河北省邯郸市2018届高三摸底考试
【河北省邯郸市】2018届高三上学期摸底考试语文试卷及答案
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河北省邯郸市2018届高三上学期摸底考试语文试卷一、现代文阅读(35分)(一)论述类文本阅读(本题共3小题,9分)阅读下面的文字,完成1〜3题。
社会治理智能化,就是在网络化和网络空间基础上,通过大数据、云计算物联网等信息技术,重构社会生产与社会组织彼此关联的形态,使社会治理层次和水平得到提升,使治逻过柱更加优化、更加科学、更加智慧。
要在促进大数据与社会治理深入融合中提升社会治理智能化水平,我们就必须准确把掘“互联网十”时代社会治理创新面临的新形势新要求,并充分应用强大数据资源。
社会治理内嵌于社会结构之中,随着社会结构的变化,社会治理的对象、主体也必然发生相应的变化。
一方面,当前社会治理所服务的对象结构发生了显著变化。
具体表现为:社会阶层结构出现新老演化,人口的年龄结构、素质结构和空间分布结构发生了很大变动,老龄化社会加速到来,受过高等教育或拥有专业技能的群体日益扩大,家庭结构呈现规模小型化、类型多样化特征,社会流动性不断增强,跨地区流动已成为常态,越来越多的人口向大城市或中心城镇集聚。
特别是思想活跃、利益诉求多样的新兴社会阶层对创新社会治理提出了很多新课題。
另一方面,参与社会治理的主体,也从政府单一主体过渡到一个由政府、非政府纽织、公众个体等构成的行动者系统。
这些变化迫切需要提升社会治理智能化水平,也为社会治理智能化创造了良好条件。
智能化意味着精准分析、精准治理、精准服务、精准反馈。
各类社会治理主体通过获取、存播、管理、分析等手段,将具有海规模、快速流梓等特征的大数据究成法数据,广泛应用于社会治理领域,更好地服务不同社会群体,将成为政府和社会组织实施智能治理的重要法宝。
长期以来,社会治理面临的最大难題就是风险的不可控性和难以预见性。
现代社会处于信息化和网络化复杂交织的图景之中,与我国经济转轨、社会转型的背景相叠加,使现代社会治理呈现出新特征。
总的来看,我国经济社会发展面临的形势是严峻复杂的,表现为短期矛矛盾和长期矛盾叠加、结构性因素和周期性因素并存、传统安全和非传统安全威胁相互交织,特别是我国基层社会治理体系较为薄弱等等。
河北省邯郸市2018届高三上学期摸底考试物理试题 精品
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一、选择题:本题共10小题,共40分。
在每小题给出的四个选项中,第1~6题只有一项符合题目要求,第7~10题有多项符合题目要求。
全部选对的得4分,选对但不全的得2分,有选错的得0分。
1.在光滑水平面上一辆平板车,一人手握大锤站在车上。
开始时人、锤和车均静止且这三者的质量依次为1m 、2m 、3m 。
人将大锤水平向左抛出后,人和车的速度大小为v ,则抛出瞬间大锤的动量大小为( )A. v m 1B. v m 3C. v m m )(31+D.v m m )(32+2.如图所示,某空间存在正交的匀强电场和匀强磁场,三个速度不同的质子沿垂直电场线和磁感线方向从同一点射入场区,其轨迹为图中Ⅰ、Ⅱ、Ⅲ进入磁场区时速度大小分别为1v 、2v 、3v ,射出磁场区时速度大小分别为1v '、2v '、3v ',不计质子所受的重力,下列判断正确的是( )A.321>>v v v ,11>v v ',33>v v 'B.321>>v v v ,11<v v ',33<v v ' C.321<<v v v ,11>v v ',33>v v ' D.321<<v v v ,11<v v ',33<v v ' 3.在某些恒星内,3个α粒子结合成1个C 126,C 126原子的质量是12.0000u ,He 42原子的质量是4.0026u ,已知kg u 271066.11-⨯=,则此核反应中释放的核能约为( )A. 7.266eVB. 5.266eVC.J 91016.1-⨯D.J 121016.1-⨯4.“天舟一号”货运飞船于2018年4月20日在文昌航天发射中心成功发射升空,并于4月22日与“天宫二号”空间实验室对接前。
