for San Francisco’s homeless.20. _____________ (hear) the news, he even lost her head, hammering the ground hard.21. _____________ (tell) many times,he finally understood it.22. The participants followed suit, _____________ (take) more food than they normally would have.23. Among those _____________ (take) part in the project is 80-year-old Ruth Xavier.24. _____________ (weigh) only 96 grams, the great all-round electronic dictionary is portable and easy to use.25. It was worth the cost, Alcaraz felt, because she could tell the staff cared about her daughter, _____________ (greet) the 16-month-old by name each morning.26. _____________ (judge) from the happy shouts outside tonight, I’m sure they have won the game.27. The classrooms are well-equipped, some _____________ (measure) 4 meters by 6 meters.28. So many problems _____________ (remain) to be settled, I’m in a tight corner.29. With no central government, the island was ruled by kings, each_____________ (control) a different region of the country.30. It_____________ (be) Sunday, we went camping and had a nice time.31. With more and more farmers_____________ (rush) into city, their children’s education is a problem.32. China’s image is improving steadily, with more countries _____________ (recognise) its role in international affairs.33. With the final exam _____________ (approach), all the students are busy going over their lessons.34. With the boy _____________ (lead) the way, the soldiers managed to walk through the forest.35. Once entering the room, people will see a rose-colored silk curtain _____________ (hang) on the wall over the fireplace.36. There are 200 boys in this primary school _____________ (range) from seven to fourteen in age.37. The old gentleman just stood there _____________ (glare) at the pickpocket.38. It is the most urgent threat _____________ (face) our entire species, and we need to work collectively together and stop procrastinating(拖延).39. _____________ (not do) anything like this before, I didn’t know what kind of reaction I might receive.40. I admit I was too scared at that moment and the uncle _____________ (stand) beside me was giving me an awkward look.写作升格专区:学以致用最有效!学语法,练写作,升句式,提档次!1.(2024新课标I卷) 听说你是个绘画爱好者,我迫不及待地想分享我在绘画课上的愉快经历。
专题06 板块模型(原卷版)
2023年高三物理二轮常见模型与方法强化专训专练专题06 板块模型【特训典例】 一、高考真题1.(2021全国卷)水平地面上有一质量为1m 的长木板,木板的左端上有一质量为2m 的物块,如图(a )所示。
用水平向右的拉力F 作用在物块上,F 随时间t 的变化关系如图(b )所示,其中1F 、2F 分别为1t 、2t 时刻F 的大小。
木板的加速度1a 随时间t 的变化关系如图(c )所示。
已知木板与地面间的动摩擦因数为1μ,物块与木板间的动摩擦因数为2μ,假设最大静摩擦力均与相应的滑动摩擦力相等,重力加速度大小为g 。
则()A .111=F m g μB .2122211()()m m m F g m μμ+=-C .22112m m m μμ+>D .在20~t 时间段物块与木板加速度相等2.(2019全国卷)如图(a ),物块和木板叠放在实验台上,物块用一不可伸长的细绳与固定在实验台上的力传感器相连,细绳水平.t =0时,木板开始受到水平外力F 的作用,在t =4s 时撤去外力.细绳对物块的拉力f 随时间t 变化的关系如图(b )所示,木板的速度v 与时间t 的关系如图(c )所示.木板与实验台之间的摩擦可以忽略.重力加速度取g =10m/s 2.由题给数据可以得出A .木板的质量为1kgB .2s~4s 内,力F 的大小为0.4NC .0~2s 内,力F 的大小保持不变D .物块与木板之间的动摩擦因数为0.23.(2022河北卷)如图,光滑水平面上有两个等高的滑板A 和B ,质量分别为1kg 和2kg ,A 右端和B 左端分别放置物块C 、D ,物块质量均为1kg ,A 和C 以相同速度010m /s v =向右运动,B 和D 以相同速度0kv 向左运动,在某时刻发生碰撞,作用时间极短,碰撞后C 与D 粘在一起形成一个新滑块,A 与B 粘在一起形成一个新滑板,物块与滑板之间的动摩擦因数均为0.1μ=。
高考数学 专题06 确定抽象函数单调性解函数不等式黄金解题模板-人教版高三全册数学试题
专题06 确定抽象函数单调性解函数不等式【高考地位】函数的单调性是函数的一个非常重要的性质,也是高中数学考查的重点内容。
而抽象函数的单调性解函数不等式问题,其构思新颖,条件隐蔽,技巧性强,解法灵活,往往让学生感觉头痛。
因此,我们应该掌握一些简单常见的几类抽象函数单调性及其应用问题的基本方法。
【方法点评】确定抽象函数单调性解函数不等式使用情景:几类特殊函数类型解题模板:第一步 (定性)确定函数)(x f 在给定区间上的单调性和奇偶性; 第二步 (转化)将函数不等式转化为)()(N f M f <的形式;第三步 (去f )运用函数的单调性“去掉”函数的抽象符号“f ”,转化成一般的不等式或不等式组;第四步 (求解)解不等式或不等式组确定解集;第五步 (反思)反思回顾,查看关键点,易错点及解题规X.例 1 已知函数()f x 是定义在R 上的奇函数,若对于任意给定的实数12,x x ,且12x x ≠,不等式()()()()11221221x f x x f x x f x x f x +<+恒成立,则不等式()()1120x f x +-<的解集为__________.【答案】11,2⎛⎫- ⎪⎝⎭. 例2.已知定义为R 的函数()f x 满足下列条件:①对任意的实数,x y 都有:()()()1f x y f x f y +=+-;②当0x >时,()1f x >.(1)求()0f ;(2)求证:()f x 在R 上增函数;(3)若()67,3f a =≤-,关于x 的不等式()()223f ax f x x -+-<对任意[)1,x ∈-+∞恒成立,某某数a 的取值X 围.【答案】(1)()01f =;(2)证明见解析;(3)(]5,3--.即()2130x a x -++>在[)1,x ∈-+∞上恒成立,令()()213g x x a x =-++,即()min 0g x >成立即可.①当112a +<-,即3a <-时,()g x 在[)1,x ∈-+∞上单调递增, 则()()()min 11130g x g a =-=+++>解得5a >-,所以53a -<<-,②当112a +≥-即3a ≥-时,有()()2min 111130222a a a g x g a +++⎛⎫⎛⎫==-++> ⎪ ⎪⎝⎭⎝⎭解得231231a -<<,而2313-<-,所以3231a -≤<, 综上,实数a 的取值X 围是(]5,3-- 【变式演练1】设奇函数()f x 在区间[1,1]-上是增函数,且(1)1f -=-.当[1,1]x ∈-时,函数2()21f x t at ≤-+,对一切[1,1]a ∈-恒成立,则实数t 的取值X 围为( ) A.22t -≤≤ B.2t ≤-或2t ≥ C.0t ≤或2t ≥ D.2t ≤-或2t ≥或0t = 【答案】D 【解析】试题分析:由奇函数()f x 在区间[1,1]-上是增函数,且(1)1f -=-,所以在区间[1,1]x ∈-的最大值为1,所以2121t at ≤-+当0t =时显然成立,当0t ≠时,则220t at -≥成立,又[1,1]a ∈-,令()22,[1,1]g a at t a =-∈-,当0t >时,()g a 是减函数,故令()10g ≥,解得2t ≥;当0t <时,()g a 是增函数,故令()10g -≥,解得2t ≤-,综上所述,2t ≥或2t ≤-或0t =,故选D. 考点:函数的单调性与函数的奇偶性的应用.【变式演练2】已知定义在R 上的函数()f x 为增函数,当121x x +=时,不等式()()()()1201f x f f x f +>+恒成立,则实数1x 的取值X 围是( )A. (),0-∞B. 10,2⎛⎫ ⎪⎝⎭C. 1,12⎛⎫⎪⎝⎭D. ()1,+∞ 【答案】D【变式演练3】定义在非零实数集上的函数()f x 满足()()()f xy f x f y =+,且()f x 是区间(0,)+∞上的递增函数.(1)求(1),(1)f f -的值; (2)求证:()()f x f x -=; (3)解不等式1(2)()02f f x +-≤.【答案】(1)(1)0f =,(1)0f -=;(2)证明见解析;(3)⎥⎦⎤ ⎝⎛⎪⎭⎫⎢⎣⎡1,2121,0 .考点:抽象函数及应用.【变式演练4】定义在(1,1)-上的函数()f x 满足下列条件:①对任意,(1,1)x y ∈-,都有()()()1x yf x f y f x y++=++;②当(1,0)x ∈-时,有()0f x >,求证:(1)()f x 是奇函数; (2)()f x 是单调递减函数; (3)21111()()()()1119553f f f f n n +++>++,其中*n N ∈. 【答案】(1)证明见解析;(2)证明见解析;(3)证明见解析.【解析】试题分析:(1)由奇函数的定义及特殊值0)0(=f 即可证明;(2)由单调性的定义,做差证明;(3)先由题(3)211()1(3)(2)23()[][]1155(2)(3)11()23n n n n f f f n n n n n n +-+-+++==++++-+-++ 1111()()()()2323f f f f n n n n =+-=-++++∴2111()()()111955f f f n n +++++111111[()()][()()][()()]344523f f f f f f n n =-+-++-++ 1111()()()()3333f f f f n n =-=+-++∵1013n <<+,∴1()03f n ->+,∴111()()()333f f f n +->+.故21111()()()()1119553f f f f n n +++>++.考点:1.抽象函数;2.函数的单调性,奇偶性;3.数列求和. 【高考再现】1.【2017全国卷一理】函数()f x 在()-∞+∞,单调递减,且为奇函数.若()11f =-,则满足()121f x --≤≤的x 的取值X 围是()A .[]22-,B .[]11-,C .[]04,D .[]13,【答案】D【解析】因为()f x 为奇函数,所以()()111f f -=-=,于是()121f x --≤≤等价于()()()121f f x f --≤≤| 【解析】又()f x 在()-∞+∞,单调递减 【解析】121x ∴--≤≤3x ∴1≤≤ 故选D2.【2017某某理】已知奇函数()f x 在R 上是增函数,()()g x xf x =.若2(log 5.1)a g =-,0.8(2)b g =,(3)c g =,则a ,b ,c 的大小关系为 (A )a b c << (B )c b a <<(C )b a c <<(D )b c a <<【答案】C3. 【2016高考新课标2理数】已知函数()()f x x ∈R 满足()2()f x f x -=-,若函数1x y x+=与()y f x =图像的交点为1122(,),(,),,(,),m m x y x y x y ⋅⋅⋅则1()miii x y =+=∑( )(A )0 (B )m (C )2m (D )4m 【答案】C 【解析】试题分析:由于()()2f x f x -+=,不妨设()1f x x =+,与函数111x y x x+==+的交点为()()1,2,1,0-,故12122x x y y +++=,故选C. 考点:函数图象的性质【名师点睛】如果函数()f x ,x D ∀∈,满足x D ∀∈,恒有()()f a x f b x +=-,那么函数的图象有对称轴2a bx +=;如果函数()f x ,x D ∀∈,满足x D ∀∈,恒有()()f a x f b x -=-+,那么函数的图象有对称中心.4.【2015高考,理7】如图,函数()f x 的图象为折线ACB ,则不等式()()2log 1f x x +≥的解集是()A .{}|10x x -<≤B .{}|11x x -≤≤C .{}|11x x -<≤D .{}|12x x -<≤【答案】C础题,首先是函数图象平移变换,把2log y x =沿x 轴向左平移2个单位,得到2log (y x =+2)的图象,要求正确画出画出图象,利用数形结合写出不等式的解集.5. 【2014高考某某版理第7题】下列函数中,满足“()()()f x y f x f y +=”的单调递增函数是( )(A )()12f x x = (B )()3f x x = (C )()12xf x ⎛⎫= ⎪⎝⎭(D )()3x f x =【答案】D6. 【2014某某理12】已知定义在[0,1]上的函数()f x 满足: ①(0)(1)0f f ==;②对所有,[0,1]x y ∈,且x y ≠,有1|()()|||2f x f y x y -<-. 若对所有,[0,1]x y ∈,|()()|f x f y k -<,则k 的最小值为( )A .12B .14C .12πD .18【答案】B 【解析】考点:1.抽象函数问题;2.绝对值不等式.【名师点睛】本题考查抽象函数问题、绝对值不等式、函数的最值等.解答本题的关键,是利用分类讨论思想、转化与化归思想,逐步转化成不含绝对值的式子,得出结论.本题属于能力题,中等难度.在考查抽象函数问题、绝对值不等式、函数的最值等基础知识的同时,考查了考生的逻辑推理能力、运算能力、分类讨论思想及转化与化归思想.7. 【2016高考某某理数】已知f (x )是定义在R 上的偶函数,且在区间(-∞,0)上单调递增.若实数a 足1(2)(2)a f f ->,则a 的取值X 围是______.【答案】13(,)22考点:利用函数性质解不等式【名师点睛】不等式中的数形结合问题,在解题时既要想形又要以形助数,常见的“以形助数”的方法有:(1)借助数轴,运用数轴的有关概念,解决与绝对值有关的问题,解决数集的交、并、补运算非常有效. (2)借助函数图象性质,利用函数图象分析问题和解决问题是数形结合的基本方法,需注意的问题是准确把握代数式的几何意义实现“数”向“形”的转化. 【反馈练习】1. 【2017-2018学年某某省某某市高一上学期第一次联考数学试题】函数()y f x =在R 上为增函数,且()()29f m f m >+,则实数m 的取值X 围是( )A. ()9+∞,B. [)9+∞,C. (),9-∞-D. (]9-∞, 【答案】A2.