2013普通高考预测题(含答案)

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2013年高考真题——文综(天津卷) Word版含答案

2013年高考真题——文综(天津卷) Word版含答案

绝密★启用前2013年普通高等学校招生全国统一考试(天津卷)文科综合政治部分试题1.党的十八大明确指出,到2020年实现城乡居民人均收入比2010年翻一番。

这一目标的提出,体现了“居民收入增长和经济发展同步”的原则,顺应了人民过上更好生活的新期盼。

实现居民收入倍增①要大力发展社会生产力②要着力调整国民收入分配格局③是全面建成小康社会的要求④是促进收入分配公平的根本措施A. ①②③B. ①②④C. ②③④D. ①③④2.在新一代移动互联网技术的浪潮下,全球许多名骚一时的大企业陷入困境;曾作为无线通信代名词的摩托罗拉被他人掌控;家电行业先驱索尼,松下和夏普巨霸亏损;原来全球最大的智能手机平台塞班系统,如今只剩下2.6%的市场份额。

这些案例表明A.拥有最先进的技术储备,是企业长盛不衰的重要保障B.缺乏国家必要的技术规划,是企业由盛而衰的主要原因C.缺乏用于技术创新的资金投入,是企业竞争失利的根本原因D.适应市场发展趋势的技术创新,是企业保持竞争力的关键因素3.下图是2004-2012年我国对外贸易指标,图中数据显示A.产品进口优化了国内的产业结构B.产品出口拉动了我国经济的增长C.我国国内市场竞争的国际化趋势增强D.我国“走出去”的步伐不断加快4.茶文化在我国有悠久的历史。

“东方美人”是台湾苗粟出产的一种名茶,它由当地客家人种植,是被小绿叶蝉咬过的乌龙茶叶。

很久以前,英国女王偶人品尝到此茶,将悬在茶杯里的茶叶赞叹为“正在跳舞的东方美女”,茶名由此而得,这种茶的曼妙之处还在于可用冰水浸泡,透过冰滴壶,一滴滴冰水滤过茶叶,茶汤香气馥郁,色若琥珀。

这表明A.文化对社会法阵有重要影响B. 人们在时间中创造并享用文化C. 文化能够增强人得精神力量D.人们的实践受到文化潜移默化的影响5.“虽然不言不语,叫人难忘记,那是你的眼,明亮有美丽……”当年蔡琴在演唱这《你的眼神》时,就曾以特有的韵味打动过无数听众,在今春“我是歌手”的电视节目上,林志炫以不同以往的方式演唱此歌,诠释经典,在次让人如痴如醉。

2013全国统一高考(浙江卷)理综试题及答案

2013全国统一高考(浙江卷)理综试题及答案

2013年普通高等学校招生全国统一考试浙江理科综合能力测试一、选择题(本题17小题.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.下列关于高等动植物连续分裂细胞的细胞周期的叙述,正确的是A.用蛋白质合成抑制剂处理G1期细胞,不影响其进入S期B.S期细胞的染色体数目已增加一倍C.G2其细胞的细胞核DNA含量已增加一倍D.用秋水仙素处理细胞群体,M期细胞的比例会减少2.某哺乳动物神经细胞内外的K+和Na+浓度见下表。

下列属于主动转运的是A.K+经钾离子通道排出细胞B.K+与有关载体蛋白结合排出细胞C.Na+经钠离子通道排出细胞D.Na+与有关载体蛋白结合排出细胞3.某生物基因表达过程如图所示.下列叙述与该图相符的是A.在RNA聚合酶作用下DNA双螺旋解开B.DNA—RNA杂交区域中A应与T配对C.mRNA翻译只能得到一条肽链D.该过程发生在真核细胞中4.下列关于出生率的叙述,正确的是A.若某一种群年初时的个体数为100,年末时为110,其中新生个体数为20,死亡个体数为10,则该种群的年出生率为10%B.若某动物的婚配制为一雌一雄,生殖期个体的雌雄比例越接近1:1,则出生率越高C.若通过调控环境条件,使某动物的性成熟推迟,则出生率会更高D.若比较三种年龄结构类型的种群,则稳定型的出生率最高5.光照、赤霉素和赤霉素合成抑制剂对某种植物茎伸长影响的实验结果如图所示.下列叙述正确的是A.茎伸长受抑制均由赤霉素合成抑制剂引起B.赤霉素是影响茎伸长的主要因素之一C.植物茎伸长与光照时间无关D.该植物是赤霉素缺失突变体6.实验小鼠皮肤细胞培养(非克隆培养)的基本过程如图所示。

下列叙述错误的是A.甲过程需要对实验小鼠进行消毒B.乙过程对皮肤组织可用胰蛋白酶消化C.丙过程得到的细胞大多具有异倍体核型D.丁过程得到的细胞具有异质性7.下列说法不正确的是A.多孔碳可用氢氧燃料电池的电极材料B.pH计不能用于酸碱中和滴定终点的判断ks5uC.科学家发现一种新细菌的DNA链中有砷(As)元素,该As元素最有可能取代了普通DNA链中的P元素D.CH3CH—CH2和CO2反应生成可降解聚合物[O—CHCH2—O—C ]n,该反应符合绿O CH3 O化学的原则8.下列说法正确的是ks5uA.实验室从海带提取单质碘的方法是:取样灼烧溶解过滤萃取B.用乙醇和浓H2SO4制备乙烯时,可用水浴加热控制反应的温度C.氯离子存在时,铝表面的氧化膜易被破坏,因此含盐腌制品不宜直接放在铝制容器中D .将(NH 4)2SO 4、CuSO 4溶液分别加入蛋白质溶液,都出现沉淀,表明二者均可使蛋白质变性 9.短周期元素X 、Y 、Z 、W 、Q 在元素周期表的位置如表所示,其中X 元素的原子内层电子数是最外层电子数的一半,则下列说法正确的是10.下列说法正确的是A .按系统命名法,化合物 的名称是2,3,5,5—四甲基-4,4—二乙基己烷B .等物质的量的苯和苯甲酸完全燃烧消耗氧气的量不相等C .苯与甲苯互为同系物,均能使KMnO 4酸性溶液褪色D .结构片段为 的高聚物,其单体是甲醛和苯酚11.电解装置如图所示,电解槽内装有KI 及淀粉溶液,中间用阴离子交换膜隔开.在一定的电压下通电,发现左侧溶液变蓝色,一段时间后,蓝色逐渐变浅。

2013年高考语文试题(新课标Ⅱ卷)—含答案

2013年高考语文试题(新课标Ⅱ卷)—含答案

2013年普通高等学校招生全国统一考试(新课标Ⅱ卷)语文试题本试卷分第Ⅰ卷和第Ⅱ卷两部分,共8页。

满分150分钟考试结束后,将本试卷和答题卡一并交回。

【注意事项】:1.答题前,考生务必用0.5毫米黑色签字将自己的姓名、座号、考生号、区县和科类填写在答题卡和卡和试卷规定的位置上。

2.第1卷每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑;如需改动,用橡皮擦干净后,再选涂其他答案标号,答案不能答在试卷上。

3.第Ⅱ卷必须用0.5毫米黑色签字笔作答.答案必须写在答题卡各题目指定区域内相应的位置,不能写在试卷上;如需改动,先划掉原来的答案,然后再写上新的答案;不能使用涂改液、胶带纸、修正带。

不按以上要求怍答的答案无效。

第Ⅰ卷阅读题甲必考题一、现代文阅读(9分,每小题3分)阅读下面的文字,完成1~3题20世纪后期,陕西凤雏村出土了刻有“凤”字的甲骨四片,这些“凤”字的形体大致相同,均为头上带有象征神权或王权的抽象化了的毛角的短尾鸟。

东汉许筷《说文解字》云:“公耸,凤属,神鸟也.……江中有公耸,似兔而大,赤目.”据此,古代传说中鸣于岐山、兆示周王朝兴起的神鸟凤凰,其原型应该是一种形象普通、类似水鸭的短尾水鸟。

那么,普通的短尾鸟“凤”为何在周代变为华冠长尾、祥瑞美丽的神鸟了呢?我们看到,在商代早期和中期的青铜器纹饰中,只有鸟纹而没有凤纹,弄正的凤形直到殷商晚期才出现,而且此时是华冠短尾鸟和华丽而饰有眼翎的长尾鸟同时出现,可见“凤”是由鸟演变而来的.综观甲骨文和商代青铜器,凤鸟的演变应该是鸟在先,凤在后,贯穿整个商代的不是凤而是鸟。

“天命玄鸟,降而生商”,在商人的历史中鸟始终扮演着图腾始祖的重要角色。

《左传》记载郯子说:“我高祖少睐挚之立也,凤鸟适至,故纪于鸟,为鸟师而鸟名。

凤鸟氏历正也,……九扈为九农正.”凤鸟氏成为“历正”之官,是由于它知天时,九扈成为“九农正”,也是由于它们带来了耕种、耘田和收获的信息.殷人先祖之所以“鸟师而鸟名”,应该是由于这些随着信风迁批的鸟,给以少昧为首的商人的农业生产带来了四季节令的消息。

2013四川高考生物试题含答案(含详细解析)

2013四川高考生物试题含答案(含详细解析)

2013四川8.(11分)将玉米的PEPC酶基因导入水 稻后,测的光照强度对转基因水稻和原种水稻的 气孔导度及光合速率的影响结果,如下图所示。 (注:气孔导度越大,气孔开放程度越高)
(1)水稻叶肉细胞进行光合作用的场所是