已知对接前“天舟一号”在距离地面高度为h 的圆轨道上做匀速圆周运动,“天宫二号”空间站】实验室在距离地面高度为h+Δh 的圆轨道上做匀速圆周运动,已知Δh>0,且h+Δh 小于地球同步卫星的高度,则( )A.二者的角速度均小于地球自转角速度B.“天舟一号”线速度小于“天宫二号”空间实验室的线速度C.二者运动周期均小于地球自转周期D.二者做圆周运动的向心加速度大于地面上物体的重力加速度。
河北省邯郸市高三英语上学期摸底考试试题
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2018届河北省邯郸市高三上学期摸底考试英语第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
例:How much is the shirt?A. £ 19.15.B. £ 9.18.C. £9.15.答案是C。
1. What will the woman do today?A. Visit a friend.B. Do an experiment.C. Attend a lecture.2. What does the man find it difficult to do?A. Fix a toy train.B. Understand the instructions.C. Put together the folding table.3. Where is the woman’s cell phone?A. In the classroom.B. In her bag.C. In the dining hall.4. When does the woman need the book?A. On April 3rd.B. On April 2nd.C. On April 1st.5. What does the man mean?A. Few people read his article.B. The woman can’t convince him.C Most readers don’t agree with him.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对活或独白。
每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
河北省邯郸市2018届高三上学期摸底考试语文试题(含答案)
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河北省邯郸市2018届高三上学期摸底考试语文试题一、现代文阅读(35分)(一)论述类文本阅读(本题共3小题,9分)阅读下面的文字,完成1?3题。
社会治理智能化,就是在网络化和网络空间基础上,通过大數据、云计算物联网等信息技术,重构社会生产与社会组织彼此关联的形态,使社会治理层次和水平得到提升,使治逻过柱更加优化、更加科学、更加智慧。
要在促进大数据与社会治理深入融合中提升社会治理智能化水平,我们就必须准确把掘“互联网十”时代社会治理创新面临的新形势新要求,并充分应用妤大數据资源。
社会治理内嵌于社会结构之中,随着社会结构的变化,社会治理的对象、主体也必然发生相应的变化。
一方面,当前社会治理所服务的对象结构发生了显著变化。
具体表现为:社会阶层结构出现新老演化,人口的年龄结构、素质结构和空间分布结构发生了很大变动,老龄化社会加速到来,受过高等教育或拥有专业技能的群体日益扩大,家庭结构呈现规模小型化、类型多样化特征,社会流动性不断增强,跨地区流动已成为常态,越来越多的人口向大城市或中心城镇集聚。
特别是思想活跃、利益诉求多样的新兴社会阶层对创新社会治理提出了很多新课題。
另一方面,参与社会治理的主体,也从政府单一主体过渡到一个由政府、非政府纽织、公众个体等构成的行动者系统。
这些变化迫切需要提升社会治理智能化水平,也为社会治理智能化创造了良好条件。
智能化意味着精准分析、精准治理、精准服务、精准反馈。
各类社会治理主体通过获取、存播、管理、分析等手段,将具有海规模、快速流梓等特征的大数据究成法數据,广泛应用于社会治理领域,更好地服务不同社会群体,将成为政府和社会组织实施智能治理的重要法宝。