【2018届某某省某某市第一中学高三10月调研数学(理)试题】设奇函数()f x 在()0,+∞上为增函数,且()20f =,则不等式()()0f x f x x--<的解集为()A. ()()2,02,-⋃+∞B. ()(),20,2-∞-⋃C. ()(),22,-∞-⋃+∞D. ()()2,00,2-⋃【答案】D 【解析】函数()f x 为奇函数,则()()f x f x -=-,()()0f x f x x--<,化为()20f x x<,等价于()0xf x <,当0x >时,解得02x <<,当0x <时,20x -<<,不等式的解集为:()()2,00,2-⋃,选D.3.【2018届某某省某某市第一中学高三上学期第三次考试数学(文)试题】已知函数是定义在上的偶函数,且在区间上单调递增.若实数满足,则的取值X 围是( )A. B. C. D.【答案】C4.【2017届某某市滨海新区高三上学期八校联考(理科)数学试卷】已知()f x 是定义在R 上的奇函数,对任意两个不相等的正数12,x x ,都有()()2112120x f x x f x x x -<-,记()0.20.24.14.1f a =, ()2.12.10.40.4f b =,()0.20.2log 4.1log 4.1f c =,则()A. a c b <<B. a b c <<C. c b a <<D. b c a << 【答案】A【解析】设120x x << ,则()()()()122112120f x f x x f x x f x x x ->⇒>所以函数()()f x g x x=在()0,+∞上单调递减,因为()f x 是定义在R 上的奇函数,所以()g x 是定义在R上的偶函数,因此()0.20.24.14.1f a =()()0.24.11gg =<, ()2.12.10.40.4f b =()()()2.120.40.40.5gg g =>> ,()0.20.2log 4.1log 4.1f c =()()()0.251log 4.1log 4.11,2g g g g ⎛⎫⎛⎫==∈ ⎪ ⎪⎝⎭⎝⎭,即a c b << ,选A.点睛:利用函数性质比较两个函数值或两个自变量的大小,首先根据函数的性质构造某个函数,然后根据函数的奇偶性转化为单调区间上函数值,最后根据单调性比较大小,要注意转化在定义域内进行 5.【2017届某某省高三教育质量诊断性联合考试数学(文)试卷】已知定义在R 上的奇函数()f x 在[)0,+∞上递减,若()()321f x x a f x -+<+对[]1,2x ∈-恒成立,则a 的取值X 围为( ) A. ()3,-+∞ B. (),3-∞- C. ()3,+∞ D. (),3-∞ 【答案】C7.【2018届某某省六校高三上学期第五次联考理数试卷】已知函数是上的奇函数,当时为减函数,且,则=( ) A. B.C.D.【答案】A【解析】∵奇函数满足f (2)=0, ∴f (−2)=−f (2)=0.对于{x |f (x −2)>0},当x −2>0时,f (x −2)>0=f (2), ∵x ∈(0,+∞)时,f (x )为减函数, ∴0<x −2<2, ∴2<x <4.当x −2<0时,不等式化为f (x −2)<0=f (−2), ∵当x ∈(0,+∞)时,f (x )为减函数, ∴函数f (x )在(−∞,0)上单调递减, ∴−2<x −2<0,∴0<x <2.综上可得:不等式的解集为{x ∣∣0<x <2或2<x <4} 故选D. 8.【2017—2018学年某某省某某市邗江区公道中学高一数学第二次学情测试题】()f x 是定义在R 上的偶函数,且对任意的(]0a b ∈-∞,,,当a b ≠时,都有()()0f a f b a b->-.若()()121f m f m +<-,则实数m 的取值X 围为_________. 【答案】(0,2)9. 【2017届某某省某某师X 大学附属中学高三高考模拟考试二数学试题】已知()f x 是定义在区间[]1,1-上的奇函数,当0x <时,()()1f x x x =-.则关于m 的不等式()()2110f m f m -+-<的解集为__________. 【答案】[)0,1【解析】当0x >时,则()()()0,11x f x x x x x -<-=---=+,即()()1f x x x -=+,所以()()1f x x x =-+,结合图像可知:函数在[]1,1-单调递减,所以不等式()()2110f m f m -+-<可化为2220{111 111m m m m -->-≤-≤-≤-≤,解之得01m ≤<,应填答案[)0,1。
专题06(完形填空新高考区) 原卷版 高三英语百所名校好题分类快递
2022届高三英语百所名校好题分类快递(10月)专题06(阅完形填空新高考区)解析版01湖南师范大学附中2021-2022学年高三上期第二次月考英语试题John Blanchard studied the crowds in the Station. He was waiting for a girl. 13 months ago, in a library, he took a book off a ___21___ and was attracted by the notes ___22___ in it. The handwriting ___23___ a thoughtful soul.From the book he discovered the ___24___ owner's name, Miss Hollis Maynell. Later he ___25___ her address and wrote her a letter, introducing himself and inviting her to correspond. The next day he was shipped for service in a war.During the following 13 months, the two wrote frequently. A romance was ___26___. John requested a photograph but she refused, explaining that if he really cared for her ___27___ wouldn't matter. When the day finally came for him to return, they ___28___ their first meeting—7:00 p. m. at Grand Central Station. "You will recognize me." she wrote. "I will be wearing a ___29___”So here he was.A young lady in a green suit came toward him, her figure long and slim and her eyes blue as flowers. Almost uncontrollably he made one step closer to her. Then he saw Hollis Maynell- a woman ____30____ past 40. She was directly behind the girl. John felt himself ____31____ in two. So keen was his desire to follow the girl, and so deep was his ____32____ for the woman whose spirit had accompanied him. John made a decision. He saluted to the woman, ____33____ the book. "I'm John Blanchard. You must be Miss Maynell. I'm so glad to meet you. May I take you to dinner?"The woman's face broadened into a ____34____ smile. "Son,” she answered, "the young lady in the green suit ____35____ me to wear this rose. She said that if you ask me out to dinner, I should tell you she is waiting for you in the big restaurant across the street. ”21. A. profile B. shelf C. device D. column22. A. penciled B. printed C. drawn D. pressed23. A. expressed B. impressed C. reflected D. witnessed24. A. previous B. additional C. obvious D. conventional25. A. called B. visited C. located D. guessed26. A. belonging B. setting C. depending D. building27. A. attitude B. appearance C. character D. dress28. A. conducted B. plotted C. scheduled D. dated29. A. dress B. book C. hat D. rose30. A. further B. well C. beyond D. straight31. A. split B. hit C. driven D. placed32. A. relationship B. thinking C. longing D. understanding33. A. holding back B. holding off C. holding up D. holding out34. A. pitiful B. forced C. nervous D. tolerant35. A. begged B. recommended C. acquired D. squeezed02广东省珠海市2022届上学期高三摸底测试英语试题In life, do you choose to take the road everyone has taken, or do you choose the path that is best for you? It might be ____21____ to choose what everyone has done, especially if it seemed to work for them. But is that what truly ____22____ for you?On a recent hiking ____23____, my partner and I decided to take a popular trail in the ____24____ direction. This was an old, well-traveled trail that people had been hiking and biking from A to Z for years. For our own reasons, we chose to travel from Z to A. Going backwards made more sense with my ____25____. The trip would take several days and going backwards ____26____ that I would end the trip closer to the airport for my flight home.As my partner and I walked, everyone crossing our path had something to say, “You're going the wrong way. Are you lost? Are you returning ____27____ you forgot something? Are you crazy? The path occasionally crossed a ____28____. When it did, even passing cars beeped to ____29____ us in the “right” direction.Why did people only see one way? Because that's what everyone does? Because that's how it's always been done? We even ____30____ to talk to a Danish woman who said, “We have a(n) ____31____ in my country: when you go backwards to everyone else, it's because you're avoiding s omething.” I couldn't believe it. We were just enjoying connecting with nature, hiking the way that best ____32____ us. Had we gone the “wrong” way? No. At least, not for us.I decided to ____33____ the path I chose. ____34____, we had an amazing experienc e. Choosing the “wrong” path was right for me. When your inner ____35____ tells you something is right (or wrong), listen to it. It's your instinct. It's speaking to you for a reason and it knows, better than anyone, what's best for you.21. A. anxious B. normal C. awkward D. impressive22. A. designs B. plans C. works D. prepares23. A. race B. test C. project D. trip24. A. opposite B. wrong C. common D. accessible25. A. emotion B. target C. schedule D. request26. A. supposed B. ensured C. provided D. permitted27. A. if B. because C. while D. yet28. A. plain B. valley C. trail D. highway29. A. point B. inspire C. protect D. support30. A. expected B. stopped C. agreed D. hesitated31. A. saying B. point C. myth D. spell32. A. instructed B. confused C. offered D. suited33. A. show off B. think over C. pick out D. stick to34. A. Probably B. Hopefully C. Eventually D. Knowingly35. A. desire B. feeling C. voice D. reaction03江苏省高邮中学2021-2022学年高三10月学情调研英语试题Leslie Nielsen's childhood was a tough one, but he had one particular shining star in his life—his uncle, who was a well-known actor. The admiration and respect his uncle earned ____21____ Nielsen to make a career in acting. Even though he often felt he would be discovered to be a no-talent, he ____22____, gaining a scholarship to the Neighborhood Playhouse and making his first television ____23____ a few years later in 1948. ____24____, becoming a full-time, successful actor would still be an