捕获叶光绿能体的色素中含量最多的
。 叶绿素 a
(2)CO2 通过气孔进入叶肉细胞后,首先 与 C5 结合而被固定,固定产物的还原需要光反
为达到筛选目的,平板内的固体培养基应以____淀___粉_____
作为唯一碳源。②、③过程需要重复几次,目的是
_进__一__步__筛__选__纯__化_。获得分解淀粉能力强的酵母菌 (2)某同学尝试过程③的操作,其中一个平板经培养后 的菌落分布如图乙所示。该同 学的接种方法是_稀_释__涂__布_平__板__法__________;推测该同学接
2013四川4.下列所采取的措施,不涉及 “降低化学反应活化能”原理的是:
A.利用果胶酶提高水果的出汁率 B.滴加肝脏研磨液促使过氧化氢的分解 C.滴加 FeCl3溶液提高过氧化氢的分解 速率
D.利用水浴加热提高胡萝卜素的萃取效 率
【答案】D
【解析】本题考查的为酶的相关 知识。“降低反应活化能”的原 理描述的是催化剂的作用 原理,A、B 项中的两种酶和 C 项中的 FeCl3均为催化剂,所以 均能降低反应活化能。利用水浴 加热提高胡萝卜素的萃取效率则 是利用温度升高提高胡萝卜素的 溶解度的原理, D项错误。
应提供 [H]和 AT。P
(3)光照强度低于 8×102μmol·s-1时,影响转基 因水稻光合速率的主要因素是 光照强;度光照强度
为 10~14×102μmol·s-1时,原种水稻的气孔导度下
降但光合速率基本不变,可能的原因

2013年高考英语试题及答案(全国卷)

2013年高考英语试题及答案(全国卷)

2013年⾼考英语试题及答案(全国卷)★启⽤前2013年普通⾼等学校招⽣全国统⼀考试(新课标Ⅱ卷) 英语本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(⾮选择题)两部分。

考试结束,将试题卷和答题卡⼀并交回。

第Ⅰ卷注意事项: 1.答第Ⅰ卷时,考⽣务必将⾃⼰的姓名、准考证号填写在答题卡上。

2. 选出每⼩题答案后,⽤2B铅笔把答题卡上对应题⽬的答案标号涂⿊,如需改动,⽤橡⽪擦⼲净后,再选涂其他答案标号框。

不能答在试卷上,否则⽆效。

第⼀部分英语知识运⽤(共两节,满分45分)单项填空(共15⼩题:每⼩题1分,满分15分)从A、B、C、D四个选项中选出可以填⼊空⽩处的选项,并在答题卡上将该选项涂⿊。

例:It is generally considered unwise to give a child he or she wants.A. howeverB. whateverC. whicheverD. whenever答案是B。

1.--I'm sorry I made a mistake!-- Nobody is perfect.A. Take your timeB. You're rightC.Whatever you sayD. Take it easy2. Would you like to with us to the film tonight?A. come alongB. come offC. come acrossD. come through3. I was glad to meet Jenny again, I didn't want to spend all day with her. A. butB. andC.soD.or4. When I arrived, Bryan took me to see the house_ I would be staying.A. whatB. whenC. whereD. which5.I got to the office earlier that day, the 7:30 train from PaddingtonA.caughlB. to have caughtC. to catchD. having caught6. Since nobody gave him any help, he have done the research on his own.A. canB. mustC. wouldD. need7. We very early so we packed the night before.A. leaveB. had leftC. were leavingD.have left8. The watch was very good, and he 20 percent down for it.A. a good oneB. a better oneC. the best oneD. a best one10. It was only after he had read the papers Mr. Gross realized the task before him was extremely difficult to complete. AwhenB. thatC. whichD. what11 A serious study of physics is impossible some knowledge of mathematics.A. againstB. beforeC. beyondD. without12. Only by increasing the number of doctors by 50 percent properly in this hospital. A. can be the patientsB. can the patients be treatedC. the patients can be treatedD.treated can be the patients13.Four and half hours of discussion took us up to midnight,and break for cheese,chocolate and tea with sugar.A.a;aB. the; theC.不填;theD.a;不填14. It's an either-or situation - we can buy a new car this year or we can go on holiday but we can’t do . A.others B. eitherC.anotherD. both 15. -Are you sure you won't come for a drink with us? - ,if you insist A. Not at all B. It depends C. All right then D.I don't care 第⼆节完形填空(共20⼩题;每⼩题1.5分,满分30分)阅读下⾯短⽂,从短⽂后⾯各题锁哥的四个选项(A、 B、C 和 D)中选出可以填⼊空⽩处的选项,并在答题卡上将该项涂⿊。

2013年全国高考理综试题及答案-新课标1卷(解析版)

2013年全国高考理综试题及答案-新课标1卷(解析版)

2013年普通高等学校招生全国统一考试(新课标Ⅰ卷)理科综合能力测试可能用到的相对原子质量:H 1 C 12 N 14 O 16 Mg24 S 32 K39 Mn55第Ⅰ卷一、选择题:本卷共13小题。

每小题6分。

在每个小题给出的四个选项中,只有一项是符合题目要求的。

1.关于蛋白质生物合成的叙述,正确的是()A.一种tRNA可以携带多种氨基酸B.DNA聚合酶是在细胞核中合成的C.反密码子是位于mRNA上相邻的三个碱基D.线粒体中的DNA能控制某些蛋白质的合成【答案】D【解析】tRNA的一端有三个碱基外露为反密码子,与mRNA上的密码子进行碱基互补配对,另一端携带一种氨基酸到达核糖体上,通过发生脱水缩合形成肽健,合成多肽链。

所以A、C错误。

DNA聚合酶是蛋白质,在核糖体上合成,而细胞核内无核糖体,不能合成蛋白质,因而DNA聚合酶是在细胞质中合成的蛋白质类酶,通过核孔进入细胞核发挥作用。

B错。

线粒体中不仅具有自己的DNA,而且还有核糖体,能够通过转录和翻译控制一部分蛋白质的合成,所以核糖体具有一定的独立性。

D正确。

【试题点评】本题不偏不难,正面考查了有关蛋白质合成的基础知识,DNA聚合酶是在细胞核内起催化作用的,部分考生可能会误选B。

本题主要涉及的知识点是化合物的本质及合成,基因控制蛋白质合成过程中有关概念及特点。

旨在考查考生对蛋白质合成过程中相关知识点的识记及初步分析问题的能力。

2.关于同一个体中细胞有丝分裂和减数第一次分裂的叙述,正确的是()A.两者前期染色体数目相同,染色体行为和DNA分子数目不同B.两者中期染色体数目不同,染色体行为和DNA分子数目相同C.两者后期染色体行为和数目不同,DNA分子数目相同D.两者后期染色体行为和数目相同,DNA分子数目不同【答案】C【解析】有丝分裂和减数分裂分裂期过程的最大区别是染色体的行为不同。

有丝分裂前期有同源染色体但不联会,中期染色体的着丝点被纺锤丝拉到赤道板位置排列整齐,后期着丝点分裂,姐妹染色单体分离并分别移向细胞的两极,此时染色体数目暂时性加倍,DNA分子数不变,分裂的结果是分裂前后染色体数与DNA分子数与分裂前一样。

2013年全国统一高考真题数学试卷(理科)(新课标ⅰ)(含答案及解析)