长期以来,社会治理面临的最大难題就是风险的不可控性和难以预见性。
现代社会处于信息化和网络化复杂交织的图景之中,与我国经济转轨、社会转型的背景相叠加,使现代社会治理呈现出新特征。
总的来看,我国经济社会发展面临的形势是严峻复杂的,表现为短期矛矛盾和长期矛盾叠加、结构性因素和周期性因素并存、传统安全和非传统安全威胁相互交织,特别是我国基层社会治理体系较为薄弱等等。
河北省邯郸市高三英语上学期摸底考试试题
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2018届河北省邯郸市高三上学期摸底考试英语第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
例:How much is the shirt?A. £ 19.15.B. £ 9.18.C. £9.15.答案是C。
1. What will the woman do today?A. Visit a friend.B. Do an experiment.C. Attend a lecture.2. What does the man find it difficult to do?A. Fix a toy train.B. Understand the instructions.C. Put together the folding table.3. Where is the woman’s cell phone?A. In the classroom.B. In her bag.C. In the dining hall.4. When does the woman need the book?A. On April 3rd.B. On April 2nd.C. On April 1st.5. What does the man mean?A. Few people read his article.B. The woman can’t convince him.C Most readers don’t agree with him.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对活或独白。
每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
河北省邯郸市2018届高三上学期摸底考试数学(理)试题(解析版)
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高三数学试卷(理科)第Ⅰ卷(共60分)一、选择题:本大题共12个小题,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. 已知集合,,则()A. B. C. D.【答案】C【解析】由题意可得:,则.本题选择C选项.2. 若复数满足,则复数的实部与虚部之和为()A. -2B. 2C. -4D. 4【答案】B【解析】由题意可得:,则实部与虚部之和为.本题选择B选项.3. 在中,若,则()A. B. C. D.【答案】A【解析】由题意可得:,则。
本题选择A选项.4. 分别是双曲线的左、右焦点,为双曲线右支上一点,且,则的周长为()A. 15B. 16C. 17D. 18【答案】D【解析】由双曲线的方程可知:,则,据此可知的周长为.本题选择D选项.点睛:双曲线定义的集合语言:P={M|||MF1|-|MF2||=2a,0<2a<|F1F2|}是解决与焦点三角形有关的计算问题的关键,切记对所求结果进行必要的检验5. 用电脑每次可以从区间内自动生成一个实数,且每次生成每个实数都是等可能性的,若用该电脑连续生成3个实数,则这3个实数都大于的概率为()A. B. C. D.【答案】C【解析】由题意可得:每个实数都大于的概率为,则3个实数都大于的概率为.本题选择C选项.6. 如图,网格纸上小正方形的边长为1,粗线画出的是某几何体的三视图,已知该几何体的各个面中有个面是矩形,体积为,则()A. B. C. D.【答案】D【解析】由三视图可知,该几何体为直五棱柱,底面为俯视图所示,高为2,故.本题选择D选项.