uphill battle for another eight years until he ____25____ a number of film roles that finally got him noticed.But even then, what he had wasn't quite what he wanted. Nielsen always felt he should be doing comedy but his good looks and ____26____ voice kept him busy in dramatic roles. It wasn't until 1980—32 years into his career—____27____ he landed the role it would seem he was made for in Airplane! That movie led him into the second half of his career ____28____ his comedic presence alone could make a movie a financial success even when movie reviewers would not ____29____ it highly.Did Nielsen then feel ______30______ in his career? Yes and no. He was thrilled to be doing the comedy that he always felt he should do, but even during his last few years he always had a sense of ______31______ wondering what new role or challenge might be just ______32______. He never stopped working, never ______33______.Leslie Nielsen's ______34______ to acting is wonderfully inspiring. He built a hugely successful career with little more than plain old hard work and determination. He showed us that even a single desire, never given up on, can ______35______ a remarkable life.21. A. involved B. inspired C. equipped D. adjusted22. A. moved forward B. got his way C. hired himself out D. gained currency23. A. headline B. adaptation C. appearance D. compromise24. A. Therefore B. Instead C. However D. Meanwhile25. A. hunted B. leaked C. owed D. landed26. A. flexible B. artificial C. distinguished D. delicate27. A. when B. where C. that D. who28. A. where B. that C. which D. what29. A. consult B. criticize C. rate D. weaken30. A. conscious B. content C. confidential D. cautious31. A. ambition B. creativity C. inspiration D. curiosity32. A. on the contrary B. around the corner C. at a crossroads D. under stress33. A. retired B. struck C. submitted D. split34. A. desire B. contribution C. devotion D. occupation35. A. push for B. make for C. cater for D. allow for04重庆市西南大学附属中学校高2021-2022学年高三上学期第二次月考英语试I was deeply moved by the kindness of an elderly flower seller at Sydney’s Central Station one evening. I was feeling as cold as the chill winds ___21___the steps of the workers who buried hands deep into coat pockets. In many ways, 2020 had proven a ___22___year. My teenage daughter had left home for university, my job of 23 years was no more, and a creative project I’d given my all to for years had also been ___23___ called off. I also found out a person who I’d thought was a friend was ___24___it.A year earlier, I’d dreamt about an earthquake. In the dream, a red light flash ed inside a building and then the earth began to ___25___. Walls collapsed and pieces of brick rained down on me as I ran into the open, feeling___26___for the people I hadn’t loved enough. ___27___I was buried completely by the falling world, I stretched out my hands and ___28___, “Abuelo, ayudame,” in my native Spanish. “Grandfather, help me!”Now, with the very real aftershocks of a number of ___29___endings, I walked into a little flower shop to buy a gift for a friend. The gentleman at the checkout looked at me and ___30___for a moment, returning with a fragrant flower. “For you. You need to smile today,” he said in a rich and beautiful accent, ___31___me the flower with eyes full of care. It took all my strength not to ___32___tears. Then he ___33___me in a grandfatherly hug that made me want to tell him everything. ___34___, I just thanked him and headed for the train. How did he know? I wondered. For the first time in weeks, I felt ___35___.21. A. following B. taking C. slowing D. quickening22. A. smooth B. splendid C. tough D. fruitful23. A. frequently B. gradually C. severely D. suddenly24. A. far from B. free from C. more than D. other than25. A. turn B. swing C. rock D. roll26. A. ashamed B. puzzled C. anxious D. sorrowful27. A. Before B. When C. After D. Until28. A. yelled B. complained C. whispered D. declared29. A. extreme B. unexpected C. alternative D. specific30. A. looked out B. thought over C. turned away D. broke down31. A. showing B. offering C. sending D. delivering32. A. leak out B. burst into C. hold back D. wipe away33. A. grasped B. surrounded C. wrapped D. protected34. A. However B. Meanwhile C. Moreover D. Thus35. A. envy B. hope C. guilt D. pity05江苏省扬州中学2021—2022学年第一学期10月份考试Juan Manuel Ballestero was a 47-year-old sailing enthusiast, who currently lived in Porto Santo, Portugal. When the COVID-19 pandemic (流行病) outbreak started to take its hold, Ballestero began to _____41_____ across the Atlantic 10 reach his 90-year-old dad in the middle of March.“I didn’t want to _____42_____ like a coward (懦弱的人) on an island where there were no _____43_____, Ballestero said. “My father is 90 years old and I’m _____44_____ his health, especially during the pandemic.”The journey wasn’t without any trouble. Although he’d _____45_____ necessities and fuel before leaving the Portuguese island in his boat, he _____46_____ to make a stop at Cape Verde in mid-April to pick up more suppliesand fuel. _____47_____, authorities in the middle of the Atlantic Ocean refused his request to dock (停靠). So, the _____48_____ sailor carried on his journey in hopes that he’d get to see his dad, unsure of what to _____49_____ as the world battled with the pandemic. He said, “The love for my father kept me _______50_______ in these situations. I learned about myself; this voyage gave me lots of _______51_______ He finally reached the port of his _______52_______ Mar del Plata and was able to _______53_______ his 90-year-old father after his COVID-19 test _______54_______ came back negative (阴性的)— just in time for Father’s Day.Ballestero proves how the love for his father has no _______55_______ even in such a difficult time.41. A. travel B. swim C. fly D. run42. A. escape B. leave C. stay D. relax43. A. resources B. comforts C. reasons D. cases44. A. aware of B. certain of C. concerned about D. satisfied with45. A. ordered B. prepared C. produced D. consumed46. A. managed B. failed C. referred D. intended47. A. Otherwise B. Anyhow C. However D. Therefore48. A. honest B. courageous C. generous D. stubborn49. A. expect B. pay C. predict D. bring50. A. rising B. changing C. wondering D. standing51. A. hope B. inspiration C. tests D. advantages52. A. unique B. native C. pretty D. great53. A. hug B. welcome C. accept D. protect54. A. result B. effect C. agenda D. standard55. A. measurements B. choices C. questions D. limits06江苏省苏州市吴中区2021-2022学年高三上期10月份调研测试题S. E. Hinton's career as an author began while she was still a high school student.