2013年全国统一高考数学试卷(理科)(新课标Ⅰ)一、选择题:本大题共12小题,每小题5分,在每小题给出的四个选项中,只有一个是符合题目要求的.1.(5分)已知集合A={x|x2﹣2x>0},B={x|﹣<x<},则()A.A∩B=∅B.A∪B=R C.B⊆A D.A⊆B2.(5分)若复数z满足(3﹣4i)z=|4+3i|,则z的虚部为()A.﹣4B.C.4D.3.(5分)为了解某地区中小学生的视力情况,拟从该地区的中小学生中抽取部分学生进行调查,事先已经了解到该地区小学、初中、高中三个学段学生的视力情况有较大差异,而男女生视力情况差异不大.在下面的抽样方法中,最合理的抽样方法是()A.简单的随机抽样B.按性别分层抽样C.按学段分层抽样D.系统抽样4.(5分)已知双曲线C:(a>0,b>0)的离心率为,则C的渐近线方程为()A.y=B.y=C.y=±x D.y=5.(5分)执行程序框图,如果输入的t∈[﹣1,3],则输出的s属于()A.[﹣3,4]B.[﹣5,2]C.[﹣4,3]D.[﹣2,5] 6.(5分)如图,有一个水平放置的透明无盖的正方体容器,容器高8cm,将一个球放在容器口,再向容器注水,当球面恰好接触水面时测得水深为6cm,如不计容器的厚度,则球的体积为()A.B.C.D.7.(5分)设等差数列{a n}的前n项和为S n,若S m﹣1=﹣2,S m=0,S m+1=3,则m=()A.3B.4C.5D.68.(5分)某几何体的三视图如图所示,则该几何体的体积为()A.16+8πB.8+8πC.16+16πD.8+16π9.(5分)设m为正整数,(x+y)2m展开式的二项式系数的最大值为a,(x+y)2m+1展开式的二项式系数的最大值为b,若13a=7b,则m=()A.5B.6C.7D.810.(5分)已知椭圆E:的右焦点为F(3,0),过点F的直线交椭圆E于A、B两点.若AB的中点坐标为(1,﹣1),则E的方程为()A.B.C.D.11.(5分)已知函数f(x)=,若|f(x)|≥ax,则a的取值范围是()A.(﹣∞,0]B.(﹣∞,1]C.[﹣2,1]D.[﹣2,0] 12.(5分)设△A n B n C n的三边长分别为a n,b n,c n,△A n B n C n的面积为S n,n=1,2,3…若b1>c1,b1+c1=2a1,a n+1=a n,,,则()A.{S n}为递减数列B.{S n}为递增数列C.{S2n﹣1}为递增数列,{S2n}为递减数列D.{S2n﹣1}为递减数列,{S2n}为递增数列二.填空题:本大题共4小题,每小题5分.13.(5分)已知两个单位向量,的夹角为60°,=t+(1﹣t).若•=0,则t=.14.(5分)若数列{a n}的前n项和为S n=a n+,则数列{a n}的通项公式是a n=.15.(5分)设当x=θ时,函数f(x)=sinx﹣2cosx取得最大值,则cosθ=.16.(5分)若函数f(x)=(1﹣x2)(x2+ax+b)的图象关于直线x=﹣2对称,则f(x)的最大值为.三、解答题:解答应写出文字说明,证明过程或演算步骤.17.(12分)如图,在△ABC中,∠ABC=90°,AB=,BC=1,P为△ABC内一点,∠BPC=90°.(1)若PB=,求PA;(2)若∠APB=150°,求tan∠PBA.18.(12分)如图,三棱柱ABC﹣A1B1C1中,CA=CB,AB=AA1,∠BAA1=60°.(Ⅰ)证明AB⊥A1C;(Ⅱ)若平面ABC⊥平面AA1B1B,AB=CB=2,求直线A1C与平面BB1C1C所成角的正弦值.19.(12分)一批产品需要进行质量检验,检验方案是:先从这批产品中任取4件作检验,这4件产品中优质品的件数记为n.如果n=3,再从这批产品中任取4件作检验,若都为优质品,则这批产品通过检验;如果n=4,再从这批产品中任取1件作检验,若为优质品,则这批产品通过检验;其他情况下,这批产品都不能通过检验.假设这批产品的优质品率为50%,即取出的产品是优质品的概率都为,且各件产品是否为优质品相互独立.(Ⅰ)求这批产品通过检验的概率;(Ⅱ)已知每件产品检验费用为100元,凡抽取的每件产品都需要检验,对这批产品作质量检验所需的费用记为X(单位:元),求X的分布列及数学期望.20.(12分)已知圆M:(x+1)2+y2=1,圆N:(x﹣1)2+y2=9,动圆P与圆M外切并与圆N内切,圆心P的轨迹为曲线C.(Ⅰ)求C的方程;(Ⅱ)l是与圆P,圆M都相切的一条直线,l与曲线C交于A,B两点,当圆P 的半径最长时,求|AB|.21.(12分)已知函数f(x)=x2+ax+b,g(x)=e x(cx+d),若曲线y=f(x)和曲线y=g(x)都过点P(0,2),且在点P处有相同的切线y=4x+2.(Ⅰ)求a,b,c,d的值;(Ⅱ)若x≥﹣2时,f(x)≤kg(x),求k的取值范围.四、请考生在第22、23、24题中任选一道作答,并用2B铅笔将答题卡上所选的题目对应的题号右侧方框涂黑,按所涂题号进行评分;多涂、多答,按所涂的首题进行评分,不涂,按本选考题的首题进行评分.22.(10分)(选修4﹣1:几何证明选讲)如图,直线AB为圆的切线,切点为B,点C在圆上,∠ABC的角平分线BE交圆于点E,DB垂直BE交圆于D.(Ⅰ)证明:DB=DC;(Ⅱ)设圆的半径为1,BC=,延长CE交AB于点F,求△BCF外接圆的半径.23.已知曲线C1的参数方程为(t为参数),以坐标原点为极点,x 轴的正半轴为极轴建立极坐标系,曲线C2的极坐标方程为ρ=2sinθ.(1)把C1的参数方程化为极坐标方程;(2)求C1与C2交点的极坐标(ρ≥0,0≤θ<2π).24.已知函数f(x)=|2x﹣1|+|2x+a|,g(x)=x+3.(Ⅰ)当a=﹣2时,求不等式f(x)<g(x)的解集;(Ⅱ)设a>﹣1,且当x∈[﹣,]时,f(x)≤g(x),求a的取值范围.2013年全国统一高考数学试卷(理科)(新课标Ⅰ)参考答案与试题解析一、选择题:本大题共12小题,每小题5分,在每小题给出的四个选项中,只有一个是符合题目要求的.1.(5分)已知集合A={x|x2﹣2x>0},B={x|﹣<x<},则()A.A∩B=∅B.A∪B=R C.B⊆A D.A⊆B【考点】1D:并集及其运算;73:一元二次不等式及其应用.【专题】59:不等式的解法及应用;5J:集合.【分析】根据一元二次不等式的解法,求出集合A,再根据的定义求出A∩B和A∪B.【解答】解:∵集合A={x|x2﹣2x>0}={x|x>2或x<0},∴A∩B={x|2<x<或﹣<x<0},A∪B=R,故选:B.【点评】本题考查一元二次不等式的解法,以及并集的定义,属于基础题.2.(5分)若复数z满足(3﹣4i)z=|4+3i|,则z的虚部为()A.﹣4B.C.4D.【考点】A5:复数的运算.【专题】5N:数系的扩充和复数.【分析】由题意可得z==,再利用两个复数代数形式的乘除法法则化简为+i,由此可得z的虚部.【解答】解:∵复数z满足(3﹣4i)z=|4+3i|,∴z====+i,故z的虚部等于,故选:D.【点评】本题主要考查复数的基本概念,两个复数代数形式的乘除法法则的应用,属于基础题.3.(5分)为了解某地区中小学生的视力情况,拟从该地区的中小学生中抽取部分学生进行调查,事先已经了解到该地区小学、初中、高中三个学段学生的视力情况有较大差异,而男女生视力情况差异不大.在下面的抽样方法中,最合理的抽样方法是()A.简单的随机抽样B.按性别分层抽样C.按学段分层抽样D.系统抽样【考点】B3:分层抽样方法.【专题】21:阅读型.【分析】若总体由差异明显的几部分组成时,经常采用分层抽样的方法进行抽样.【解答】解:我们常用的抽样方法有:简单随机抽样、分层抽样和系统抽样,而事先已经了解到该地区小学、初中、高中三个学段学生的视力情况有较大差异,而男女生视力情况差异不大.了解某地区中小学生的视力情况,按学段分层抽样,这种方式具有代表性,比较合理.故选:C.【点评】本小题考查抽样方法,主要考查抽样方法,属基本题.4.(5分)已知双曲线C:(a>0,b>0)的离心率为,则C的渐近线方程为()A.y=B.y=C.y=±x D.y=【考点】KC:双曲线的性质.【专题】5D:圆锥曲线的定义、性质与方程.【分析】由离心率和abc的关系可得b2=4a2,而渐近线方程为y=±x,代入可得答案.【解答】解:由双曲线C:(a>0,b>0),则离心率e===,即4b2=a2,故渐近线方程为y=±x=x,故选:D.【点评】本题考查双曲线的简单性质,涉及的渐近线方程,属基础题.5.(5分)执行程序框图,如果输入的t∈[﹣1,3],则输出的s属于()A.[﹣3,4]B.[﹣5,2]C.[﹣4,3]D.[﹣2,5]【考点】3B:分段函数的解析式求法及其图象的作法;EF:程序框图.【专题】27:图表型;5K:算法和程序框图.【分析】本题考查的知识点是程序框图,分析程序中各变量、各语句的作用,再根据流程图所示的顺序,可知:该程序的作用是计算一个分段函数的函数值,由条件为t<1我们可得,分段函数的分类标准,由分支结构中是否两条分支上对应的语句行,我们易得函数的解析式.【解答】解:由判断框中的条件为t<1,可得:函数分为两段,即t<1与t≥1,又由满足条件时函数的解析式为:s=3t;不满足条件时,即t≥1时,函数的解析式为:s=4t﹣t2故分段函数的解析式为:s=,如果输入的t∈[﹣1,3],画出此分段函数在t∈[﹣1,3]时的图象,则输出的s属于[﹣3,4].故选:A.【点评】要求条件结构对应的函数解析式,要分如下几个步骤:①分析流程图的结构,分析条件结构是如何嵌套的,以确定函数所分的段数;②根据判断框中的条件,设置分类标准;③根据判断框的“是”与“否”分支对应的操作,分析函数各段的解析式;④对前面的分类进行总结,写出分段函数的解析式.6.(5分)如图,有一个水平放置的透明无盖的正方体容器,容器高8cm,将一个球放在容器口,再向容器注水,当球面恰好接触水面时测得水深为6cm,如不计容器的厚度,则球的体积为()A.B.C.D.【考点】LG:球的体积和表面积.【专题】11:计算题;5F:空间位置关系与距离.【分析】设正方体上底面所在平面截球得小圆M,可得圆心M为正方体上底面正方形的中心.设球的半径为R,根据题意得球心到上底面的距离等于(R﹣2)cm,而圆M的半径为4,由球的截面圆性质建立关于R的方程并解出R=5,用球的体积公式即可算出该球的体积.【解答】解:设正方体上底面所在平面截球得小圆M,则圆心M为正方体上底面正方形的中心.如图.设球的半径为R,根据题意得球心到上底面的距离等于(R﹣2)cm,而圆M的半径为4,由球的截面圆性质,得R2=(R﹣2)2+42,解出R=5,∴根据球的体积公式,该球的体积V===.故选:A.【点评】本题给出球与正方体相切的问题,求球的体积,着重考查了正方体的性质、球的截面圆性质和球的体积公式等知识,属于中档题.7.(5分)设等差数列{a n}的前n项和为S n,若S m﹣1=﹣2,S m=0,S m+1=3,则m=()A.3B.4C.5D.6【考点】83:等差数列的性质;85:等差数列的前n项和.【专题】11:计算题;54:等差数列与等比数列.【分析】由a n与S n的关系可求得a m+1与a m,进而得到公差d,由前n项和公式及S m=0可求得a1,再由通项公式及a m=2可得m值.【解答】解:a m=S m﹣S m﹣1=2,a m+1=S m+1﹣S m=3,所以公差d=a m﹣a m=1,+1S m==0,m﹣1>0,m>1,因此m不能为0,得a1=﹣2,所以a m=﹣2+(m﹣1)•1=2,解得m=5,另解:等差数列{a n}的前n项和为S n,即有数列{}成等差数列,则,,成等差数列,可得2•=+,即有0=+,解得m=5.