点睛:在由三视图还原为空间几何体的实际形状时,要从三个视图综合考虑,根据三视图的规则,空间几何体的可见轮廓线在三视图中为实线,不可见轮廓线在三视图中为虚线.在还原空间几何体实际形状时,一般是以正视图和俯视图为主,结合侧视图进行综合考虑.7. 若,则()A. B. C. D.【答案】C【解析】本题选择C选项.8. 设函数的导函数为,若为偶函数,且在上存在极大值,则的图像可能为()A. B.C. D.【答案】C【解析】若为偶函数,则为奇函数,故排除B、D.又在上存在极大值,故排除A选项,本题选择C选项.9. 我国古代名著《庄子·天下篇》中有一句名言“一尺之棰,日取其半,万世不竭”,其意思为:一尺的木棍,每天截取一半,永远都截不完,现将该木棍依此规律截取,如图所示的程序框图的功能就是计算截取7天后所剩木棍的长度(单位:尺),则①②③处可分别填入的是()【答案】B【解析】一共取了7次,,A、C、D不能完成功能,B能完成功能.本题选择B选项.点睛:识别、运行程序框图和完善程序框图的思路(1)要明确程序框图的顺序结构、条件结构和循环结构.(2)要识别、运行程序框图,理解框图所解决的实际问题.(3)按照题目的要求完成解答并验证.10. 已知函数,点是平面区域内的任意一点,若的最小值为,则的值为()A. -1B. 0C. 1D. 2【答案】A【解析】由函数的解析式可得:,结合题意可得目标函数在给定的可行域内的最小值为,可行域的顶点坐标为,结合目标函数的几何意义可得目标函数在点处取得最小值,即:,解得:.本题选择A选项.点睛:由于约束条件中存在参数,所以可行域无法确定,此时一般是依据所提供的可行域的面积或目标函数的最值,来确定含有参数的某不等式所表示的坐标系中的某区域,从而确定参数的值11. 若函数恰有4个零点,则的取值范围为()A. B.C. D.【答案】B【解析】设,作出这两个函数在上的图像,如图所示,在上的零点为,在上的零点为,数形结合可得,.本题选择B选项.点睛:(1)问题中参数值影响变形时,往往要分类讨论,需有明确的标准、全面的考虑;(2)求解过程中,求出的参数的值或范围并不一定符合题意,因此要检验结果是否符合要求.12. 直线与抛物线相交于两点,,给出下列4个命题:的重心在定直线上;的最大值为;的重心在定直线上;的最大值为.其中的真命题为()A. B. C. D.【答案】A【解析】联立直线与抛物线的方程整理可得:,结合题意可得:,且:,则△ABC的重心坐标为,的重心在定直线上;由弦长公式可得:,据此可得:,令,则,据此可得函数在区间上单调递增,在区间上单调递减,,据此可得:的最大值为;本题选择A选项.第Ⅱ卷(共90分)二、填空题(每题5分,满分20分,将答案填在答题纸上)13. 在中,若,则__________.【答案】【解析】由正弦定理可得:,不妨设,则.14. 若,则__________.【答案】【解析】由对数的运算法则可得:,且:,据此可得:.15. 若的展开式中的系数为20,则__________.【答案】【解析】由题意可得:则含有的项为:,则的系数为:,解得:.16. 已知一个四面体的每个顶点都在表面积为的球的表面上,且,,则__________.【答案】【解析】由题意可得,该四面体的四个顶点位于一个长方体的四个顶点上,设长方体的长宽高为,由题意可得:,据此可得:,则球的表面积:,结合解得:.三、解答题(本大题共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. 在等差数列中,,公差.记数列的前项和为. (1)求;(2)设数列的前项和为,若成等比数列,求.【答案】(1);(2) .【解析】试题分析:(1)由题意可求得数列的首项为1,则数列的前n项和.(2)裂项可得,且,据此可得.试题解析:(1)∵,∴,∴,∴,∴,.(2)若成等比数列,则,即,∴,∵,∴.点睛:使用裂项法求和时,要注意正负项相消时消去了哪些项,保留了哪些项,切不可漏写未被消去的项,未被消去的项有前后对称的特点,实质上造成正负相消是此法的根源与目的.18. 如图,在底面为矩形的四棱锥中,.(1)证明:平面平面;(2)若异面直线与所成角为,,,求二面角的大小.【答案】(1)证明见解析;(2) .【解析】试题分析:(1)由题意结合几何关系可证得平面,结合面面垂直的判断定理即可证得平面平面.