____21____ by the fights of the two gangs in her high school, Hinton wrote The Outsiders, an honest, sometimes shocking novel told from the point of ____22____ of a fourteen-year-old boy.The Outsiders was ____23____ during Hinton's freshman year at the University of Tulsa, and was an immediate ____24____. Today, with more than fourteen million ____25____ in print, the book is the bestselling young adult novel of all time. The book was also made into a film in 1983.The Outsiders brought with it publicity and 4____26____. S.E. Hinton became known as “The V o ice of the Youth.” This ____27____ success also brought a lot of pressure,____28____ a three-year-long writer's block. Her boyfriend (now husband)____29____ helped break this block by suggesting she write two pages a ____30____ before going anywhere. This finally led to her second ____31____, That Was Then, This Is Now.In 1988 she was awarded the first annual Margaret A. Edwards Award, given in ____32____ of “an author whose book or books, over a period of time, have been accepted by ____33____ adults as an authentic voice that ____34____ to illuminate their experiences and emotions,____35____ insight into their lives.”21. A. Delighted B. Disturbed C. Discouraged D. Defeated22. A. view B. sight C. dream D. fear23. A. designed B. started C. published D. rejected24. A. hit B. loss C. aim D. beat25.A. lines B. pages C. words D. copies26. A. shame B. fame C. failure D. challenge27. A. longstanding B. impossible C. overnight D. imaginary28. A. resulting in B. suffering from C. giving up D. preparing for29. A. suddenly B. usually C. permanently D. eventually30. A. year B. month C. week D. day31. A. fight B. novel C. film D. block32. A. regret B. mercy C. honor D. favor33. A. wise B. silly C. aged D. young34. A. struggles B. continues C. interrupts D. hesitates35. A. giving B. breaking C. witnessing D. overlooking07湖北省黄冈市2022届高三上学期9月调研考试英语试题As a kid, I remember that whenever my dad was repairing something, he would ask me to hold the hammer, just so we'd have time to talk with each other.When I left for college, my dad would 41 me every Sunday morning. Years later, when I bought a house, my dad 42 it in the summer heat. All he asked me to do was 43 his brush and talk to him. But I was too 44 those days to talk with him.Four years ago, my dad was putting together a swing set for my daughter. He asked me to bring him a cup oftea and have a 45 with him, but I had to prepare for a trip that weekend.One Sunday morning we were talking on the phone 46 , and I noticed my dad had 47 some things we discussed recently. I was in a hurry, so our conversation was 48 . Later that day came a call. My father had had an aneurysm (动脉瘤) and was in the hospital. 49 I flew back and on my way I was thinking about all the 50 opportunities to talk with my dad.By the time 1 arrived, my father had 51 . Now it was he who didn't have time for a (n) 52 with me. After his death I 53 much more about him. All he 54 was only my time. And now I 55 that nothing is more important than my company with my dear parents.41. A. witness B. call C. instruct D. discover42. A. painted B. visited C. repaired D. tidied43. A. find B. hold C. bring D. clean.44. A. upset B. happy C. busy D. disappointed45. A. talk B. meal C. game D. walk46. A. on time B、all the time C. as well D. as usual47. A. understood B. promoted C. remembered D. forgotten48. A. joyful B. annoying C. short D. delayed49. A. Suddenly B. Immediately C. Personally D. Fortunately50. A. missed. B. valuable C. rare D. good51. A. passed by. B. passed away C. died out D. set off52. A. laugh B. trip C. adventure D. conversation53. A. complained B. learned C. cared D. guessed54. A. commanded B. expressed C. requested D. reminded55. A. concerned B. ignored C. conquered D. recognized08重庆市顶级名校2021-2022学年高三上学期第二次月考英语试题I was deeply moved by the kindness of an elderly flower seller at Sydney’s Central Station one evening. I was feeling as cold as the chill winds 41 the steps of the workers who buried hands deep into coat pockets. In many ways, 2020 had proven a 42 year. My teenage daughter had left home for university, my job of 23 years was no more, and a creative project I’d given my all to for years had also been 43 called off. I also found out a person who I’d thought was a friend was44 it.A year earlier, I’d dreamt about an earthquake. In the dream, a red light flashed inside a building and then the earth began to 45 . Walls collapsed and pieces of brick rained down on me as I ran into the open, feeling 46 for the people I hadn’t loved enough. 47 I was buried completely by the falling world, I stretched out my hands and 48 , “Abuelo, ayudame,” in my native Spanish. “Grandfather, help me!”Now, with the very real aftershocks of a number of 49 endings, I walked into a little flower shop to buy a gift for a friend. The gentleman at the checkout looked at me and 50 for a moment, returning with a fragrant flower. “For you. You need to smile today,” he said in a rich and beautiful accent, 51 me the flower with eyes full of care. It took all my strength not to 52 tears. Then he 53 me in a grandfatherly hug that made me want to tell him everything. 54 , I just thanked him and headed for the train. How did he know? I wondered. For the first time in weeks, I felt 55 .41. A. following B. taking C. slowing D. quickening42. A. smooth B. splendid C. tough D. fruitful43. A. frequently B. gradually C. severely D. suddenly44. A. far from B. free from C. more than D. other than45. A. turn B. swing C. rock D. roll46. A. ashamed B. puzzled C. anxious D. sorrowful47. A. Before B. When C. After D. Until48. A. yelled B. complained C. whispered D. declared49. A. extreme B. unexpected C. alternative D. specific50. A. looked out B. thought over C. turned away D. broke down51. A. showing B. offering C. sending D. delivering52. A. leak out B. burst into C. hold back D. wipe away53. A. grasped B. surrounded C. wrapped D. protected54. A. However B. Meanwhile C. Moreover D. Thus55. A. envy B. hope C. guilt D. pity09广东省深圳外国语学校2022届高三上学期第一次月考英语试卷Country diary: a chainsaw massacre in the alder woodsO n an overcast, drizzly afternoon at Durham Wildlife Trust’s low Barns nature 21 alder provided thebrightest splash of color in the 22 .A tree had been felled and sawn into 23 . Chainsaw wounds on this species can look like a massacre, because soon after the timber is cut, it turns a lurid shade of red, almost like blood, in stark contrast to the battleship-grey bark, 24 those wounds, which briefly 25 raw meat, fade to orange and finally to chestnut brown.When this reserve was established half a century ago, around old gravel pits, some moisture-loving alders were planted to help 26 a bare, windswept site. Alder wood is one of the finest sources of charcoal, and the plantation trees are old enough now to be coppiced, to produce barbecue fuel.There is also an important natural alder wood here, created by a cataclysm almost two and a half centuries ago, which led to the designation of the reserve as a site of special scientific interest.The Great Flood of 1771 27 Weardale, washing away bridges all the way to the coast. When the water subsided, the course of the River Wear had 28 half a mile south, and the old riverbed became what is now the reserve’s Long Alder Wood, the finest example of its kind in the region.When it sometimes floods, this tangle of gnarled (苍劲的) trees has a 29 of the Florida Everglades about it, with mossy, fallen trunks sinking back into the ooze. Year round, there are wonderful 30 to watch birds from an embankment level with the tree canopy. This afternoon an acrobatic flock of about 30 goldfinches (金翅雀) 31 and chattered through the twigs, feeding on tiny seeds that fall from the woody cones.Sadly, since the mid-1990s, another 32 has befallen this locally 33 woodland: alder dieback disease has killed around half the mature trees. Coppicing is leading to some regeneration, though in this precious 34 dead timber is allowed to lay where it falls, reserved for the needs of a 35 community of fungi, invertebrates and woodpeckers, rather than back-garden burger-flippers on summer evenings.21. A. reserve B. preserve C. conserve D. deserve22. A. scene B. landscape C. scope D. view23. A. logs B. materials C. resources D. sources24. A. Everlastingly B. Sustainably C. Continually D. Eventually25. A. assemble B. present C. overcast D. resemble26. A. rejuvenate B. remain C. reform D. revenge27. A. gone through B. got through C. swept through D. cut through28. A. changed B. revised C. reversed D. shifted29. A. clue B. plot C. evidence D. hint30. A. alternatives B. possibilities C. opportunities D. probabilities31. A. schemed B. crawled C. bounced D. scattered32. A. misfortune B. adversity C. setback D. catastrophe33. A. unique B. peculiar C. especial D. particular34. A. habitat B. territory C. frontier D. boundary35. A. various B. versatile C. multiple D. diverse21-25 ABADD 26-30 DCDDC 31-35 CDAAD。