又一解:由等差数列的求和公式可得(m﹣1)(a1+a m﹣1)=﹣2,m(a1+a m)=0,(m+1)(a1+a m+1)=3,可得a1=﹣a m,﹣2a m+a m+1+a m+1=+=0,解得m=5.故选:C.【点评】本题考查等差数列的通项公式、前n项和公式及通项a n与S n的关系,考查学生的计算能力.8.(5分)某几何体的三视图如图所示,则该几何体的体积为()A.16+8πB.8+8πC.16+16πD.8+16π【考点】L!:由三视图求面积、体积.【专题】16:压轴题;27:图表型.【分析】三视图复原的几何体是一个长方体与半个圆柱的组合体,依据三视图的数据,得出组合体长、宽、高,即可求出几何体的体积.【解答】解:三视图复原的几何体是一个长方体与半个圆柱的组合体,如图,其中长方体长、宽、高分别是:4,2,2,半个圆柱的底面半径为2,母线长为4.∴长方体的体积=4×2×2=16,半个圆柱的体积=×22×π×4=8π所以这个几何体的体积是16+8π;故选:A.【点评】本题考查了几何体的三视图及直观图的画法,三视图与直观图的关系,柱体体积计算公式,空间想象能力9.(5分)设m为正整数,(x+y)2m展开式的二项式系数的最大值为a,(x+y)2m+1展开式的二项式系数的最大值为b,若13a=7b,则m=()A.5B.6C.7D.8【考点】DA:二项式定理.【专题】5P:二项式定理.【分析】根据二项式系数的性质求得a和b,再利用组合数的计算公式,解方程13a=7b求得m的值.【解答】解:∵m为正整数,由(x+y)2m展开式的二项式系数的最大值为a,以及二项式系数的性质可得a=,同理,由(x+y)2m+1展开式的二项式系数的最大值为b,可得b==.再由13a=7b,可得13=7,即13×=7×,即13=7×,即13(m+1)=7(2m+1),解得m=6,故选:B.【点评】本题主要考查二项式系数的性质的应用,组合数的计算公式,属于中档题.10.(5分)已知椭圆E:的右焦点为F(3,0),过点F的直线交椭圆E于A、B两点.若AB的中点坐标为(1,﹣1),则E的方程为()A.B.C.D.【考点】K3:椭圆的标准方程.【专题】5D:圆锥曲线的定义、性质与方程.【分析】设A(x1,y1),B(x2,y2),代入椭圆方程得,利用“点差法”可得.利用中点坐标公式可得x1+x2=2,y1+y2=﹣2,利用斜率计算公式可得==.于是得到,化为a2=2b2,再利用c=3=,即可解得a2,b2.进而得到椭圆的方程.【解答】解:设A(x1,y1),B(x2,y2),代入椭圆方程得,相减得,∴.∵x1+x2=2,y1+y2=﹣2,==.∴,化为a2=2b2,又c=3=,解得a2=18,b2=9.∴椭圆E的方程为.故选:D.【点评】熟练掌握“点差法”和中点坐标公式、斜率的计算公式是解题的关键.11.(5分)已知函数f(x)=,若|f(x)|≥ax,则a的取值范围是()A.(﹣∞,0]B.(﹣∞,1]C.[﹣2,1]D.[﹣2,0]【考点】7E:其他不等式的解法.【专题】16:压轴题;59:不等式的解法及应用.【分析】由函数图象的变换,结合基本初等函数的图象可作出函数y=|f(x)|的图象,和函数y=ax的图象,由导数求切线斜率可得l的斜率,进而数形结合可得a的范围.【解答】解:由题意可作出函数y=|f(x)|的图象,和函数y=ax的图象,由图象可知:函数y=ax的图象为过原点的直线,当直线介于l和x轴之间符合题意,直线l为曲线的切线,且此时函数y=|f(x)|在第二象限的部分解析式为y=x2﹣2x,求其导数可得y′=2x﹣2,因为x≤0,故y′≤﹣2,故直线l的斜率为﹣2,故只需直线y=ax的斜率a介于﹣2与0之间即可,即a∈[﹣2,0]故选:D.【点评】本题考查其它不等式的解法,数形结合是解决问题的关键,属中档题.12.(5分)设△A n B n C n的三边长分别为a n,b n,c n,△A n B n C n的面积为S n,n=1,2,3…若b1>c1,b1+c1=2a1,a n+1=a n,,,则()A.{S n}为递减数列B.{S n}为递增数列C.{S2n﹣1}为递增数列,{S2n}为递减数列D.{S2n﹣1}为递减数列,{S2n}为递增数列【考点】82:数列的函数特性;8H:数列递推式.【专题】16:压轴题;54:等差数列与等比数列;55:点列、递归数列与数学归纳法.=a n可知△A n B n C n的边B n C n为定值a1,由b n+1+c n+1﹣【分析】由a n+12a1=及b1+c1=2a1得b n+c n=2a1,则在△A n B n C n中边长B n C n=a1为定值,另两边A n C n、A n B n的长度之和b n+c n=2a1为定值,由此可知顶点A n在以B n、C n为焦点的椭圆上,根据b n+1﹣c n+1=,得b n﹣c n=,可知n→+∞时b n→c n,据此可判断△A n B n C n的边B nC n的高h n随着n的增大而增大,再由三角形面积公式可得到答案.【解答】解:b1=2a1﹣c1且b1>c1,∴2a1﹣c1>c1,∴a1>c1,∴b1﹣a1=2a1﹣c1﹣a1=a1﹣c1>0,∴b1>a1>c1,又b1﹣c1<a1,∴2a1﹣c1﹣c1<a1,∴2c1>a1,∴,由题意,+a n,∴b n+1+c n+1﹣2a n=(b n+c n﹣2a n),∴b n+c n﹣2a n=0,∴b n+c n=2a n=2a1,∴b n+c n=2a1,由此可知顶点A n在以B n、C n为焦点的椭圆上,﹣c n+1=,∴=a1﹣b n,又由题意,b n+1﹣a1=,∴b n﹣a1=,∴b n+1∴,c n=2a1﹣b n=,∴[][]=[﹣]单调递增(可证当n=1时>0)故选:B.【点评】本题主要考查由数列递推式求数列通项、三角形面积海伦公式,综合考查学生分析解决问题的能力,有较高的思维抽象度,是本年度全国高考试题中的“亮点”之一.二.填空题:本大题共4小题,每小题5分.13.(5分)已知两个单位向量,的夹角为60°,=t+(1﹣t).若•=0,则t=2.【考点】9H:平面向量的基本定理;9O:平面向量数量积的性质及其运算.【专题】5A:平面向量及应用.【分析】由于•=0,对式子=t+(1﹣t)两边与作数量积可得=0,经过化简即可得出.【解答】解:∵,,∴=0,∴tcos60°+1﹣t=0,∴1=0,解得t=2.故答案为2.【点评】熟练掌握向量的数量积运算是解题的关键.14.(5分)若数列{a n}的前n项和为S n=a n+,则数列{a n}的通项公式是a n=(﹣2)n﹣1.【考点】88:等比数列的通项公式.【专题】54:等差数列与等比数列.【分析】把n=1代入已知式子可得数列的首项,由n≥2时,a n=S n﹣S n﹣1,可得数列为等比数列,且公比为﹣2,代入等比数列的通项公式分段可得答案.【解答】解:当n=1时,a1=S1=,解得a1=1当n≥2时,a n=S n﹣S n﹣1=()﹣()=,整理可得,即=﹣2,故数列{a n}从第二项开始是以﹣2为首项,﹣2为公比的等比数列,故当n≥2时,a n=(﹣2)n﹣1,经验证当n=1时,上式也适合,故答案为:(﹣2)n﹣1【点评】本题考查等比数列的通项公式,涉及等比数列的判定,属基础题.15.(5分)设当x=θ时,函数f(x)=sinx﹣2cosx取得最大值,则cosθ=﹣.【考点】GP:两角和与差的三角函数;H4:正弦函数的定义域和值域.【专题】16:压轴题;56:三角函数的求值.【分析】f(x)解析式提取,利用两角和与差的正弦函数公式化为一个角的正弦函数,由x=θ时,函数f(x)取得最大值,得到sinθ﹣2cosθ=,与sin2θ+cos2θ=1联立即可求出cosθ的值.【解答】解:f(x)=sinx﹣2cosx=(sinx﹣cosx)=sin(x﹣α)(其中cosα=,sinα=),∵x=θ时,函数f(x)取得最大值,∴sin(θ﹣α)=1,即sinθ﹣2cosθ=,又sin2θ+cos2θ=1,联立得(2cosθ+)2+cos2θ=1,解得cosθ=﹣.故答案为:﹣【点评】此题考查了两角和与差的正弦函数公式,同角三角函数间的基本关系,以及正弦函数的定义域与值域,熟练掌握公式是解本题的关键.16.(5分)若函数f(x)=(1﹣x2)(x2+ax+b)的图象关于直线x=﹣2对称,则f(x)的最大值为16.【考点】57:函数与方程的综合运用;6E:利用导数研究函数的最值.【专题】11:计算题;16:压轴题;51:函数的性质及应用;53:导数的综合应用.【分析】由题意得f(﹣1)=f(﹣3)=0且f(1)=f(﹣5)=0,由此求出a=8且b=15,由此可得f(x)=﹣x4﹣8x3﹣14x2+8x+15.利用导数研究f(x)的单调性,可得f(x)在区间(﹣∞,﹣2﹣)、(﹣2,﹣2+)上是增函数,在区间(﹣2﹣,﹣2)、(﹣2+,+∞)上是减函数,结合f(﹣2﹣)=f(﹣2+)=16,即可得到f(x)的最大值.【解答】解:∵函数f(x)=(1﹣x2)(x2+ax+b)的图象关于直线x=﹣2对称,∴f(﹣1)=f(﹣3)=0且f(1)=f(﹣5)=0,即[1﹣(﹣3)2][(﹣3)2+a•(﹣3)+b]=0且[1﹣(﹣5)2][(﹣5)2+a•(﹣5)+b]=0,解之得,因此,f(x)=(1﹣x2)(x2+8x+15)=﹣x4﹣8x3﹣14x2+8x+15,求导数,得f′(x)=﹣4x3﹣24x2﹣28x+8,令f′(x)=0,得x1=﹣2﹣,x2=﹣2,x3=﹣2+,当x∈(﹣∞,﹣2﹣)时,f′(x)>0;当x∈(﹣2﹣,﹣2)时,f′(x)<0;当x∈(﹣2,﹣2+)时,f′(x)>0;当x∈(﹣2+,+∞)时,f′(x)<0∴f(x)在区间(﹣∞,﹣2﹣)、(﹣2,﹣2+)上是增函数,在区间(﹣2﹣,﹣2)、(﹣2+,+∞)上是减函数.又∵f(﹣2﹣)=f(﹣2+)=16,∴f(x)的最大值为16.故答案为:16.【点评】本题给出多项式函数的图象关于x=﹣2对称,求函数的最大值.着重考查了函数的奇偶性、利用导数研究函数的单调性和函数的最值求法等知识,属于中档题.三、解答题:解答应写出文字说明,证明过程或演算步骤.17.(12分)如图,在△ABC中,∠ABC=90°,AB=,BC=1,P为△ABC内一点,∠BPC=90°.(1)若PB=,求PA;(2)若∠APB=150°,求tan∠PBA.【考点】HP:正弦定理;HR:余弦定理.【专题】58:解三角形.【分析】(I)在Rt△PBC,利用边角关系即可得到∠PBC=60°,得到∠PBA=30°.在△PBA中,利用余弦定理即可求得PA.(II)设∠PBA=α,在Rt△PBC中,可得PB=sinα.在△PBA中,由正弦定理得,即,化简即可求出.【解答】解:(I)在Rt△PBC中,=,∴∠PBC=60°,∴∠PBA=30°.在△PBA中,由余弦定理得PA2=PB2+AB2﹣2PB•ABcos30°==.∴PA=.(II)设∠PBA=α,在Rt△PBC中,PB=BCcos(90°﹣α)=sinα.在△PBA中,由正弦定理得,即,化为.∴.【点评】熟练掌握直角三角形的边角关系、正弦定理和余弦定理是解题的关键.18.(12分)如图,三棱柱ABC﹣A1B1C1中,CA=CB,AB=AA1,∠BAA1=60°.(Ⅰ)证明AB⊥A1C;(Ⅱ)若平面ABC⊥平面AA1B1B,AB=CB=2,求直线A1C与平面BB1C1C所成角的正弦值.