(2)建立空间直角坐标系,结合半平面的法向量可得二面角的大小是.试题解析:(1)证明:由已知四边形为矩形,得,∵,,∴平面.又,∴平面.∵平面,∴平面平面.(2)解:以为坐标原点,建立如图所示的空间直角坐标系.设,,则,,,,所以,,则,即,解得(舍去).设是平面的法向量,则,即,可取.设是平面的法向量,则即,可取,所以,由图可知二面角为锐角,所以二面角的大小为.19. 共享单车是指企业在校园、地铁站点、公交站点、居民区、商业区、公共服务区等提供自行车单车共享服务,是共享经济的一种新形态,一个共享单车企业在某个城市就“一天中一辆单车的平均成本(单位:元)与租用单车的数量(单位:车辆)之间的关系”进行调查研究,在调查过程中进行了统计,得出相关数据见下表:每天一辆车平均成本根据以上数据,研究人员分别借助甲、乙两种不同的回归模型,得到两个回归方程,方程甲:,方程乙:.(1)为了评价两种模型的拟合效果,完成以下任务:①完成下表(计算结果精确到0.1)(备注:,称为相应于点的残差(也叫随机误差));每天一辆车平均成本估计值估计值②分别计算模型甲与模型乙的残差平方和及,并通过比较,的大小,判断哪个模型拟合效果更好.(2)这个公司在该城市投放共享单车后,受到广大市民的热烈欢迎,共享单车常常供不应求,于是该公司研究是否增加投放,根据市场调查,这个城市投放8千辆时,该公司平均一辆单车一天能收入10元,6元收入的概率分别为0.6,0.4;投放1万辆时,该公司平均一辆单车一天能收入10元,6元收入的概率分别为0.4,0.6,问该公司应该投放8千辆还是1万辆能获得更多利润?(按(1)中拟合效果较好的模型计算一天中一辆单车的平均成本,利润=收入-成本). 【答案】(1)模型乙的拟合效果更好;(2) 1万辆. 【解析】试题分析:(1)由题意完成表格,计算残差平方和可得,,则模型乙的拟合效果更好.(2)分别计算投放量为8千辆和1万辆时公司一天获得的总利润可得投放1万辆能获得更多利润,应该增加到投放1万辆. 试题解析:(1)①经计算,可得下表:②,,,故模型乙的拟合效果更好.(2)若投放量为8千辆,则公司一天获得的总利润为元,若投放量为1万辆,由(1)可知,每辆车的成本为(元)所以公司一天获得的总利润为(元)因为,所以投放1万辆能获得更多利润,应该增加到投放1万辆.学+科+网...学+科+网...学+科+网...学+科+网...学+科+网...学+科+网...学+科+网...20. 如图,设椭圆:的离心率为,分别为椭圆的左、右顶点,为右焦点,直线与的交点到轴的距离为,过点作轴的垂线,为上异于点的一点,以为直径作圆.(1)求的方程;(2)若直线与的另一个交点为,证明:直线与圆相切.【答案】(1) ;(2)证明见解析.【解析】试题分析:(1)结合题意可求得,,则的方程为.(2)由题意可得,直线与圆相切时,直线的斜率为,结合(1)中求得的椭圆方程即可证得题中的结论. 试题解析:(1)解:由题可知,,∴,,设椭圆的方程为,由,得,∴,,,故的方程为.(2)证明:由(1)可得:,设圆的圆心为,则,圆的半径为,直线的方程为.设过与圆相切的直线方程为,则,整理得:,由,得,又∵,∴直线与圆相切.21. 已知函数的图象在处的切线过点.(1)若函数,求的最大值(用表示);(2)若,证明:.【答案】(1) ;(2)证明见解析.【解析】试题分析:(1)由题意可得:.结合导函数研究函数的单调性可得.(2)由题意结合(1)的结论有,构造函数,结合函数的特征即可证得题中的结论.试题解析:(1)由,得,的方程为,又过点,∴,解得.∵,∴,当时,,单调递增;当时,,单调递减.故.(2)证明:∵,∴,,∴令,,,令得;令得.∴在上递减,在上递增,∴,∴,,解得:.请考生在22、23两题中任选一题作答,如果多做,则按所做的第一题记分.22. 选修4-4:坐标系与参数方程在极坐标系中,曲线的极坐标方程为,点,以极点为原点,以极轴为轴的正半轴建立平面直角坐标系,已知直线(为参数)与曲线交于两点,且.(1)若为曲线上任意一点,求的最大值,并求此时点的极坐标;(2)求.【答案】(1) ,;(2) .【解析】试题分析:(1)利用题意结合辅助角公式可得当时,取得最大值,此时,的极坐标为.