高考英语 必考点 专题06 词性转化(构词法)(高效演练)(含解析)-人教版高三全册英语试题
专题06 词性转化〔构词法〕——高效演练一.单句语法填空1. The boy ran(quick)to school.【答案】quickly【解析】考查词性转化之形容词转化为副词。
空格前的ran需用副词修饰,故填quickly。
2. "What's that? "Father shouted (angry).【答案】angrily【解析】考查词性转化之形容词转化为副词。
空格前的shouted需用副词修饰,故填angrily。
3. The little girl is(extreme)eager to know the result of the exam.【答案】extremely【解析】考查词性转化之形容词转化为副词。
空格后的形容词eager需用副词修饰,故填extremely。
句意:这个小女孩迫切想知道考试的结果。
4. Your composition is (bad)organized. Please do your writing exercise more attentively next time.【答案】badly【解析】考查词性转化之形容词转化为副词。
空格后的形容词bad需用副词修饰,故填badly。
句意:你的作文组织结构太差。
下次请务必多加注意。
5. He is rather(self) so that nobody prefers to have a talk with him.【答案】selfish【解析】考查词性转化之名词转化为形容词。
空格前的rather是副词,要修饰形容词,故填selfish。
句意:他这个人太自私,没有人愿意跟他说话。
6. The good working condition in the factory is(attract).【答案】attractive【解析】考查词性转化之动词转化为形容词。
空格处是表语位置,需用形容词,故填attractive。
专题06 重力和弹力(解析版)
2023届高三物理一轮复习重点热点难点专题特训专题06 重力和弹力特训目标特训内容目标1 重力和重心(1T—4T)目标2 判定弹力有无(5T—8T)目标3 弹力的方向(9T—12T)目标4弹力的大小(13T—16T)一、重力和重心1.2022年春晚舞蹈《只此青绿》表演中,需要舞者两脚前后分开,以胯部为轴,上半身后躺,与地面近乎平行,在舞者缓慢后躺的过程中,下列说法不正确的是()A.人对地面的压力和地面对人的支持力是一对平衡力B.舞者所受的支持力始终等于舞者的重力C.舞者越向下躺,重心越低D.舞者两脚间距越大,下弯时越稳定【答案】A【详解】A.人对地面的压力和地面对人的支持力是一对作用力与反作用力,选项A错误,符合题意;B.舞者缓慢后躺,可认为都是平衡状态,则舞者所受的支持力始终等于舞者的重力,选项B正确,不符合题意;C.舞者越向下躺,整个身体的重心越低,选项C正确,不符合题意;D.舞者两脚间距越大,重心越向下,下弯时越稳定,选项D正确,不符合题意。
故选A。
2.如图所示,浙江某中学学生做化学实验时,水平放置的圆形铁环上放入分液漏斗,两者均处于静止状态,则()A.分液漏斗受到的重力方向一定指向地心B.圆形铁环对分液漏斗的作用力与底角θ无关C.圆形铁环对分液漏斗的弹力是由分液漏斗形变产生的D.若分液漏斗中溶液不断流出,分液漏斗的重心不断下降【答案】B【详解】A.重力是由地球的吸引而产生的,方向竖直向下,而不一定指向地心,A错误;B.圆形铁环对分液漏斗的作用力与分液漏斗的重力平衡,故圆形铁环对分液漏斗的作用力与底角θ无关,B正确;C.圆形铁环对分液漏斗的弹力是由圆形铁环形变产生的,C错误;D.分液漏斗装满溶液时,重心在几何中心处,溶液不断流出的过程重心下降,当溶液流完后重心又回到球心处,故重心先下降后上升,D错误。
故选B。
3.如图所示,质量为m的小玉坠用一根长为L的不可伸长的轻绳穿过后悬挂于货架水平杆上M、N两点并处于静止状态;如右图所示,现将长度也为L,质量也为m且分布均匀的项链悬挂于M、N两点并处于静止状态,小玉坠的重心和项链的重心到水平杆的距离分别是h1和h2,M点对轻绳和项链的拉力分别是F1和F2,则()A.h1>h2,F1>F2B.h1<h2,F1>F2C.h1<h2,F1<F2D.h1>h2,F1=F2【答案】A【详解】如题右图中的重力全部集中在玉坠上,重心在玉坠上,图中项链的质量均匀分布于整条项链,其重心一定在最低点的上方,所以玉坠重心到水平杆的距离比项链重心到水平杆的距离大,即h1>h2;对玉坠和项链的受力分析如图1和图2所示,图1中M、N两点对绳子的拉力方向沿着绳子的方向,图2中M、N两点对项链的拉力方向沿着项链的切线方向,图1中M、N两点对绳子拉力的夹角,比图2中M、N两点对项链拉力的夹角大,根据力的合成可知,图1中M 点对绳子的拉力F1比图2中M点对项链的拉力F2要大,即F1>F2。
专题06 等差数列、等比数列及数列的求和-高考数学试题探源与变式(解析版)
专题六 等差数列、等比数列及数列的求和【母题原题1】【2019浙江,10】设,a b R ∈,数列{}n a 中,21,n n n a a a a b +==+,b N *∈ ,则( ) A. 当101,102b a => B. 当101,104b a => C. 当102,10b a =-> D. 当104,10b a =->【答案】A 【解析】选项B :不动点满足2211042x x x ⎛⎫-+=-= ⎪⎝⎭时,如图,若1110,,22n a a a ⎛⎫=∈< ⎪⎝⎭,排除如图,若a 为不动点12则12n a = 选项C :不动点满足22192024x x x ⎛⎫--=--= ⎪⎝⎭,不动点为ax 12-,令2a =,则210n a =<,排除选项D :不动点满足221174024x x x ⎛⎫--=--= ⎪⎝⎭,不动点为122x =±,令122a =±,则11022n a =±<,排除.选项A :证明:当12b =时,2222132431113117,,12224216a a a a a a =+≥=+≥=+≥≥, 处理一:可依次迭代到10a ;处理二:当4n ≥时,221112n nn a a a +=+≥≥,则117117171161616log 2log log 2n n n n a a a -++>⇒>则12117(4)16n n a n -+⎛⎫≥≥ ⎪⎝⎭,则626410217164646311114710161616216a ⨯⎛⎫⎛⎫≥=+=++⨯+⋯⋯>++> ⎪ ⎪⎝⎭⎝⎭.故选A【母题原题2】【2018浙江,10】已知成等比数列,且.若,则A.B.C.D.【答案】B 【解析】 令则,令得,所以当时,,当时,,因此,若公比,则,不合题意;若公比,则但, 即,不合题意;因此,,选B.点睛:构造函数对不等式进行放缩,进而限制参数取值范围,是一个有效方法.如【母题原题3】【2017浙江,6】已知等差数列{}n a 的公差为d,前n 项和为n S ,则“d>0”是465"+2"S S S >的A. 充分不必要条件B. 必要不充分条件C. 充分必要条件D. 既不充分也不必要条件 【答案】C【解析】由()46511210212510S S S a d a d d +-=+-+=,可知当0d >时,有46520S S S +->,即4652S S S +>,反之,若4652S S S +>,则0d >,所以“d>0”是“S 4 + S 6>2S 5”的充要条件,选C .【名师点睛】本题考查等差数列的前n 项和公式,通过套入公式与简单运算,可知4652S S S d +-=, 结合充分必要性的判断,若p q ⇒,则p 是q 的充分条件,若p q ⇐,则p 是q 的必要条件,该题“0d >” ⇔ “46520S S S +->”,故互为充要条件. 【母题原题4】【2016浙江,文8理6】如图,点列{}{},n n A B 分别在某锐角的两边上,且*1122,,n n n n n n A A A A A A n ++++=≠∈N ,*1122,,n n n n n n B B B B B B n ++++=≠∈N .(P≠Q 表示点P 与Q 不重合)若n n n d A B =,n S 为1n n n A B B +△的面积,则A .{}n S 是等差数列B .{}2n S 是等差数列C .{}n d 是等差数列D .{}2n d 是等差数列 【答案】A【解析】S n 表示点n A 到对面直线的距离(设为n h )乘以1n n B B +长度的一半,即112n n n n S h B B +=,由题目中条件可知1n n B B +的长度为定值,那么我们需要知道n h 的关系式,由于1,n A A 和两个垂足构成了直角梯形,那么11sin n n h h A A θ=+⋅,其中θ为两条线的夹角,即为定值,那么1111(sin )2n n n n S h A A B B θ+=+⋅,111111(||sin )2n n n n S h A A B B θ+++=+⋅,作差后:1111(sin )2n n n n n n S S A A B B θ+++-=⋅,都为定值,所以1n n S S +-为定值.故选A.【母题原题5】【2019浙江,20】设等差数列{}n a 的前n 项和为n S ,34a =,43a S =,数列{}n b 满足:对每12,,,n n n n n n n S b S b S b *++∈+++N 成等比数列.(1)求数列{},{}n n a b 的通项公式;(2)记,n C n *=∈N证明:12+.n C C C n *++<∈N【答案】(1)()21n a n =-,()1n b n n =+;(2)证明见解析. 【解析】(1)由题意可得:1112432332a d a d a d +=⎧⎪⎨⨯+=+⎪⎩,解得:102a d =⎧⎨=⎩, 则数列{}n a 的通项公式为22n a n =-.其前n 项和()()02212n n n S nn +-⨯==-.则()()()()1,1,12n n n n n b n n b n n b -++++++成等比数列,即:()()()()21112n n n n n b n n b n n b ++=-+⨯+++⎡⎤⎡⎤⎡⎤⎣⎦⎣⎦⎣⎦,据此有:()()()()()()()()2222121112121n n n n nn n n n b b n n n n n n b n n b b ++++=-++++++-+,故()()()()()22112121(1)(1)(1)(2)n n n n n n b n n n n n n n n n +--++==++++--+.(2)结合(1)中的通项公式可得:2nC==<=<=,则()()()12210221212nC C C n n n+++<-+-++--=【母题原题6】【2018浙江,20】已知等比数列{a n}的公比q>1,且a3+a 4+a 5=28,a4+2是a3,a5的等差中项.数列{b n }满足b 1=1,数列{(b n +1−b n )a n}的前n 项和为2n 2+n.(Ⅰ)求q的值;(Ⅱ)求数列{b n }的通项公式.【答案】(Ⅰ)(Ⅱ)【解析】(Ⅰ)由是的等差中项得,所以,解得.由得,因为,所以.(Ⅱ)设,数列前n项和为.由解得.由(Ⅰ)可知,所以,故,.设,所以,因此,又,所以.点睛:用错位相减法求和应注意的问题:(1)要善于识别题目类型,特别是等比数列公比为负数的情形;(2)在写出“”与“”的表达式时应特别注意将两式“错项对齐”以便下一步准确写出“”的表达式;(3)在应用错位相减法求和时,若等比数列的公比为参数,应分公比等于1和不等于1两种情况求解.【命题意图】1.考查等差数列、等比数列的通项公式及求和公式;2.考查数列的求和方法;3.考查运算求解能力、转化与化归思想以及分析问题解决问题的能力.【命题规律】数列是高考重点考查的内容之一,命题形式多种多样,大小均有.其中,小题重点考查等差数列、等比数列基础知识以及数列的递推关系,和其它知识综合考查的趋势明显,小题难度加大趋势明显;解答题的难度中等或稍难,随着文理同卷的实施,数列与不等式综合热门难题(压轴题),有所降温,难度趋减,将稳定在中等变难程度.往往在解决数列基本问题后考查数列求和,在求和后往往与不等式、函数、最值等问题综合.在考查等差数列、等比数列的求和基础上,进一步考查“裂项相消法”、“错位相减法”等,与不等式结合,“放缩”思想及方法尤为重要.【答题模板】解答数列大题,一般考虑如下三步:第一步:确定数列的基本量.即根据通项公式、求和公式,通过布列方程或方程组,求得进一步解题所需的基本量;第二步:确定数列特征,选择求和方法.根据已有数据,研究送来的的特征,选择“分组求和法”“错位相减法”“裂项相消法”等求和方法;第三步:解答综合问题.根据题目要求,利用函数、导数、不等式等,进一步求解.【方法总结】1.公式法:如果一个数列是等差、等比数列或者是可以转化为等差、等比数列的数列,我们可以运用等差、等比数列的前n项和的公式来求和.对于一些特殊的数列(正整数数列、正整数的平方和立方数列等)也可以直接使用公式求和.2.倒序相加法:类似于等差数列的前n项和的公式的推导方法,如果一个数列{}n a的前n 项中首末两端等“距离”的两项的和相等或等于同一个常数,那么求这个数列的前n项和即可用倒序相加法,如等差数列的前n项和公式即是用此法推导的.3.错位相减法:如果一个数列的各项是由一个等差数列和一个等比数列的对应项之积构成的,那么这个数列的前n 项和即可用此法来求,如等比数列的前n 项和公式就是用此法推导的.若n n n a b c =∙,其中{}n b 是等差数列,{}n c 是公比为q 等比数列,令112211n n n n n S b c b c b c b c --=++++,则n qS =122311n n n n b c b c b c b c -+++++两式错位相减并整理即得.4.裂项相消法:把数列的通项拆成两项之差,即数列的每一项都可按此法拆成两项之差,在求和时一些正负项相互抵消,于是前n 项的和变成首尾若干少数项之和,这一求和方法称为裂项相消法.适用于类似1n n c a a +⎧⎫⎨⎬⎩⎭(其中{}n a 是各项不为零的等差数列,c 为常数)的数列、部分无理数列等.用裂项相消法求和,需要掌握一些常见的裂项方法: (1)()1111n n k k n n k ⎛⎫=- ⎪++⎝⎭,特别地当1k =时,()11111n n n n =-++; (21k=,特别地当1k ==(3)()()221111212122121n n a n n n n ⎛⎫==+- ⎪-+-+⎝⎭(4)()()()()()1111122112n a n n n n n n n ⎛⎫==- ⎪ ⎪+++++⎝⎭(5))()11(11q p qp p q pq <--= 5.分组转化求和法:有一类数列{}n n a b +,它既不是等差数列,也不是等比数列,但是数列{},{}n n a b 是等差数列或等比数列或常见特殊数列,则可以将这类数列适当拆开,可分为几个等差、等比数列或常见的特殊数列,然后分别求和,再将其合并即可.6.并项求和法:一个数列的前n 项和中,可两两结合求解,则称之为并项求和.形如()()1nn a f n =-类型,可采用两项合并求解.例如,22222210099989721n S =-+-++-()()()100999897215050=++++++=.7. [特别提醒]:在利用裂项相消法求和时应注意:(1)在把通项裂开后,是否恰好等于相应的两项之差;(2)在正负项抵消后,是否只剩下了第一项和最后一项,或有时前面剩下两项,后面也剩下两项.(3)裂项过程中易忽视常数,如)211(21)2(1+-=+n n n n 容易误裂为112n n -+,漏掉前面的系数12; (4)裂项之后相消的过程中容易出现丢项或添项的问题,导致计算结果错误. 8. [特别提醒]:用错位相减法求和时,应注意(1)要善于识别题目类型,特别是等比数列公比为负数的情形;(2)在写出“n S ”与“n qS ”的表达式时应特别注意将两式“错项对齐”以便下一步准确写出“n n S qS -”的表达式.(3)给数列和S n 的等式两边所乘的常数应不为零,否则需讨论;(4)在转化为等比数列的和后,求其和时需看准项数,不一定为n .一、选择题1.【上海市虹口区2019届高三二模】已知等比数列的首项为2,公比为,其前项和记为,若对任意的,均有恒成立,则的最小值为( )A .B .C .D .