【考点】LW:直线与平面垂直;LY:平面与平面垂直;MI:直线与平面所成的角.【专题】5F:空间位置关系与距离;5G:空间角.【分析】(Ⅰ)取AB的中点O,连接OC,OA1,A1B,由已知可证OA1⊥AB,AB ⊥平面OA1C,进而可得AB⊥A1C;(Ⅱ)易证OA,OA1,OC两两垂直.以O为坐标原点,的方向为x轴的正向,||为单位长,建立坐标系,可得,,的坐标,设=(x,y,z)为平面BB1C1C的法向量,则,可解得=(,1,﹣1),可求|cos <,>|,即为所求正弦值.【解答】解:(Ⅰ)取AB的中点O,连接OC,OA1,A1B,因为CA=CB,所以OC⊥AB,由于AB=AA1,∠BAA1=60°,所以△AA1B为等边三角形,所以OA1⊥AB,又因为OC∩OA1=O,所以AB⊥平面OA1C,又A1C⊂平面OA1C,故AB⊥A1C;(Ⅱ)由(Ⅰ)知OC⊥AB,OA1⊥AB,又平面ABC⊥平面AA1B1B,交线为AB,所以OC⊥平面AA1B1B,故OA,OA1,OC两两垂直.以O为坐标原点,的方向为x轴的正向,||为单位长,建立如图所示的坐标系,可得A(1,0,0),A1(0,,0),C(0,0,),B(﹣1,0,0),则=(1,0,),=(﹣1,,0),=(0,﹣,),设=(x,y,z)为平面BB1C1C的法向量,则,即,可取y=1,可得=(,1,﹣1),故cos<,>==,又因为直线与法向量的余弦值的绝对值等于直线与平面的正弦值,故直线A1C与平面BB1C1C所成角的正弦值为:.【点评】本题考查直线与平面所成的角,涉及直线与平面垂直的性质和平面与平面垂直的判定,属难题.19.(12分)一批产品需要进行质量检验,检验方案是:先从这批产品中任取4件作检验,这4件产品中优质品的件数记为n.如果n=3,再从这批产品中任取4件作检验,若都为优质品,则这批产品通过检验;如果n=4,再从这批产品中任取1件作检验,若为优质品,则这批产品通过检验;其他情况下,这批产品都不能通过检验.假设这批产品的优质品率为50%,即取出的产品是优质品的概率都为,且各件产品是否为优质品相互独立.(Ⅰ)求这批产品通过检验的概率;(Ⅱ)已知每件产品检验费用为100元,凡抽取的每件产品都需要检验,对这批产品作质量检验所需的费用记为X(单位:元),求X的分布列及数学期望.【考点】CG:离散型随机变量及其分布列;CH:离散型随机变量的期望与方差.【专题】5I:概率与统计.【分析】(Ⅰ)设第一次取出的4件产品中恰有3件优质品为事件A1,第一次取出的4件产品全是优质品为事件A2,第二次取出的4件产品全是优质品为事件B1,第二次取出的1件产品是优质品为事件B2,这批产品通过检验为事件A,依题意有A=(A1B1)∪(A2B2),且A1B1与A2B2互斥,由概率得加法公式和条件概率,代入数据计算可得;(Ⅱ)X可能的取值为400,500,800,分别求其概率,可得分布列,进而可得期望值.【解答】解:(Ⅰ)设第一次取出的4件产品中恰有3件优质品为事件A1,第一次取出的4件产品全是优质品为事件A2,第二次取出的4件产品全是优质品为事件B1,第二次取出的1件产品是优质品为事件B2,这批产品通过检验为事件A,依题意有A=(A1B1)∪(A2B2),且A1B1与A2B2互斥,所以P(A)=P(A1B1)+P(A2B2)=P(A1)P(B1|A1)+P(A2)P(B2|A2)==(Ⅱ)X可能的取值为400,500,800,并且P(X=800)=,P(X=500)=,P(X=400)=1﹣﹣=,故X的分布列如下:X 400 500 800P故EX=400×+500×+800×=506.25【点评】本题考查离散型随机变量及其分布列涉及数学期望的求解,属中档题.20.(12分)已知圆M:(x+1)2+y2=1,圆N:(x﹣1)2+y2=9,动圆P与圆M外切并与圆N内切,圆心P的轨迹为曲线C.(Ⅰ)求C的方程;(Ⅱ)l是与圆P,圆M都相切的一条直线,l与曲线C交于A,B两点,当圆P 的半径最长时,求|AB|.【考点】J3:轨迹方程;J9:直线与圆的位置关系.【专题】5B:直线与圆.【分析】(I)设动圆的半径为R,由已知动圆P与圆M外切并与圆N内切,可得|PM|+|PN|=R+1+(3﹣R)=4,而|NM|=2,由椭圆的定义可知:动点P的轨迹是以M,N为焦点,4为长轴长的椭圆,求出即可;(II)设曲线C上任意一点P(x,y),由于|PM|﹣|PN|=2R﹣2≤4﹣2=2,所以R ≤2,当且仅当⊙P的圆心为(2,0)R=2时,其半径最大,其方程为(x﹣2)2+y2=4.分①l的倾斜角为90°,此时l与y轴重合,可得|AB|.②若l的倾斜角不为90°,由于⊙M的半径1≠R,可知l与x轴不平行,设l与x轴的交点为Q,根据,可得Q(﹣4,0),所以可设l:y=k(x+4),与椭圆的方程联立,得到根与系数的关系利用弦长公式即可得出.【解答】解:(I)由圆M:(x+1)2+y2=1,可知圆心M(﹣1,0);圆N:(x﹣1)2+y2=9,圆心N(1,0),半径3.设动圆的半径为R,∵动圆P与圆M外切并与圆N内切,∴|PM|+|PN|=R+1+(3﹣R)=4,而|NM|=2,由椭圆的定义可知:动点P的轨迹是以M,N为焦点,4为长轴长的椭圆,∴a=2,c=1,b2=a2﹣c2=3.∴曲线C的方程为(x≠﹣2).(II)设曲线C上任意一点P(x,y),由于|PM|﹣|PN|=2R﹣2≤3﹣1=2,所以R≤2,当且仅当⊙P的圆心为(2,0)R=2时,其半径最大,其方程为(x﹣2)2+y2=4.①l的倾斜角为90°,则l与y轴重合,可得|AB|=.②若l的倾斜角不为90°,由于⊙M的半径1≠R,可知l与x轴不平行,设l与x轴的交点为Q,则,可得Q(﹣4,0),所以可设l:y=k(x+4),由l于M相切可得:,解得.当时,联立,得到7x2+8x﹣8=0.∴,.∴|AB|===由于对称性可知:当时,也有|AB|=.综上可知:|AB|=或.【点评】本题综合考查了两圆的相切关系、直线与圆相切问题、椭圆的定义及其性质、直线与椭圆相交问题转化为方程联立得到根与系数的关系、弦长公式等基础知识,需要较强的推理能力和计算能力及其分类讨论的思想方法.21.(12分)已知函数f(x)=x2+ax+b,g(x)=e x(cx+d),若曲线y=f(x)和曲线y=g(x)都过点P(0,2),且在点P处有相同的切线y=4x+2.(Ⅰ)求a,b,c,d的值;(Ⅱ)若x≥﹣2时,f(x)≤kg(x),求k的取值范围.【考点】3R:函数恒成立问题;6H:利用导数研究曲线上某点切线方程.【专题】16:压轴题;53:导数的综合应用.【分析】(Ⅰ)对f(x),g(x)进行求导,已知在交点处有相同的切线及曲线y=f (x)和曲线y=g(x)都过点P(0,2),从而解出a,b,c,d的值;(Ⅱ)由(I)得出f(x),g(x)的解析式,再求出F(x)及它的导函数,通过对k的讨论,判断出F(x)的最值,从而判断出f(x)≤kg(x)恒成立,从而求出k的范围.【解答】解:(Ⅰ)由题意知f(0)=2,g(0)=2,f′(0)=4,g′(0)=4,而f′(x)=2x+a,g′(x)=e x(cx+d+c),故b=2,d=2,a=4,d+c=4,从而a=4,b=2,c=2,d=2;(Ⅱ)由(I)知,f(x)=x2+4x+2,g(x)=2e x(x+1)设F(x)=kg(x)﹣f(x)=2ke x(x+1)﹣x2﹣4x﹣2,则F′(x)=2ke x(x+2)﹣2x﹣4=2(x+2)(ke x﹣1),由题设得F(0)≥0,即k≥1,令F′(x)=0,得x1=﹣lnk,x2=﹣2,①若1≤k<e2,则﹣2<x1≤0,从而当x∈(﹣2,x1)时,F′(x)<0,当x∈(x1,+∞)时,F′(x)>0,即F(x)在(﹣2,x1)上减,在(x1,+∞)上是增,故F(x)在[﹣2,+∞)上的最小值为F(x1),而F(x1)=﹣x1(x1+2)≥0,x≥﹣2时F(x)≥0,即f(x)≤kg(x)恒成立.②若k=e2,则F′(x)=2e2(x+2)(e x﹣e﹣2),从而当x∈(﹣2,+∞)时,F′(x)>0,即F(x)在(﹣2,+∞)上是增,而F(﹣2)=0,故当x≥﹣2时,F(x)≥0,即f(x)≤kg(x)恒成立.③若k>e2时,F′(x)>2e2(x+2)(e x﹣e﹣2),而F(﹣2)=﹣2ke﹣2+2<0,所以当x>﹣2时,f(x)≤kg(x)不恒成立,综上,k的取值范围是[1,e2].【点评】此题主要考查利用导数研究曲线上某点切线方程,函数恒成立问题,考查分类讨论思想,解题的关键是能够利用导数工具研究函数的性质,此题是一道中档题.四、请考生在第22、23、24题中任选一道作答,并用2B铅笔将答题卡上所选的题目对应的题号右侧方框涂黑,按所涂题号进行评分;多涂、多答,按所涂的首题进行评分,不涂,按本选考题的首题进行评分.22.(10分)(选修4﹣1:几何证明选讲)如图,直线AB为圆的切线,切点为B,点C在圆上,∠ABC的角平分线BE交圆于点E,DB垂直BE交圆于D.(Ⅰ)证明:DB=DC;(Ⅱ)设圆的半径为1,BC=,延长CE交AB于点F,求△BCF外接圆的半径.【考点】NC:与圆有关的比例线段.【专题】5B:直线与圆.【分析】(I)连接DE交BC于点G,由弦切角定理可得∠ABE=∠BCE,由已知角平分线可得∠ABE=∠CBE,于是得到∠CBE=∠BCE,BE=CE.由已知DB⊥BE,可知DE为⊙O的直径,Rt△DBE≌Rt△DCE,利用三角形全等的性质即可得到DC=DB.(II)由(I)可知:DG是BC的垂直平分线,即可得到BG=.设DE的中点为O,连接BO,可得∠BOG=60°.从而∠ABE=∠BCE=∠CBE=30°.得到CF⊥BF.进而得到Rt△BCF的外接圆的半径=.【解答】(I)证明:连接DE交BC于点G.由弦切角定理可得∠ABE=∠BCE,而∠ABE=∠CBE,∴∠CBE=∠BCE,BE=CE.又∵DB⊥BE,∴DE为⊙O的直径,∠DCE=90°.∴△DBE≌△DCE,∴DC=DB.(II)由(I)可知:∠CDE=∠BDE,DB=DC.故DG是BC的垂直平分线,∴BG=.设DE的中点为O,连接BO,则∠BOG=60°.从而∠ABE=∠BCE=∠CBE=30°.∴CF⊥BF.∴Rt△BCF的外接圆的半径=.【点评】本题综合考查了圆的性质、弦切角定理、等边三角形的性质、三角形全等、三角形的外接圆的半径等知识,需要较强的推理能力、分析问题和解决问题的能力.23.已知曲线C1的参数方程为(t为参数),以坐标原点为极点,x 轴的正半轴为极轴建立极坐标系,曲线C2的极坐标方程为ρ=2sinθ.(1)把C1的参数方程化为极坐标方程;(2)求C1与C2交点的极坐标(ρ≥0,0≤θ<2π).【考点】Q4:简单曲线的极坐标方程;QH:参数方程化成普通方程.【专题】11:计算题;35:转化思想;4R:转化法;5S:坐标系和参数方程.【分析】(1)曲线C1的参数方程消去参数t,得到普通方程,再由,能求出C1的极坐标方程.(2)曲线C2的极坐标方程化为直角坐标方程,与C1的普通方程联立,求出C1与C2交点的直角坐标,由此能求出C1与C2交点的极坐标.【解答】解:(1)将,消去参数t,化为普通方程(x﹣4)2+(y﹣5)2=25,即C1:x2+y2﹣8x﹣10y+16=0,将代入x2+y2﹣8x﹣10y+16=0,得ρ2﹣8ρcosθ﹣10ρsinθ+16=0.∴C1的极坐标方程为ρ2﹣8ρcosθ﹣10ρsinθ+16=0.(2)∵曲线C2的极坐标方程为ρ=2sinθ.∴曲线C2的直角坐标方程为x2+y2﹣2y=0,。