(2)联立直线的参数方程和圆的直角坐标方程,结合韦达定理可得的值是.试题解析:(1),,∴当时,取得最大值,此时,的极坐标为.(2)由,得,即,故曲线的直角坐标方程为.将代入并整理得:,解得,∵,∴由的几何意义得,,,故.23. 选修4-5:不等式选讲已知函数.(1)求不等式的解集;(2)若函数的图像在上与轴有3个不同的交点,求的取值范围.【答案】(1) ;(2) .【解析】试题分析:(1)利用不等式的特点零点分段可得不等式的解集为(2)令,结合函数的图象和题意可得的取值范围是.试题解析:(1)由,得,∴或或,解得,故不等式的解集为.(2),当时,,当且仅当,即时取等号,∴,当时,递减,由,得,又,结合的图像可得.。
【全国大联考】【河北(新课标Ⅰ)】河北省邯郸市2018届高三上学期摸底考试化学试题(解析版)
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题型考点分布:河北省邯郸市2018届高三上学期摸底考试化学试题1.化学与生活密切相关,下列说法不正确的是A.将燃油车改为燃气车有利于减少污染物排放B.含氟牙膏能有效预防龋齿,且其中所加氟化物的量宜多不宜少C.糖类、油脂及蛋白质都含有C、H、O元素D.酸雨浸蚀的土壤中加入熟石灰,可对该土壤进行修复2.下列生活用品的主要成分是蛋白质的是A.羽绒被B.有机玻璃C.面巾纸D.涤纶布【答案】A【解析】生活用品主要成分是蛋白质是羽绒被,故A正确。
3.在水溶液中能大量共存的一组粒子是A.S2O32-、Na+、Cl-、H+B.Na+、Mg2+、Cl--、SO42-C.Ba2+、Na+、SO42-、OH-D.K+、C2H5OH、MnO4-、H+【答案】B【解析】A、S2O32-与H+发生:S2O32-+2H+=S↓+SO2↑+H2O,不能大量共存,故A错误;B、能大量共存,故B正确;C、Ba2+与SO42-发生Ba2++SO42-=BaSO4↓,不能大量共存,故C错误;D、乙醇被酸性高锰酸钾溶液氧化成乙酸,不能大量共存,故D错误。
4.《验方新编》中有如下叙述:“凡暑月身热昏沉,未明症候,恐是出丹。
用生白扁豆数粒食之.如不知腥味,则以生白扁豆、水泡湿,研汁一小杯,调水一盏服之。
即愈。
”文中涉及的操作方法是A.蒸馏B.升华C.浸取D.分液【答案】C【解析】以生白扁豆、水泡湿,这是浸取的方法,故C正确。
5.下列关于有机物的说法正确的是A.石油的催化重整、煤的干馏均可以得到芳香烃B.乙酸、草酸均不能使酸性高锰酸钾溶液褪色C.新戊烷的所有碳原子位于同一平面D.奶油、鸡蛋清均属于高分子化合物且均能发生水解反应【答案】A【解析】A、石油催化重整,可以使链状的烃转化为环状烃,煤的干馏得到出炉煤气、煤焦油、焦炭,其中出炉煤气和煤焦油含有芳香烃,故A正确;B、乙酸不能使酸性高锰酸钾溶液褪色,但草酸可以,发生5H2C2O4+2MnO4-+6H+=10CO2↑+2Mn2++8H2O,故B错误;C、新戊烷的结构简式为C(CH3)4,甲烷空间结构为正四面体,4个-CH3取代甲烷4个H的位置,因此新戊烷中所有碳原子没有共面,故C 错误;D、奶油的成分是油脂,不属于高分子化合物,故D错误。
河北省邯郸市2018届高三上学期摸底考试数学理试题Word版含答案
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河北省邯郸市2018届高三上学期摸底考试数学理试题Word版含答案高三数学试卷(理科)第Ⅰ卷(共60分)一、选择题:本大题共12个小题,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.已知集合2{|20}A x x x =-->,{|0}B x x =>,则AB =()A .(1,2)B .(0,2)C .(2,)+∞D .(1,)+∞ 2.若复数z 满足(1)23i z i -=+,则复数z 的实部与虚部之和为() A .-2 B .2 C .-4 D .43.在ABC ?中,若4AB AC AP +=,则PB =()A .3144AB AC - B .3144AB AC -+ C .1344AB AC -+ D .1344AB AC -4. 