【答案】B 【解析】S n•,①n 为奇数时,S n •,可知:S n 单调递减,且•,∴S n ≤S 1=2; ②n 为偶数时,S n•,可知:S n 单调递增,且•,∴S 2≤S n.∴S n 的最大值与最小值分别为:2,. 考虑到函数y =3t在(0,+∞)上单调递增,∴A .B .∴B﹣A的最小值.故选:B.2.【浙江省三校2019年5月份第二次联考】已知数列满足,若存在实数,使单调递增,则的取值范围是()A.B.C.D.【答案】A【解析】由单调递增,可得,由,可得,所以.时,可得.①时,可得,即.②若,②式不成立,不合题意;若,②式等价为,与①式矛盾,不合题意.排除B,C,D,故选A.3.【浙江省2019年高考模拟训练卷(三)】已知数列满足,,,数列满足,,,若存在正整数,使得,则()A. B. C. D.【答案】D【解析】因为,,则有,,且函数在上单调递增,故有,得,同理有,又因为,故,所以.故选D.4.【广东省韶关市2019届高考模拟测试(4月)】已知数列{}n a 满足2*123111()23n a a a a n n n N n ++++=+∈,设数列{}n b 满足:121n n n n b a a ++=,数列{}n b 的前n 项和为n T ,若*()1n n N T n nλ<∈+恒成立,则实数λ的取值范围为( ) A .1[,)4+∞ B .1(,)4+∞ C .3[,)8+∞ D .3(,)8+∞【答案】D 【解析】数列{}n a 满足212311123n a a a a n n n ++++=+,① 当2n ≥时,21231111(1)(1)231n a a a a n n n -+++⋯+=-+--,② ①﹣②得:12n a n n=,故:22n a n =,数列{}n b 满足:22121214(1)n n n n n b a a n n +++==+221114(1)n n ⎡⎤=-⎢⎥+⎣⎦, 则:2222211111114223(1)n T n n ⎡⎤⎛⎫⎛⎫⎛⎫=-+-++-⎢⎥ ⎪ ⎪ ⎪+⎝⎭⎝⎭⎝⎭⎢⎥⎣⎦21114(1)n ⎛⎫=- ⎪+⎝⎭, 由于*()1n n N T n nλ<∈+恒成立, 故:21114(1)1n n n λ⎛⎫-< ⎪++⎝⎭, 整理得:244n n λ+>+,因为211(1)4441n y n n +==+++在*n N ∈上单调递减,故当1n =时,max213448n n +⎛⎫= ⎪+⎝⎭ 所以38λ>. 故选:D .5.【浙江省温州市2019届高三2月高考适应】已知数列{} 满足0<<<π,且,则( )A .B .C .D .【答案】A 【解析】 由,取特殊值:,,得:=,=,排除C 、D ;==,=>;且,,均小于,猜测,下面由图说明:当时,由迭代蛛网图:当时,由迭代蛛网图:可得,当n分别为奇数、偶数时,单调递增,且都趋向于不动点,由图像得,综上可得,故选A.6.【浙江省湖州三校2019年普通高等学校招生全国统一考试】已知数列满足,,则使的正整数的最小值是()A.2018 B.2019 C.2020 D.2021【答案】C【解析】令,则,所以,从而,因为,所以数列单调递增,设当时, 当时,所以当时,,,从而,因此,选C.二、解答题7.【天津市部分区2019年高三质量调查试题(二)】各项均为正数的等比数列满足,.(1)求数列的通项公式;(2)设,数列的前项和为,证明:.【答案】(1) (2)见证明【解析】解:(1)设等比数列的公比为,由得,解得或.因为数列为正项数列,所以,所以,首项,故其通项公式为.(2)由(Ⅰ)得所以,所以.8.【浙江省浙南名校联盟2019届高三上学期期末】已知等比数列的公比,前项和为.若,且是与的等差中项.(I)求;(II)设数列满足,,数列的前项和为.求证:.【答案】(Ⅰ)(II)见证明【解析】(I)由,得①.再由是,的等差中项,得,即②.由①②,得,即,亦即,解得或,又,故.代入①,得,所以,即;(II)证明:对任意,,,即.又,若规定,则.于是,从而,即.8.9.【浙江省嘉兴市2019届高三上期末】在数列、中,设是数列的前项和,已知,,,.(Ⅰ)求和;(Ⅱ)若时,恒成立,求整数的最小值.【答案】(1),(2)整数的最小值是11.【解析】 (Ⅰ)因为,即,所以是等差数列,又,所以,从而.(Ⅱ)因为,所以,当时,①②①-②可得,,即,而也满足,故. 令,则,即,因为,,依据指数增长性质,整数的最小值是11.10.【河南省濮阳市2019届高三5月模拟】已知数列}{n b 的前n 项和为n S ,2n n S b +=,等差数列}{n a 满足123b a =,157b a += (Ⅰ)求数列{}n a ,{}n b 的通项公式; (Ⅱ)证明:122313n n a b a b a b ++++<.【答案】(Ⅰ)1n a n =+,112n n b -⎛⎫= ⎪⎝⎭;(Ⅱ)详见解析.【解析】 (Ⅰ)2n n S b += ∴当1n =时,1112b S b ==- 11b ∴=当2n ≥时,1122n n n n n b S S b b --=-=--+,整理得:112n n b b -=∴数列{}n b 是以1为首项,12为公比的等比数列 112n n b -⎛⎫∴= ⎪⎝⎭设等差数列{}n a 的公差为d123b a =,157b a += 11346a d a d +=⎧∴⎨+=⎩,解得:121a d =⎧⎨=⎩()()112111n a a n d n n ∴=+-=+-⨯=+(Ⅱ)证明:设()212231111231222nn n n T a b a b a b n -⎛⎫⎛⎫=++⋅⋅⋅+=⨯+⨯+⋅⋅⋅++⋅ ⎪ ⎪⎝⎭⎝⎭()23111112312222n n T n +⎛⎫⎛⎫⎛⎫∴=⨯+⨯+⋅⋅⋅++⋅ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭两式相减可得:()()23111111111111421111122222212n n n n n T n n ++-⎛⎫- ⎪⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫⎝⎭=+++⋅⋅⋅+-+⋅=-+⋅+⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭-13322n n ++=- 332n n n T +=-即12231332n n nn a b a b a b -+++⋅⋅⋅+=-302n n +> 122313n n a b a b a b -∴++⋅⋅⋅+< 11.【浙江省金华十校2019届下学期高考模拟】已知数列{}n a 中,14a =,n a >,1314n n n n a a a a +=-+,记22212111...n nT a a a =+++. (1)证明:2n a >;(2)证明:115116n na a +≤<; (3)证明:8454n n n T -<<. 【答案】(1)见解析;(2)见解析;(2)见解析 【解析】(1)∵3133(2)(2)1422n n n n n n n na a a a a a a a +---=-+-=-,∴31323221212n n n n n n na a a a a a a +---==---,令1n t a =,则2312()122n n a m t t t a +-==---,∵n a >t ∈,∴'2()260m t t t =--<,∴()m t在单调递减,∴16()()10339m t m ->=-=>,即n a 时,1202n n a a +->-恒成立, ∴12n a +-与2n a -同号,又1220a -=>.∴2n a >成立.(2)2124214111514816n n n n n a a a a a +⎛⎫=-+=-+ ⎪⎝⎭221115412816⎛⎫<-+= ⎪⎝⎭,又212111515481616n n n a a a +⎛⎫=-+ ⎪⎝⎭…,∴115116n n a a +≤<. (3)先证4n nT <,因为2n a >,所以2114n a <,所以222121111...44n n n T n a a a =+++<⋅=,再证845n n T >-,∵1314n n n na a a a +=-+,∴()121144n n n n a a a a +-=+, 又21232141115151481616n n n n n a a a a a +⎛⎫=-+=-+> ⎪⎝⎭,∴11615n n a a +>,∴116()31n n n a a a +<+,又10n n a a +-<,∴2211()4()431n n n n n a a a a a ++->-,所以221222121114...()314n n n n n T a a a a a +=+++>-+4488(416)31443145n n n >-+=->-, 故8454n n n T -<<. 12.【浙北四校2019届高三12月模拟】已知数列满足,().(Ⅰ)证明数列为等差数列,并求的通项公式;(Ⅱ)设数列的前项和为,若数列满足,且对任意的恒成立,求的最小值.【答案】(Ⅰ)证明见解析,;(Ⅱ).【解析】∵(n+1)a n+1﹣(n+2)a n=2,∴﹣==2(﹣),又∵=1,∴当n≥2时,=+(﹣)+(﹣)+…+(﹣)=1+2(﹣+﹣+…+﹣)=,又∵=1满足上式,∴=,即a n=2n,∴数列{a n}是首项、公差均为2的等差数列;(Ⅱ)解:由(I)可知==n+1,∴b n=n•=n•,令f(x)=x•,则f′(x)=+x••ln,令f′(x)=0,即1+x•ln=0,解得:x0≈4.95,则f(x)在(0, x0)上单调递增,在(x0,+单调递减.∴0<f(x)≤max{f(4),f(5),f(6)},又∵b5=5•=,b4=4•=﹣,b6=6•=﹣,∴M的最小值为.。
专题06 函数的定义域、值域--《2023年高考数学命题热点聚焦与扩展》【原卷版】
【热点聚焦】函数的定义域作为函数的要素之一,是研究函数的基础,函数的定义域问题也是高考的热点.函数的值域(最值)也是高考中的一个重要考点,并且值域(最值)问题通常会渗透在各类题目之中,成为解题过程的一部分.【重点知识回眸】1.函数的有关概念(1)函数的定义域、值域:在函数y=f(x),x∈A中,x叫做自变量,x的取值范围A叫做函数的定义域;与x的值相对应的y值叫做函数值,函数值的集合{f(x)|x∈A}叫做函数的值域.显然,值域是集合B的子集.(2)函数的三要素:定义域、值域和对应关系.(3)相等函数:如果两个函数的定义域和对应关系完全一致,则这两个函数相等,这是判断两函数相等的依据.(4)函数的表示法表示函数的常用方法有:解析法、图象法、列表法.提醒:两个函数的值域和对应关系相同,但两个函数不一定相同,例如,函数f(x)=|x|,x ∈[0,2]与函数f(x)=|x|,x∈[-2,0].2.分段函数若函数在其定义域内,对于定义域内的不同取值区间,有着不同的对应关系,这样的函数通常叫做分段函数.分段函数虽然由几部分组成,但它表示的是一个函数.提醒:分段函数是一个函数,而不是几个函数,分段函数的定义域是各段定义域的并集,值域是各段值域的并集.3.常见函数定义域的求法类型x满足的条件n f x(n∈N*)f(x)≥02()(n∈N*)f(x)有意义21()n f x1与[f(x)]0f(x)≠0f x()log a f(x)(a>0且a≠1)f(x)>0a f(x)(a>0且a≠1)f(x)有意义tan[f (x )]f (x )≠π2+k π,k ∈Z四则运算组成的函数 各个函数定义域的交集实际问题使实际问题有意义4.①若()y f x =的定义域为(),a b ,则不等式()a g x b <<的解集即为函数()()y f g x =的定义域;②若()()y f g x =的定义域为(),a b ,则函数()g x 在(),a b 上的的值域即为函数()y f x =的定义域.5.常见函数的值域:在处理常见函数的值域时,通常可以通过数形结合,利用函数图像将值域解出,熟练处理常见函数的值域也便于将复杂的解析式通过变形与换元向常见函数进行化归.(1)一次函数(y kx b =+):一次函数为单调函数,图像为一条直线,所以可利用边界点来确定值域.(2)二次函数(2y ax bx c =++),给定区间.二次函数的图像为抛物线,通常可进行配方确定函数的对称轴,然后利用图像进行求解.(关键点:①抛物线开口方向,②顶点是否在区间内).(3)反比例函数:1y x=(1)图像关于原点中心对称(2)当,0x y →+∞→ ,当,0x y →-∞→. (4)对勾函数:()0ay x a x=+> ① 解析式特点:x 的系数为1;0a >注:因为此类函数的值域与a 相关,求a 的值时要先保证x 的系数为1,再去确定a 的值 例:42y x x =+,并不能直接确定4a =,而是先要变形为22y x x ⎛⎫=+ ⎪⎝⎭,再求得2a =② 极值点:,x a x a ==③ 极值点坐标:(,2,,2a a a a --④ 定义域:()(),00,-∞+∞⑤ 自然定义域下的值域:(),22,a a ⎡-∞-+∞⎣(5)函数:()0ay x a x=-> 注意与对勾函数进行对比① 解析式特点:x 的系数为1;0a > ② 函数的零点:x a =③ 值域:R(5)指数函数(xy a =):其函数图像分为1a >与01a <<两种情况,可根据图像求得值域,在自然定义域下的值域为()0,+∞(6)对数函数(log a y x =)其函数图像分为1a >与01a <<两种情况,可根据图像求得值域,在自然定义域下的值域为()0,+∞(7)三角函数的有界性,如sin [1,1],x ∈-cos [1,1]x ∈-. 6.函数值域问题处理策略 (1)换元法:① ()()(),log ,sin f x a y ay f x y f x ===⎡⎤⎡⎤⎣⎦⎣⎦:此类问题在求值域时可先确定()f x 的范围,再求出函数的范围.② ()()(),log ,sin x a y f a y f x y f x ===:此类函数可利用换元将解析式转为()y f t =的形式,然后求值域即可.③形如y ax b cx d =++(2)均值不等式法:特别注意“一正、二定、三相等”.(3)判别式法:若原函数的定义域不是实数集时,应结合函数的定义域,将扩大的部分剔除.(4)分离常数法:一般地, ① ax by cx d+=+:换元→分离常数→反比例函数模型② 2ax bx c y dx e ++=+:换元→分离常数→ay x x=±模型③ 2dx ey ax bx c+=++:同时除以分子:21y ax bx c dx e=+++→②的模型 ④ 22ax bx cy dx ex f++=++:分离常数→③的模型(5)单调性性质法:利用函数的单调性(6)导数法:利用导数与函数的连续性求图复杂函数的极值和最值, 然后求出值域 (7)数形结合法【典型考题解析】热点一已知函数解析式求定义域【典例1】(广东·高考真题(文))函数f (x )=11x-+lg(1+x )的定义域是( ) A .(-∞,-1)B .(1,+∞)C .(-1,1)∪(1,+∞)D .(-∞,+∞) 【典例2】(山东·高考真题(文))函数21()4ln(1)f x x x =-+( )A .[-2,0)∪(0,2]B .(-1,0)∪(0,2]C .[-2,2]D .(-1,2]【典例3】(2019·江苏·高考真题)函数276y x x =+-_____. 【典例4】(2022·北京·高考真题)函数1()1f x x x=-_________. 【总结提升】已知函数的具体解析式求定义域的方法(1)简单函数的定义域:若f(x)是由一些基本初等函数通过四则运算构成的,则它的定义域为各基本初等函数的定义域的交集.(2)复合函数的定义域:先由外层函数的定义域确定内层函数的值域,从而确定对应的内层函数自变量的取值范围,还需要确定内层函数的定义域,两者取交集即可. 热点二 求抽象函数的定义域【典例5】(全国·高考真题(理))已知()f x 的定义域为(1,0)-,则函数(21)f x +的定义域为 ( ) A .(1,1)-B .1(1,)2--C .(1,0)-D .1(,1)2【典例6】(2023·全国·高三专题练习)已知函数()31f x +的定义域为[]1,7,求函数()f x 的定义域.【典例7】(2022·全国·高三专题练习)已知函数(1)y f x +=的定义域为112⎡⎤-⎢⎥⎣⎦,,则函数2(log )y f x =的定义域为( )A .(0,)+∞B .(0,1)C .222⎤⎢⎥⎣⎦D .2⎡⎤⎣⎦,【总结提升】(1)若已知函数f (x )的定义域为[a ,b ],则复合函数f (g (x ))的定义域由a ≤g (x )≤b 求出. (2)若已知函数f (g (x ))的定义域为[a ,b ],则f (x )的定义域为g (x )在x ∈[a ,b ]时的值域. 