2013高考语文试题及答案全国卷(word版)

2013年普通高等学校招生全国统一考试语文(全国卷)注意事项:1. 本试卷分第Ⅰ卷(阅读题)和第Ⅱ卷(表达题)两部分。

第Ⅰ卷1至8页,第Ⅱ卷9至10页。

2. 答题前,考生务必将自己的姓名、准考证号填写在本试题相应的位置。

3. 全部答案在答题卡上完成,答在本试题上无效。

4. 考试结束,将本试题和答题卡一并交回。

第Ⅰ卷阅读题甲必考题一、现代文阅读(9分,每小题3分)阅读下面的文字,完成1~3题老子其人其书的时代,自司马迁《史记》以来只有异说,清代学者崇尚考据,对此议论纷纷,如汪中作《老子考异》,力主老子为战国时人,益启争端。

钱穆先生说:“老子伪迹不彰,真相大白,则先秦诸子学术思想之系统条贯始终不明,其源流派别终无可言.”大家都期待这个问题有新的解决线索.过去对于古书真伪及年代的讨论,只能以材料证明纸上材料,没有其它的衡量标准,因而难有定论。

用来印证《老子》的古书,大多收到辨伪家的怀疑,年代确不可移的,恐怕要数到《林非子》。

《吕氏春秋》和《淮南子》,但这几木书戍书太晚,没有多少作用.近年战国秦汉简帛侠籍大黄出上,为学术界提供了许多前所未见的地下材料,这使我们有可能重新考虑《老子》的时代问题。

1973牛长沙马王堆三亏汉基出土的串书,内有《老子》两种版本,甲本字体比较早,不避汉高祖讳,应抄写于高祖即帝位前,乙本避高祖讳,可以抄写于文帝初。

这两本《老子》抄写年代都晚,无益于《老子》著作年代的推定,但乙本前面有《黄帝书》四篇,系。

黄”、“老”合抄之本,则从根本上改变了学术界对早期道家的认识。

郭沐若先生曾指出,道家都是以“发明黄老道德意”为其指归,故也可称之为黄老学派.《老子》和《黄帝书》是道家的经典,在汉初被妙写在《老子》前面的《黄帝书》显然在当时公众心目中已据有崇高位置,不会是刚刚撰就的作品。

同时,《黄帝书》与《申子》、《慎子》、《韩非子》等有许多共通文句,而申不害、慎到、韩非二人均曾学黄老之术,这些共通之处可认作对《皇帝书》的引用阐发。

天津2013年高考英语试题及答案(Word版)

2013年普通高等学校招生全国统一考试(天津卷)英语笔试一、单项选择1.————I’m going to Venice next week。

—--- 。

Carnival will be held then。

Have fun!A. You're crazyB. You're lucky C。

You’d better not D。

You never know2.If you are in trouble,Mike is always willing to a band.A。

lend B。

shake C。

wave D.want3。

I think watching TV every evening is a waste of time—-—-there are meaningful things to do。