热点三 求函数的值域(最值)【典例8】(江西·高考真题(理))若函数()y f x =的值域是1[,3]2,则函数1()()()F x f x f x =+的值域是( )A .1[,3]2B .10[2,]3 C .510[,]23D .10[3,]3【典例9】(2023·全国·高三专题练习)已知函数()y f x =的定义域是R ,值域为[]1,2,则下列四个函数①()21y f x =-;①()21y f x =-;①()12f x y -=;①()2log 11y f x =++,其中值域也为[]1,2的函数个数是( ) A .4B .3C .2D .1【典例10】(2023·全国·高三专题练习)已知函数2()(2)sin(1)1xf x x x x x =--+-在[1,1)-(1,3]⋃上的最大值为M ,最小值为N ,则M N +=( )A .1B .2C .3D .4【典例11】(2022·河南·郑州四中高三阶段练习(文))高斯是德国著名的数学家,近代数学奠基者之一,享有“数学王子”的美誉,用其名字命名的“高斯函数”:设x ∈R ,用[]x 表示不超过x 的最大整数,则[]y x =称为高斯函数,也称取整函数,例如:[]1.32-=-,[]3.43=,已知()11313xf x =-+,则函数()y f x ⎡⎤=⎣⎦的值域为______. 【典例12】(2023·全国·高三专题练习)函数()21f x x x =+-________;函数24y x x =-________.【典例13】(2023·河南·洛宁县第一高级中学一模(文))已知函数()211122f x x x =++. (1)求()f x 的图像在点()()22f ,处的切线方程; (2)求()f x 在1,22⎡⎤⎢⎥⎣⎦上的值域.热点四 求参数的值或取值范围【典例14】(2023·全国·高三专题练习)设a R ∈,函数()2229,1163,1x ax x f x x a x x ⎧-+≤⎪=⎨+->⎪⎩,若()f x 的最小值为()1f ,则实数a 的取值范围为( ) A .[]1,2B .[]1,3C .[]0,2D .[]2,3【典例15】(2022·全国·高三专题练习)已知函数()221f x ax x =++R ,则实数a 的取值范围是__.【典例16】(2016·北京·高考真题(理))设函数33,(){2,x x x a f x x x a -≤=->. ①若0a =,则()f x 的最大值为____________________; ②若()f x 无最大值,则实数a 的取值范围是_________________.【精选精练】1.(2023·全国·高三专题练习)若集合-1|2M x y x ==⎧⎨⎩,{}2|N y y x -==,则( )A .M N ⋂=∅B .M N ⊆C .N M ⊆D .M =N2.(2022·全国·高三专题练习)下列函数中,其定义域和值域分别与函数lg 10x y =的定义域和值域相同的是( ) A .y =xB .y =lg xC .y =2xD .y x3.(2022·全国·高三专题练习)若函数()21f x ax ax =-+R ,则a 的范围是( ) A .()0,4 B .[)0,4 C .(]0,4D .[]0,44.(2023·全国·高三专题练习)已知函数()f x 的定义域为[]0,1,值域为[]1,2,那么函数()2f x +的定义域和值域分别是( )A .[]0,1,[]1,2B .[]2,3,[]3,4C .[]2,1--,[]1,2D .[]1,2-,[]3,45.(2022·江西·高三阶段练习(文))函数()s 2π2inx f x x =+在[0,1]上的值域为( ) A .[1,2] B .[1,3] C .[2,3] D .[2,4]6.(2022·全国·高三专题练习)已知(12)3,1()ln ,1a x a x f x x x -+<⎧=⎨≥⎩的值域为R ,那么a 的取值范围是( ) A .(﹣∞,﹣1]B .(﹣1,12)C .[﹣1,12)D .(0,1)7.(2023·全国·高三专题练习)函数f (x 2sin 12x π- )A .54,433k k πππ⎡⎤++⎢⎥⎣⎦(k ∈Z ) B .154,433k k ⎡⎤++⎢⎥⎣⎦(k ∈Z )C .54,466k k πππ⎡⎤++⎢⎥⎣⎦ (k ∈Z ) D .154,466k k ⎡⎤++⎢⎥⎣⎦(k ∈Z )8.(2023·山西大同·高三阶段练习)函数6()e 1||1x mx f x x =+++的最大值为M ,最小值为N ,则M N +=( ) A .3B .4C .6D .与m 值有关9.(2022·江苏南京·高三开学考试)已知函数()()()()5sin sin ,99f x x x g x f f x ππ⎛⎫⎛⎫=++-= ⎪ ⎪⎝⎭⎝⎭,则()g x 的最大值为( )A 2B 3C .32D .210.(2022·广东·石门高级中学高二阶段练习)函数()12cos f x x x x =+-的最小值为( ) A .1ππ B .22ππC .-1D .0二、多选题11.(2023·全国·高三专题练习)已知函数122()log (2)log (4)f x x x =--+,则下列结论中正确的是( )A .函数()f x 的定义域是[4,2]-B .函数(1)=-y f x 是偶函数C .函数()f x 在区间[1,2)-上是减函数D .函数()f x 的图象关于直线1x =-对称 三、双空题12.(2023·全国·高三专题练习)已知函数()ln ,1e 2,1xx b x f x x +>⎧=⎨-≤⎩,若(e)3(0)f f =-,则b =_____,函数()f x 的值域为____.13.(2023·全国·高三专题练习)已知函数()121xf x a =+-为奇函数,则实数a =__,函数f (x )在[1,3]上的值域为__. 四、填空题14.(2022·全国·高三专题练习)函数()02112y x x x =++-的定义域是________.15.(2022·上海闵行·二模)已知函数()()41log 42xf x m x =+-的定义域为R ,且对任意实数a ,都满足()()f a f a ≥-,则实数m =___________;16.(2022·上海市嘉定区第二中学模拟预测)已知函数()y f x =是定义域为R 的奇函数,且当0x <时,()1af x x x=++.若函数()y f x =在[)3,+∞上的最小值为3,则实数a 的值为________.17.(2022·北京·清华附中模拟预测)已知函数()()2ln ,1,1x a x f x x a x +≥⎧⎪=⎨+<⎪⎩,下列说法正确的是___________.①当0a ≥时,()f x 的值域为[0,)+∞; ②a ∀∈R ,()f x 有最小值;③R a ∃∈,()f x 在(0,)+∞上单调递增: ④若方程1f x有唯一解,则a 的取值范围是(,2)-∞-.18.(2022·全国·高三专题练习)已知函数f (x )()221mx m x m =--+-的值域是[0,+∞),则实数m 的取值范围是__.。
专题06动量冲量及动量定理和动量守恒(讲义)(原卷版)
专题06 动量、冲量及动量定理和动量守恒01专题网络·思维脑图 02考情分析·解密高考 03高频考点·以考定法 04核心素养·难点突破 05创新好题·轻松练考点内容 学习目标动量与动量定理 1.掌握动量的含义及动量的计算;2.掌握冲量的含义及计算;3.掌握动量守恒定律的条件及基础计算,如爆炸、反冲及人船模型的应用;4.掌握两类碰撞模型及应用冲量及计算 动量守恒定律 爆炸、反冲及人船模型碰撞模型一、动量与动量定理及应用1.动量与动能的比较234.流体作用的柱状模型:对于流体运动,可沿流速v 的方向选取一段柱形流体,设在极短的时间Δt 内通过某一截面积为S 的横截面的柱形流体的长度为Δl ,如图所示.设流体的密度为ρ,则在Δt 的时间内流过该横截面的流体的质量为Δm =ρS Δl =ρSv Δt ,根据动量定理,流体微元所受的合外力的冲量等于该流体微元动量的变化量,即F Δt =Δm Δv ,分两种情况:(以原来流速v 的方向为正方向)① 作用后流体微元停止,有Δv =-v ,代入上式有F =-ρSv 2;② 作用后流体微元以速率v 反弹,有Δv =-2v ,代入上式有F =-2ρSv 2. 二、动量守恒定律及应用 1.判断守恒的三种方法① 理想守恒:不受外力或所受外力的合力为0,如光滑水平面上的板-块模型、电磁感应中光滑导轨上的双杆模型.② 近似守恒:系统内力远大于外力,如爆炸、反冲.③ 某一方向守恒:系统在某一方向上不受外力或所受外力的合力为0,则在该方向上动量守恒,如滑块-斜面(曲面)模型.2.动量守恒定律的三种表达形式① m 1v 1+m 2v 2=m 1v 1′+m 2v 2′,作用前的动量之和等于作用后的动量之和(常用). ② Δp 1=-Δp 2,相互作用的两个物体动量的增量等大反向. ③ Δp =0,系统总动量的增量为零. 三、碰撞模型及拓展1.碰撞问题遵循的三条原则 ① 动量守恒:p 1+p 2=p 1′+p 2′. ② 动能不增加:E k1+E k2≥E k1′+E k2′.③ 速度要符合实际情况:若碰后同向,后方物体速度不大于前方物体速度. 2.弹性碰撞:两球发生弹性碰撞时应满足动量守恒定律和机械能守恒定律.以质量为m 1、速度为v 1的小球与质量为m 2的静止小球发生弹性正碰为例,有m 1v 1=m 1v 1′+m 2v 2′ 12m 1v 12=12m 1v 1′2+12m 2v 2′2 解得v 1′=m 1-m 2v 1m 1+m 2,v 2′=2m 1v 1m 1+m 2.结论:① 当m 1=m 2时,v 1′=0,v 2′=v 1,两球碰撞后交换了速度.② 当m 1>m 2时,v 1′>0,v 2′>0,碰撞后两球都沿速度v 1的方向运动. ③ 当m 1<m 2时,v 1′<0,v 2′>0,碰撞后质量小的球被反弹回来. ④ 当m 1≫m 2时,v 1′=v 1,v 2′=2v 1.3.完全非弹性碰撞:动量守恒、末速度相同:m 1v 1+m 2v 2=(m 1+m 2)v 共,机械能损失最多,机械能的损失:ΔE =12m 1v 12+12m 2v 22-12(m 1+m 2)v 共2. 4.碰撞拓展① “保守型”碰撞拓展模型图例(水平面光滑)小球-弹簧模型小球-曲面模型相当于完全非弹性碰撞,动量满足mv=(m+M)v,损失的动能最大,分别转化为考向一:动量与动量定理【探究重点】1.研究对象可以是单一物体,也可以是物体系统.2.表达式是矢量式,需要规定正方向.3.匀变速直线运动,如果题目不涉及加速度和位移,用动量定理比用牛顿第二定律求解更简捷.4.在变加速运动中F为Δt时间内的平均冲力.5.电磁感应问题中,利用动量定理可以求解时间、电荷量或导体棒的位移.【高考解密】1.(2022·湖北卷·7)一质点做曲线运动,在前一段时间内速度大小由v增大到2v,在随后的一段时间内速度大小由2v 增大到5v.前后两段时间内,合外力对质点做功分别为W1和W2,合外力的冲量大小分别为I1和I2.下列关系式一定成立的是()A.W2=3W1,I2≤3I1B.W2=3W1,I2≥I1C.W2=7W1,I2≤3I1D.W2=7W1,I2≥I1【考向预测】2.(2022·湖南衡阳市一模)飞船在进行星际飞行时,使用离子发动机作为动力,这种发动机工作时,由电极发射的电子射入稀有气体(如氙气),使气体离子化,电离后形成的离子由静止开始在电场中加速并从飞船尾部高速连续喷出,利用反冲使飞船本身得到加速.已知一个氙离子质量为m,电荷量为q,加速电压为U,飞船单位时间内向后喷射出的氙离子的个数为N,从飞船尾部高速连续喷出氙离子的质量远小于飞船的质量,则飞船获得的反冲推力大小为()A.1N2qUm B.1NqUm2C.N2qUm D.N qUm 2考向二:冲量及计算【探究重点】1.对动量定理的理解①动量定理中的冲量是合力的冲量,而不是某一个力的冲量,它可以是合力的冲量,可以是各力冲量的矢量和,也可以是外力在不同阶段冲量的矢量和.②Ft=p′-p是矢量式,两边不仅大小相等,而且方向相同.式中Ft是物体所受的合外力的冲量.③ Ft =p ′-p 除表明两边大小、方向的关系外,还说明了两边的因果关系,即合外力的冲量是动量变化的原因.④ 由Ft =p ′-p ,得F =p ′-p t =Δpt,即物体所受的合外力等于物体动量的变化率.⑤ 当物体运动包含多个不同过程时,可分段应用动量定理求解,也可以全过程应用动量定理. 2.解题基本思路 【高考解密】3. (2022·海南)在冰上接力比赛时,甲推乙的作用力是1F ,乙对甲的作用力是2F ,则这两个力( )A .大小相等,方向相反B .大小相等,方向相同C .1F 的冲量大于2F 的冲量D .1F 的冲量小于2F 的冲量【考向预测】4. (2023·黑龙江八校高三模拟)如图所示,表面光滑的楔形物块ABC 固定在水平地面上,≫ABC <≫ACB ,质量相同的物块a 和b 分别从斜面顶端沿AB 、AC 由静止自由滑下.在两物块到达斜面底端的过程中,下列说法正确的是( ) A .两物块所受重力冲量相同 B .两物块的动量改变量相同 C .两物块的动能改变量相同D .两物块到达斜面底端时重力的瞬时功率相同 考向三:动量守恒 【探究重点】动量守恒定律的三种表达形式① m 1v 1+m 2v 2=m 1v 1′+m 2v 2′,作用前的动量之和等于作用后的动量之和(用的最多). ② Δp 1=-Δp 2,相互作用的两个物体动量的增量等大反向. ③ Δp =0,系统总动量的增量为零.【高考解密】5. (2020·全国卷≫·21)水平冰面上有一固定的竖直挡板,一滑冰运动员面对挡板静止在冰面上,他把一质量为4.0 kg 的静止物块以大小为5.0 m/s 的速度沿与挡板垂直的方向推向挡板,运动员获得退行速度;物块与挡板弹性碰撞,速度反向,追上运动员时,运动员又把物块推向挡板,使其再一次以大小为5.0 m/s 的速度与挡板弹性碰撞.总共经过8次这样推物块后,运动员退行速度的大小大于5.0 m/s ,反弹的物块不能再追上运动员.不计冰面的摩擦力,该运动员的质量不可能为( )A .48 kgB .53 kgC .55 kgD .58 kg【考向预测】6. (2023·江苏南京市第十三中学高三检测)如图所示,光滑水平面上静止着一长为L 的平板车,一人站在车的右端,他将一质量为m 的小球水平抛出,抛出点位于车右端点的正上方h 处,小球恰好落在车的左端点.已知人和车的总质量为M ,不计空气阻力,重力加速度为g .则( ) A .平板车、人和小球组成的系统动量守恒 B .小球的初速度大小为L2h2ghC .小球落到车上后与车共速,速度方向向左D .抛出小球的过程人做功为mgML 24h M +m考向四:爆炸、反冲运动及人船模型 【探究重点】1.爆炸现象的三个规律动量守恒爆炸物体间的相互作用力远远大于受到的外力,所以在爆炸过程中,系统的总动量守恒动能增加在爆炸过程中,有其他形式的能量(如化学能)转化为机械能,所以系统的机械能增加位置不变爆炸的时间极短,因而作用过程中物体产生的位移很小,可以认为爆炸后各部分仍然从爆炸前的位置以新的动量开始运动2作用原理 反冲运动是系统内两物体之间的作用力和反作用力产生的效果 动量守恒反冲运动中系统不受外力或内力远大于外力,所以反冲运动遵循动量守恒定律机械能增加反冲运动中,由于有其他形式的能转化为机械能,所以系统的总机械能增加3① 两物体满足动量守恒定律:mv 人-Mv 船=0② 两物体的位移大小满足:m x 人t -M x 船t =0,x 人+x 船=L ,得x 人=M M +m L ,x 船=mM +m L③ 人动则船动,人静则船静,人快船快,人慢船慢,人左船右;④ 人船位移比等于它们质量的反比;人船平均速度(瞬时速度)比等于它们质量的反比,即x 人x 船=v 人v 船=Mm .【高考解密】7. (2023·河南省模拟)发射导弹过程可以简化为:将静止的质量为M (含燃料)的导弹点火升空,在极短时间内以相对地面的速度v 0竖直向下喷出质量为m 的炽热气体,忽略喷气过程中重力和空气阻力的影响,则喷气结束时导弹获得的速度大小是( ) A.mMv 0 B.M m v 0 C.M M -m v 0D.m M -m v 0【考向预测】8. (2023·江苏徐州市模拟)如图所示,有一质量M =6 kg 、棱长为0.2 m 的正方体木块,静止于光滑水平面上,木块内部有一从顶面贯通至底面的通道,一个质量为m =2 kg 的小球由静止开始从通道的左端运动到右端,在该过程中木块的位移大小为( ) A .0.05 m B .0.10 m C .0.15 m D .0.5 m考向五:碰撞模型 【探究重点】1.碰撞:碰撞是指物体间的相互作用持续时间很短,而物体间相互作用力很大的现象. 2.特点:在碰撞现象中,一般都满足内力远大于外力,可认为相互碰撞的系统动量守恒. 3.两类碰撞模型【高考解密】 9. (2022·北京)质量为1m 和2m 的两个物体在光滑水平面上正碰,其位置坐标x 随时间t 变化的图像如图所示。
专题06 化学反应与能量(练习)-2024年高考化学二轮复习讲练测(新教材新高考)(原
专题05 化学反应与能量目录01 反应热及其表示方法02 热化学方程式盖斯定律03 原电池化学电源04 电解原理05 金属的腐蚀与防护01 反应热及其表示方法1.(2024·河南省部分名校高三联考)丙烯是合成聚丙烯的单体。
在催化剂作用下,丙烷脱氢生成丙烯的能量变化如图所示。
下列叙述错误的是( )A.上述反应的正反应是吸热反应B.催化剂的作用是降低反应的活化能C.升温,该反应的平衡常数K 增大D.使用不同催化剂,该反应的△H不同2.某反应2A=3B,它的反应能量变化曲线如图所示,下列有关叙述正确的是()A .该反应为吸热反应B .A 比B 更稳定C .加入催化剂会改变反应的焓变D .整个反应的ΔH=E 1-E 23.(2024·山东潍坊市五县区高三阶段监测)化学反应H 2+Cl 2=2HCl 的过程及能量变化如图所示。