A。

less B. more C。

the least D。

the most 4。

-———Mary’s been offered a job in a university,but she doesn’t want to take it。

--—- ? It’s a very good chance。

A。

Guess what B。

So what C. Who cares D。

How why 5. small, the company has about 1,000 buyers in over 30 countries。

A。

As B.If C。

Although D。

Once6。

We have launched another man-made satellite, is announced in today’s newspaper。

A。

that B.which C。

who D。

what7.While she was in Paris,she developed a for fine art。

2013高考语文试题汇编(答案)

2013年普通高等学校招生全国统一考试(新课标Ⅰ) (1)2013年普通高等学校招生全国统一考试(大纲卷) (2)2013年普通高等学校招生全国统一考试(新课标Ⅱ卷) (3)2013年北京市高考语文试题解析 (5)2013年普通高等学校招生全国统一考试(天津卷) (8)2013年普通高等学校招生全国统一考试(天津卷) ........................................................................................................... 错误!未定义书签。

2013年普通高等学校招生全国统一考试(安徽卷) (9)2013年普通高等学校招生全国统一考试(安徽卷) (10)2013年普通高等学校招生全国统一考试(福建卷) (11)2013年普通高等学校招生全国统一考试(广东卷) (12)2013年湖北省高考语文试卷及答案 (14)2013年湖南高考语文试卷及答案 (15)2013年普通高等学校招生全国统一考试(江苏) (16)2013高考语文江西卷(详解版) (18)2013年普通高等学校招生全国统一考试(辽宁卷) (20)2013高考语文山东卷(详解版) (22)2013年高考语文试题(四川卷)及答案 (24)2013年普通高等学校招生全国统一考试(浙江卷) (25)2013年普通高等学校招生全国统一考试(重庆卷) ...................................................................................................... 错误!未定义书签。

2013年普通高等学校招生全国统一考试(新课标Ⅰ)一、现代文阅读(9分,每小题3分)B 2. D 3.C二、古诗文阅读。

(36分)(一)文言文阅读(19分)4.A5.B6. D7.(1)在朝廷官员中最是年高德劭,皇上也诚心诚意任用他,诸位大臣没有敢望其项背的。

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2013普通高考预测题英语试题试卷说明:1.本试卷满分150分,分两个部分。

第一部分为选择题共90分,请在机读答题卡上作答,第二部分为主关题部分,请直接在试卷上书写答案;2.本试卷请在120分钟完成;第一部分选择题共90分第一节单项选择(10分每小题1分)请在A B C D四个选项中选择最能符合空白处的选项1. Not only polluted but crowded.A. was the city; were the streetsB. the city was; were the streetsC. was the city; the streets wereD. the city was; the streets were2. — What should I wear to attend his wedding party?— Dress you like.A. whatB. howeverC. whateverD. how3. The discovery of new evidence led to ________.A. the thief having caughtB. catch the thiefC. the thief being caughtD. the thief to be caught4.I think the reason why they play smart phone whenever they are the smart phone is getting cheaper and cheaper.A.are isB.不填 areC.is areD.不填 is5. — Would you mind if I park my car here?— _____A. Yes, please.B. Of course not. It’s not allowed here.C. I’d rather you didn’t actually.D. No, you’d better not.6. He stayed there for quite a long time, during ________ time he learned much spoken English.A. thatB. whenC. whichD.what7. in the crashed car, he couldn't move his legs even a bit.A. To trapB. TrappedC. TrappingD. To be trapped8. What _______ disappointment it is to see him lose the game since he spent quite ______ little time practising.A. a; aB. 不填;aC. a;不填D. 不填; 不填9. He has _____ his dream of becoming a scientist.A. come trueB. realisedC. madeD. recognised10. “Did Jim come?” “I don’t know. He _______ while I was out.”A. might have comeB. might comeC. mush have comeD. should have come第二节完型填空(30分每个1.5分)联系上下文选择最适合空白处的最佳选项Love cures people—both the ones who give it and the ones who receive it. ——Dr. Karl MenningerI was being interviewed by a senior manager for a major insurance company.I told him 11 that the main reason why I was 12 with them was my13 to keep my family in Boston. My wife of 26 years old had recentlydied of a heart attack. A(n) 14 in Boston would also help me reduce some of the pain of the loss. Bruce, the interviewer, was politely sympathetic, and didn’t probe(调查)any 15 .He acknowledged my loss and,with great respect, moved on to another 16 .After the next 17 of interview, Bruce took me to lunch with another manager. Then he asked me to take a 18 with him. He told methat he, 19 , had lost his wife. And he had also been married 20 yearsand had three children. In his sharing, I realized that he had 20 the samepain as I had, a pain that was almost 21 to explain to someone whohad not lost a loved 22 . He offered his business card and home phone number and 23 that, if I needed help or just wanted someone to 24 ,I should feel 25 to give him a call, no matter 26 I got the jobor not. He wanted me to know that he was there if I 27 needed any help.From that one act of kindness, when he had no 28 whether we wouldever see each other again, he helped me 29 one of life’s greatest losses. He turned the normal 30 business interview process into an actof caring and support for another person in a time of need11.A.friendly B.patiently C.carefully D.honestly 12.A.interviewing B.speaking C.employing D.knowing13.A.right B.need C.worry D.agreement14.A.wife B.period C.job D.experience 15.A.farther B.other C.further D.ahead16.A.puzzle B.thing C.subject D.object17.A.speaking B.round C.moment D.term18.A.rest B.break C.walk D.chance19.A.either B.too C.still D.once20.A.received B.gave C.wanted D.experienced 21.A.possible B.impossible C.unimaginable D.unthinkable 22.A.one B.girl C.wife D.lady23.A.suggested B.offered C.spoke D.said24.A.listen to B.refer to C.talk to D.explain to 25.A.sure B.glad C.convenient D.free26.A.when B.how C.why D.whether27.A.ever B.once C.never D.still28.A.idea B.news C.promise D.answer29.A.work out B.deal with C.figure out D.put out30.A.long B.short C.warm D.cold第三节阅读理解(共40分每小题2分)阅读下列短文,从每题所给的四个选项中,选出最佳选项。

(共20题,每小题2分,满分40分(A)The Museum of Childhood is spread over 4 floors, with the first floor acting like a balcony around the edge of the building, so you can look down over the ground floor central hall with the shop. Information Desk and Benugo Café.The museum of Childhood is housed in a large Victorian building in east London. It has been on this site since 1872. it has undergone refurbishment (翻新) and reopened in 2006 and an elevator now makes all floors accessible. This is not a quiet place and children are allowed to have fun here. Child safety is extremely important and a member of staff remains by the front door at all times. Also, mote the “Code of Behavior” notice, which include: children under 12 must be supervised by an adult; no eating in the galleries and no running.Toy exhibits are in glass cases and there are plenty of low –level exhibits for younger children to see. The glass cases have lots of thought –inspiring questions on them to encourage discussion between adults and children. When you or the children need some quiet time, there are sofas at either end of the second floor with reading books available.Pros: Many free activities for kidsCons: Can be too warm insideVisit Duration: 1.5 hours Opening Hours: 10:00 a.m. – 5:45 p.m.Last admission is 5:30 p.m.The Museum is closed on 25 and 26 December and 1 January every year. Admission: Admission the Museum is free. There is a small charge for some activities. 31.In the Museum of Childhood, .A.people can reach any floor by elevatorB.there are sofas at either end of the first floorC.there are exhibits on the museum historyD.the Information Desk can be found on each floor32.The Museum of Childhood may be attractive to .A.parents who only have children under 12B.children who are fond of toy exhibitsC.parents and children who need quiet timeD.kids who like playing games with their parents33.What is the disadvantage of the museum?A.Low – level exhibits are too boring.B.Children may feel slightly hot in it.C.Parents have to stay with their kids.D.No staff members attend to the kids.34.What information can we get about the museum?A.All the activities for children are free.B.The museum is located in west London.C.Children are allowed to enter after 5:30.D.The museum is unavailable on Christmas Day(B)Just as the English language has changed quickly in the previous century, so has the use of it.After the British Broadcasting Corporation (BBC) was founded in 1927, the particular style of speech of the BBC announcers was recognized as standard English or Received Pronunciation (RP) English. Now, most people still consider that the pronunciation and delivery of BBC announcers is the clearest and most understandable spoken English.English has had a strong relationship with classes and social positions. However, since the Second World War there has been a clear change of attitude towards speech snobbery(自我优越感), and marks of class distinction(区别) such as styles of speech have been gradually diminishing, especially in the younger generation.By the end of the 1960s, it had become clear that it was not necessary to speak standard English or even correct grammar to become popular, successful and rich. The fashionable speech of the day was no longer the special right of a special class but rather a defiant(反抗的) expression of classlessness.The greatest single influence of the shaping of the English language in modern times is the American English. Over the last 25 years the English used by many people, particularly by those in the media, advertising and show business, has become more and more mid-Atlantic in style, delivery and accent.In the 1970s, fashion favored careless pronunciation and a language full of jargon, slang and “in” words, a great quantity of which couldn’t be understood by the outside world. What is considered modern and fashionable in Britain today is often not the kind of English taught in schools and colleges.35. Which of the following is NOT true?A. The use of the English language has not changed much in the previous century.B. BBC announcers speak standard English.C. English has no relationship with classes and social positions now.D. Young people don’t like class distinction.36. What does the author imply by saying “there has been a clear change of attitude towards speech snobbery”?A. Everyone speaks English as the BBC announcers do.B. There has been a great change of attitudes towards the status of English.C. Most people don’t believe the way of speaking by BBC announcers is much better than that of other people.D. It is necessary to speak Standard English with correct grammar.37.According to the author, there was a trend in the US for young people ________.A. to speak Standard EnglishB. to speak English without class distinctionC. to speak old EnglishD. to speak English with grammatical mistakes38. It can be concluded from the passage that________.A. Standard English is taught only in schools and collegesB. young people are defiant because they refuse to speak Standard EnglishC. the English language has not been influenced by the American English in the last25 yearsD. there has been a great change in the English language in the previous century(C)The law has a great many rules, showing when and how far a man is to be punished, or if he should be made to hand over money or property to his neighbors, and so on. These rules are contained in books. A lawyer learns them mainly by reading books. He begins by doing little else than reading, and after he has prepared himself by three years’ study, still, he has to, almost every day, read more about some new questions which he has to answer.The power to use books, then, is a special skill which a would-be lawyer ought to possess. He ought to have enough flexibility(灵活性)to make it easy for him to collect ideas from printed words. He ought to have some readiness to find what a book contains, and something of an instinct(直觉)for where to look for what he wants.But although this is the power which he will first feel in need of, it is not the most important. A lawyer does not study law to recite it; he studies it to use it and act upon the rules which he has learned in real life. His business is to try cases in court and to advise people what to do in order to keep out or get out of trouble.49. The first thing a law student has to do is to _______.A. read booksB. hand over moneyC. practice lawD. answer questions40. The major business of a lawyer is ________.A. to discuss the material he has readB. to advise people who have law problemsC. to learn about real lifeD. to study the law41. What is the most important to a lawyer?A. To possess a lot of books.B. To have enough flexibility when collecting ideas from printed words.C. To be able to quickly find out what a book contains.D. To be able to use his book knowledge in the right way in his future business.42. According to the passage, a good lawyer should know how to ________.A. understand and use what he readsB. be effective in everythingC. collect ideas from different sourcesD. be effective in court(D)Have you ever heard about “cyberbullying”? This is a new kind of bullying. Teenage cyberbullies post lies and hateful things about other young people on the Internet in order to embarrass them and make them feel bad.It is getting to be a serious problem, and it is spreading very fast. The victims don't want to go out or go to only school and avoid talking to everyone because they worry that people have already seen the lies on the Internet.Teenagers bully other teenagers online, by sending insults or threats(侮辱或威胁)directly to them in emails or instant messages. Also, they spread hateful comments about a person through e-mail, instant messaging, or by posting on blogs or Web sites teenager often read.A teenage student from Canada was physically bullied for years. Finally, people began to bully him online too, sending him hateful e-mails telling him no one liked him. He didn't know how to get away from the bullies, so he left school and isolated (隔离)himself from everyone.Another example is when a group of middle school students bullied a teenage girl and started sending her unpleasant messages about her physical appearance. They called her names online and said many things to hurt her. She was very badly emotionally wounded.Some parents, teachers, and other adults are trying to stop it by talking to victims they know about. If victims can talk about these problems with someone, they can find ways to deal with cyberbullying.Parents of students need to check and see what their children are doing online. If they see their children bullying someone online, they need to help their children understand that they are seriously hurting their victims and see that it is wrong.Recently I read that thousands of students and teachers in Michigan attended a conference about stopping bullying. They all worked together and discussed ways to stop bullying in their schools.43. Cyber bullying is__________.A. a kind of net cheat and jokesB. a way of telling lies merelyC. an online insult or threatD. a new method of grading44. What can we learn about the two teenagers mentioned in Paragraphs 4 and 5?A. They both suffered from online hateful insults.B. They were both physically bullied for years.C. They both had to leave school for a while.D. They were both bullied by their classmates.45. Cyberbullying can’t possibly arise_________.A. so long as schools don’t give free access to the InternetB. if teachers charge the students with adequate schoolworkC. if cyber users raise their awareness of their respect for othersD. if parents are strict enough with their children in their online time..46. What can you learn from the passage?A. Cyberbullying has been under control.B. The influence of Cyberbullying is getting around fast.C. Teenagers mean no harm by cyberbullying.D. Victims are ready to turn to other people for help.(E)Kids will often ignore your requests for them to shut off the TV, start their chores(杂事),or do their homework as a way to avoid following your directions. Before you know it, you’ve started to sound like a broken record as you repeatedly ask them to do their assignments, clean their room, or take out the trash. Rather than saying “Do your chores now.”you’ll be more effective if you set a target time for when the chores have to be completed. So instead of arguing about starting chores, just say,“If chor es aren’t done by 4 pm, here are the consequences.”Then it’s up to your child to complete the chores. Put the ball back in their court. Don’t argue or fight with them, just say,“That’s the way it’s going to be.”It shouldn’t be punitive(惩罚性的)as much as it should be persuasive.“If your chores aren’t done by 4 pm, then no video game time until chores are done. And if finishing those chores runs into homework time, that’s going to be your loss.”On the other hand, when dealing with homework, keep it very simple. Have a time when homework starts, and at that time, all electronics go off and do not go back on until you see that their homework is done. If your kids say they have no homework, then they should use that time to study or read. Either way, there should be a time set aside when the electronics are off.When a kid wears his iPod or headphones when you’re trying to talk to him,make no bones about it;he is not ignoring you, he is disrespecting you. At that point, everything else should stop until he takes the earplugs out of his ears. Don’t try to communicate with him when he’s wearing headphones—even if he tells you he can hear you. Wearing them whil e you’re talking to him is a sign of disrespect.Parents should be very tough about this kind of thing. Remember, mutual respect becomes more important as children mature.47.According to the passage, it seldom happens that ________.A.kids turn a deaf ear to their parents’ requestsB.parents’ directions sound like a broken recordC.children are ready to follow their parents’ directionsD.parents are unaware of what they are repeating to their kids 48.Parents will be able to deal with their children more effectively if they ________.A.avoid direct ways of punishmentB.make them do things at their requestC.argue and fight with their childrenD.allow their children to behave in their own way49.It can be inferred from the passage that ________.A.parents should take off his headphones when trying to have a talk with their childB.it will make no difference that a kid is wearing his earplugs while talking to his parentsC.parents shouldn’t give in to their kid when he shows no sign of respect D.kids’ purposely talking to their parents with iPod gives them a sense of power andcontrol50.The main idea of the passage is ________.A.that respecting each other is more important than anything elseB.how kids behave to ignore and disrespect their parentsC.that children should make choices and decisions on their ownD.how parents can deal with their kids’ behavior effectively第四节补全对话(10分每小题2分)读下面一段对话,从所给出的选项中选出最符合语境的一项Man:The time has come to say goodbye.Woman:So soon. 51 .Man: 52 .Woman:It certainly has been a pleasure seeing you again and talking about old days. Man: 53 . And I really want to thank you for spending so much time showing me the sights,Woman: 54 .It gave me a chance to get away from my everyday work and do something a little different.Man: 55 .Woman:Oh, yes, That’s our present plan unless something bad comes up. I should be there in early September.Man:I’ll be expecting you.A.I’ve had a nice time.B.It seems as if you just got here.C.But I have to leave here now.D.Oh, it was fun for me , too.E.Are you sure you can make it?F.I feel that way, too.G.Will you be out to see me next year as you promised?第二部分作文与表达(共60分)第一节阅读表达(共10分每小题2分)阅读下面一段文字,回答下列问题Environmentally friendly gifts are perfect for birthdays, anniversaries, and other holidays. The quality of environmentally friendly gifts is usually comparable to or even better than the quality of gifts that are not as environmentally friendly. Because there are such a wide variety of environmentally friendly gifts available on the market, you have many great ideas to choose from. You’ll be able to buy gifts for everyone on your list and do your part to .You can actually give many gifts that have been made from recycled paper. The quality of recycled paper is just as good as the paper made from raw1 materials, so you can give a quality gift and still do your part to protect the environment. This recycled paper is turned into gifts like journals, diaries, note cards, and similar items. Just because they’re made from recycled paper doesn’t mean that they’re going to be plain or boring. Recycled paper can be dyed many colors and gifts can be made in a wide variety of styles and textures. With the wide variety of gifts available, you’re sure to find something that your fri ends and family members will love.Clothing featuring plants, animals, and flowers is popular among nature lovers. If you are able to visit a store offering nature clothing, you’ll have a wide selection to choose from. Jackets, shirts, and other clothing items can have nature scenes printed on them for your favorite nature lover. If you want to personalize2 these items, you can even have the receiver’s name printed on each item to make an extra-special gift. Many of these items can be made with natural fibers for an environmentally friendly gift that your loved one will enjoy for years.Giving environmentally friendly gifts is a great way to show friends and family members with a love of nature that you care about their interests. Even the simplest gift will be appreciated if it shows that you thought about what your loved one enjoys and gave them something that fits in with their environmentally friendly lifestyle. Not only will you be giving a wonderful gift that your loved ones can enjoy for a long time, you’ll also be doing your part to protect the environment.提示:1. raw adj未加工的, 处于自然状态的2. personalize v.为个人特制1. What is the best title for the passage? (Please answer within 10 words.)_______________________________2. Which sentence in the passage can be replaced by the following one? There are so many gifts of this kind that you can surely find something perfect for your family or your friends._______________________________3. Please fill in the blank with the proper words or phrases to complete the sentence. (Please answer within 10 words.)_______________________________4. Who should the environmentally friendly gifts be given to? Why? (Please answer within 30 words.)_______________________________5. Translate the underlined sentence in the last paragraph into Chinese.第二节短文改错假如在英语课上,英语老师要求你修改如下文章,请你按照规范进行修改The National People's Congress (NPC) of People's Republic of China is the higher organ of the state power. The Standed Committee of the NPC is permanent organ of the NPC. The terms of office of the NPC and its Standing Committee are five year. The NPC and its Standing Committee was empowering with the rights of legislation, decision, supervision, election and removal第三节作文请就学生目前过重的课业负担写一篇120字左右的文章。

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