下列说法错误的是( )A .光照和点燃条件下的反应热相同B .H -Cl 键的键能为862 kJ·mol -1C .热化学方程式:H 2(g)+Cl 2(g)=2HCl(g) ΔH =-183 kJ·mol -1D .2mol 气态氢原子的能量高于1mol 氢气的能量4.化学反应中的能量变化,通常主要表现为热量的变化。
下列相关表述正确的是( )A .一定条件下,将0.5molH 2(g)和0.5molI 2(g)置于密闭容器中充分反应生成HI 放热akJ ,其热化学方程式为:I 2(g )+H 2(g)2HI(g) △H=-2akJ•mol -1B .在101kPa 时,2 g H 2完全燃烧生成液态水,放出285.8kJ 热量,表示氢气燃烧热的热化学方程式表示为:2221H (g)+O (g)=H O(1)ΔH=-285.8kJ/mol 2C .S(g)+ O 2(g)=SO 2(g) ΔH 1 S(g)+ O 2(g)=SO 2(g) ΔH 2 ΔH 1>ΔH 2D .HCl 和NaOH 反应的中和热ΔH=-57.3kJ/mol ,则0.5molH 2SO 4和足量Ba(OH)2反应的ΔH=-57.3kJ/mol 5.(2023·北京市东城区第一六六中学高三期中)以CO 和H 2为原料合成甲醇是工业上的成熟方法,直接以CO 2为原料生产甲醇是目前的研究热点。
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第 1 页 共 10 页 专题06 高三▪数学卷(理B )三角函数的图象与性质注意事项:1.答题前,先将自己的姓名、准考证号填写在试题卷和答题卡上,并将准考证号条形码粘贴在答题卡上的指定位置。
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一、选择题(本大题共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.在()02π,内,使tan 1x >成立的x 的取值范围为( ) A .ππ,42⎛⎫ ⎪⎝⎭B .5π3π,42⎛⎫ ⎪⎝⎭C .ππ5π3π,,4242⎛⎫⎛⎫ ⎪ ⎪⎝⎭⎝⎭D .ππ5π3π,,4242⎛⎫⎛⎫ ⎪ ⎪⎝⎭⎝⎭【答案】D【解析】结合正切函数tan y x =的图象,可得使tan 1x >成立的x 的取值范围πππ,π42k k ⎛⎫++ ⎪⎝⎭,k ∈Z .结合()02πx ∈,,可得在()02π,内,使tan 1x >成立的x 的取值范围为ππ5π3π,,4242⎛⎫⎛⎫⎪ ⎪⎝⎭⎝⎭,故选D .2.角α的终边过点()43P a a -,(0a ≠),则2sin cos αα+=( )A .25 B .25-C .25或25- D .与α的值有关【答案】C【解析】由题意得5r a ,根据正弦函数值、余弦函数值的定义,当0a >时,3sin 5α=,4cos 5α=-,则22sin cos 5αα+=;当0a <时,3sin 5α=-,4cos 5α=, 则22sin cos 5αα+=-,故选C .3.若θ是第三象限角,且coscos 22θθ=-,则2θ是( )此卷只装订不密封班级 姓名 准考证号 考场号座位号A .第一象限角B .第二象限角C .第三象限角D .第四象限角【答案】B【解析】∵θ是第三象限角,∴π(21)π(21)π2k k θ+<<++()k ∈Z , ∴π3πππ224k k θ+<<+()k ∈Z ,则2θ是第二或第四象限角,又∵cos cos 22θθ=-,∴cos 02θ<,∴2θ必为第二象限角,故选B . 4.已知函数()()πsin 03f x x ωω⎛⎫=+> ⎪⎝⎭的最小正周期为π,则该函数图象( ).A .关于点π,03⎛⎫⎪⎝⎭对称 B .关于直线π4x =对称,C .关于点π,04⎛⎫⎪⎝⎭对称D .关于直线π3x =对称, 【答案】A【解析】∵2=ω,∴π()sin(2)3f x x =+,∴π()03f =,故选A .5.设32sin ,2x ⎛⎫= ⎪⎝⎭a ,11,cos 64x ⎛⎫= ⎪⎝⎭b ,且∥a b ,则x tan 的值为( ). A .41B .21 C .1 D .2【答案】C【解析】∵∥a b ,∴06123)cos 41()sin 2(=⨯-⋅x x ,∴21cos sin =x x ,∴21cos sin cos sin 22=+x x x x ,∴211tan tan 2=+x x,解得1tan =x ,故选C . 6.已知函数()()2sin f x x ωϕ=+的图象如图所示,则=)0(f ( ). A .1-B .1C .2-D .2【答案】C【解析】由图象知最小正周期25ππ2π2π3443T ω⎛⎫=-== ⎪⎝⎭,故3ω=,又π4x =时,0)(=x f ,第 3 页 共 10 页即π2sin(3)04ϕ⨯+=,可得32ππ4k ϕ=-,所以()302sin 2ππ4f k ⎛⎫=-= ⎪⎝⎭C . 7.为了得到函数=y x x x cos sin 3sin 2+的图象,可以将函数x y 2sin =的图象( ).A .向左平移6π个单位长度,再向下平移21个单位长度B .向右平移6π个单位长度,再向上平移21个单位长度C .向左平移12π个单位长度,再向下平移21个单位长度D .向右平移12π个单位长度,再向上平移21个单位长度【答案】D 【解析】x x y 2sin 23)2cos 1(21+-=π1sin(2)62x =-+,故选D .8.已知sin(2π)2cos(π)αα-=-,则3333sin (π)5cos (4π)7π4cos ()3sin (7π)2αααα-+-=--+( ) A .38-B .32C .1356-D .356【答案】A【解析】∵sin(2π)2cos(π)αα-=-,∴sin(2π)2cos αα--=-,sin 2cos αα=-,则tan 2α=-,∴33333333sin (π)5cos (4π)sin 5cos 7ππ4cos ()3sin (7π)4cos ()3sin 22αααααααα-+-+=--+--+33333333sin 5cos sin 5cos 553114sin 3sin sin tan 88αααααααα++===--=--=--+--,故选A . 9.若1)sin()(-+=ϕωx A x f (0>ω),对任意实数t ,都有ππ()()33f t f t +=-+,记()cos()g x A x ωϕ=+,则π()3g 的值为( ). A .0 B .1-C .A -D .A【答案】A 【解析】由题意π3x =是1)sin()(-+=ϕωx A x f 的一条对称轴,∴πsin 13ωϕ⎛⎫+=-⎪⎝⎭或1,∴πcos 03ωϕ⎛⎫+=⎪⎝⎭,∴π()03g =,故选A . 10.已知函数)tan()(ϕω+=x A x f (0>ω,2πϕ<)的部分图象如图所示,则=)32(πf ( ).A .31-B .13-C .32--D .32-【答案】C【解析】根据图象得)883(2ππ-=T 2π=,∴2==Tπω,∵0)43tan()83(=+=ϕππA f ,∴πϕπk =+43,∴43ππϕ-=k (k ∈Z ). ∵2πϕ<,∴4πϕ=.又ϕtan )0(A f =14tan ==πA ,∴1=A . ∴)42tan()(π+=x x f .∴=)32(πf )434tan(ππ+)34tan(ππ+=3131-+=32--=.故选C . 11.函数)sin()(ϕω+=x x f (其中ω是正数)的图象向右平移16π个单位后对应一个偶函数,向左平移163π个单位后对应一个奇函数,则ω的最小值为( ). A .21 B .1C .2D .4【答案】C【解析】函数)(x f 最小正周期的最大值为π,此时ω最小,且ω的最小值为2.故选C .第 5 页 共 10 页12.已知函数x x f ωsin 2)(=在区间ππ[,]34-上递增,则正实数ω的最大值为( ).A .2B .23 C .43 D .32 【答案】B 【解析】令ππ22x ω-≤≤,∵0>ω,∴ππ22x ωω-≤≤,∴ππ[,]34-⊆ππ,22ωω⎡⎤-⎢⎥⎣⎦, ∴ππ42ππ32ωω⎧≤⎪⎪⎨⎪-≥-⎪⎩解得230≤<ω,故选B .二、填空题(本大题有4小题,每小题5分,共20分.请把答案填在题中横线上) 13.已知扇形的周长为20cm ,当扇形的面积最大时,扇形圆心角为弧度________. 【答案】2【解析】设扇形的弧长为l ,半径为R ,圆心角为α,由已知条件220l R +=, 得202l R =-,由02l R π<<,得02022R R π<-<,∴10101R π<<+; 扇形的面积为2211(202)10(5)2522S lR R R R R R ==-=-+=--+扇, 当5R =时,S 最大,此时10l =,2lRα==,故当扇形所对的圆心角为2时,扇形有最大面积.14.已知关于x 的方程2sin 2sin 0x x a ++=有解,则a 的取值范围是______.【答案】[3,1]-【解析】∵关于的方程2sin 2sin 0x x a ++=有解,∴存在x 使2sin 2sin a x x =--,而22sin 2sin 1(sin 1)x x x --=-+,且[]sin 1,1x ∈-,∴[]3,1a ∈-.15.若动直线x a =与函数x x f sin )(=和x x g cos )(=的图像分别交于N M ,两点,则MN 的最大值为_________.【答案】2【解析】x x MN cos sin -=)4sin(2π-=x ,∴MN 的最大值为2.16.已知函数πsin 6xy =的定义域为[]b a ,,值域为]1,21[-,则a b -的取值范围是_______. 【答案】[]8,4【解析】结合函数6sinxy π=的图象可知a b -的取值范围是[]8,4.三、解答题(本大题有6小题,共70分.解答应写出文字说明、证明过程或演算步骤) 17.(10分)已知向量1(2=a ,(cos ,sin )x x =b ,3π(π,)2x ∈;(1)若∥a b ,求sin x 和cos x 的值; (2)若12π13π2cos()()6k x k +⋅=+∈Z a b ,求5πtan()12x +的值. 【答案】(1)见解析;(2)3-. 【解析】(1)∵∥a b,1sin 2x x ∴=,于是sin x x ,∴3tan =x , 又3π(π,)2x ∈,∴4π3x =,∴sin x 4πsin 3==πsin 3-=; cos x 4πcos 3==π1cos 32-=-.(2)1πππcos cos sin sin cos sin()2666x x x x x ⋅==+=+a b ,而12π13πππ2cos()2cos(2π2)2cos()()666k x k x x k π++=+++=+∈Z , 于是ππsin()2cos()66x x +=+,即πtan()26x +=;ππtan()tan5πππ2164tan()tan[()]3ππ12641211tan()tan 64x x x x +++∴+=++===--⨯-+. 18.(12分)已知函数)(),0( )2sin()(x f y x x f =<<-+=ϕπϕ图象的一条对称轴是直线8π=x .第 7 页 共 10 页(1)求ϕ的值;(2)画出函数)(x f y =在区间],0[π上的图象. 【答案】(1)3π4ϕ=-;(2)见解析. 【解析】(1)π8x =是函数图象的对称轴,∴1)82sin(±=+⨯ϕπ, ∴πππ42k ϕ+=+,k ∈Z ,π0ϕ-<<,∴3π4ϕ=-. (2)由知)432sin(π-=x y :故函数)(x f y =在区间],0[π上的图象是:19.(12分)已知函数π()2sin(2)6f x a x b =-+的定义域为π[0,]2,函数的最大值为1,最小值为5-.(1)求a ,b 的值;(2)如何由()2sin 2g x a x =的图象得到函数()f x 的图象. 【答案】(1)见解析;(2)见解析. 【解析】(1)∵π02x ≤≤,∴ππ5π2666x -≤-≤,则1πsin(2)123x -≤-≤; 若0a >,则由题设知215a b a b +=⎧⎨-+=-⎩,解得2a =,3b =-;若0a <,则由题设知251a b a b +=-⎧⎨-+=⎩解得2a =-,1b =-.(2)当0a >时,由(1)知,π()4sin(2)36f x x =--,()4sin 2g x x =,∵π()4sin[2()]312f x x =--, ∴将函数()4sin 2g x x =的图象先向右平移π12个单位,再向下平移3个单位即可; 当0a <时,由(1)知,π()4sin(2)16f x x =---,()4sin 2g x x =-,∵π()4sin[2()]112f x x =---, ∴将函数()4sin 2g x x =-的图象先向右平移12π个单位,再向下平移1个单位即可. 20.(12分)已知函数2()2sin cos 2cos f x x x x =+(x ∈R ).(1)求)(x f 的最小正周期,并求)(x f 的最小值及取得最小值时x 的集合;(2)令π()()18g x f x =+-,若2)(-<a x g 对于ππ,63x ⎡⎤∈-⎢⎥⎣⎦恒成立,求实数a 的取值范围.【答案】(1)见解析;(2)22+>a .【解析】(1)π()sin 2cos21)14f x x x x =++++,其最小正周期是2ππ2T ==, 又当ππ22π42x k +=-+,即()38x k k π=π-∈Z 时, ∴函数)(x f 的最小值为21-. 此时x 的集合为()38x x k k ⎧π⎫=π-∈⎨⎬⎩⎭Z .(2)π()()18g x f x =+-ππ)44x x =++.由ππ,63x ⎡⎤∈-⎢⎥⎣⎦得π2π2,33x ⎡⎤∈-⎢⎥⎣⎦,则⎥⎦⎤⎢⎣⎡-∈1,212cos x ,第 9 页 共 10 页∴x x g 2cos 2)(=⎥⎦⎤⎢⎣⎡-∈2,22. 若2)(-<a x g 对于ππ,63x ⎡⎤∈-⎢⎥⎣⎦恒成立,则2)(2max =>-x g a ,∴22+>a .21.(12分)在某个旅游业为主的地区,每年各个月份从事旅游服务工作的人数会发生周期性的变化.现假设该地区每年各个月份从事旅游服务工作的人数()f n 可近似地用函数()2100cos π3f n A n k ω⎛⎫⎛⎫=++ ⎪ ⎪⎝⎭⎝⎭来刻画.其中正整数n 表示月份且[]1,12n ∈,例如1n =时表示1月份;A 和k 是正整数;0ω>.统计发现,该地区每年各个月份从事旅游服务工作的人数有以下规律: ①各年相同的月份,该地区从事旅游服务工作的人数基本相同;②该地区从事旅游服务工作的人数最多的8月份和最少的2月份相差约400人;③2月份该地区从事旅游服务工作的人数约为100人,随后逐月递增直到8月份达到最多. (1)试根据已知信息,确定一个符合条件的()f n 的表达式;(2)一般地,当该地区从事旅游服务工作的人数超过400人时,该地区也进入了一年中的旅游“旺季”.那么,一年中的哪几个月是该地区的旅游“旺季”?请说明理由. 【答案】(1)π2()200cos π30063f n n ⎛⎫=++ ⎪⎝⎭;(2)见解析.【解析】(1)根据三条规律,可知该函数为周期函数,且周期为12. 由此可得,2126T ππωω==⇒=;由规律②可知,max ()(8)100100f n f A k ==+, min ()(2)100100f n f A k ==-+,∴400200)2()8(==-A f f ,∴2=A .又当2n =时,()2100100100f A k =-+=,所以3k =. 综上可得,π2()200cos π30063f n n ⎛⎫=++ ⎪⎝⎭符合条件.(2)由题意,π2()200cos π30040063f n n ⎛⎫=++> ⎪⎝⎭,可得π21cos π632n ⎛⎫+> ⎪⎝⎭,∴ππ2π2ππ2π3633k n k -<+<+,k ∈Z , ∴126122k n k -<<-,k ∈Z .因为[]1,12n ∈,n *∈N ,所以当1k =时,610n <<,故7n =,8,9,即一年中的7,8,9四个月是该地区的旅游“旺季”.22.(12分)已知函数)12(cos )(,cos sin 1)(2π+=+=x x g x x x f .(1)设0x x =是函数)(x f y =的图象上一条对称轴,求)(0x g 的值. (2)若函数)0(),2()2()(>+=ωωωxg xf x h 在区间]3,32[ππ-上是增函数,求ω的最大值. 【答案】(1)43)(0=x g ;(2)12.【解析】(1)由已知,x x x f cos sin 1)(+=x 2sin 211+=, 0x x =是函数)(x f y =图象的一条对称轴,∴0π2π2x k =+()k ∈Z , ∴200π()cos ()12g x x =+01π[1cos(2)]26x =++12π[1cos(π)]23k =++, 当k 为偶数时,41)(0=x g ;当k 为奇数时,43)(0=x g . (2)x x h ωsin 211)(+=1π[1cos()]26x ω+++)sin 21(sin 21x x ωω-=x ωcos 43+23)3sin(2123++=+πωx , 当]33,332[3,]3,32[πωππωππωππ++-∈+-∈x x 时, ]3,32[)(ππ-∈x x h 在 上是增函数,且0>ω,21,21],2,2[]33,332[最大值为ωωπππωππωπ≤-⊆++-∴.。