12,F F 分别是双曲线C :22197x y -=的左、右焦点,P 为双曲线C 右支上一点,且1||8PF =,则12PF F ?的周长为()A . 15B .16 C. 17 D .185.用电脑每次可以从区间(0,1)内自动生成一个实数,且每次生成每个实数都是等可能性的,若用该电脑连续生成3个实数,则这3个实数都大于13的概率为() A .127 B .23 C. 827 D .496.如图,网格纸上小正方形的边长为1,粗线画出的是某几何体的三视图,已知该几何体的各个面中有n 个面是矩形,体积为V ,则()A .4,10n V ==B .5,12n V == C. 4,12n V == D .5,10n V ==7.若sin()2(sin 2cos )4πααα+=+,则sin2α=()A .45-B .45 C. 35- D .358. 设函数()f x 的导函数为'()f x ,若()f x 为偶函数,且在(0,1)上存在极大值,则'()f x 的图像可能为()A .B .C. D .9. 我国古代名著《庄子·天下篇》中有一句名言“一尺之棰,日取其半,万世不竭”,其意思为:一尺的木棍,每天截取一半,永远都截不完,现将该木棍依此规律截取,如图所示的程序框图的功能就是计算截取7天后所剩木棍的长度(单位:尺),则①②③处可分别填入的是()10.已知函数2()1f x ax bx =-+,点(,)a b 是平面区域201x y x m y +-≤??≥??≥-?内的任意一点,若(2)(1)f f -的最小值为-6,则m 的值为()A . -1B . 0 C. 1 D .211. 若函数sin(2),6()cos(2),62x x m f x x m x ππππ?--≤恰有4个零点,则m 的取值范围为()A .11(,](,]126123ππππ-- B .1125(,](,](,]123126123ππππππ---- C. 11[,)[,)126123ππππ-- D .1125[,)[,)[,)123126123ππππππ----12.直线y x a =+与抛物线25(0)y ax a =>相交于,A B 两点,(0,2)C a ,给出下列4个命题:1P :ABC ?的重心在定直线730x y -=上;2p :||3AB a -2103p :ABC ?的重心在定直线370x y -=上;4p :||3AB a -5其中的真命题为()A .12,p pB .14,p p C. 23,p p D .34,p p第Ⅱ卷(共90分)二、填空题(每题5分,满分20分,将答案填在答题纸上)13.在ABC ?中,若sin :sin :sin 3:4:6A B C =,则cos B = . 14.若2332log (log )log (log )2x y ==,则x y += . 15.若5 ()(12)x a x ++的展开式中3x 的系数为20,则a = .16.已知一个四面体ABCD的每个顶点都在表面积为9π的球O的表面积,且AB CD a==,5AC AD BC BD====,则a=.三、解答题(本大题共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. 在等差数列{}na中,3412a a+=,公差2d=,记数列21{}na-的前n项和为nS.(1)求nS;(2)设数列1{}n nna S+的前n项和为nT,若25,,ma a a成等比数列,求mT.18. 如图,在底面为矩形的四棱锥P ABCD-中,PB AB⊥.(1)证明:平面PBC⊥平面PCD;(2)若异面直线PC与BD所成角为60,PB AB=,PB BC⊥,求二面角B PD C--的大小.19. 共享单车是指企业在校园、地铁站点、公交站点、居民区、商业区、公共服务区等提供自行车单车共享服务,是共享经济的一种新形态,一个共享单车企业在某个城市就“一天中一辆单车的平均成本(单位:元)与租用单车的数量(单位:车辆)之间的关系”进行调查研究,在调查过程中进行了统计,得出相关数据见下表:租用单车数量x(千辆) 2 3 4 5 8每天一辆车平均成本y(元) 2根据以上数据,研究人员分别借助甲、乙两种不同的回归模型,得到两个回归方程,方程甲:(1)4 1.1yx=+,方程乙:(2)26.41.6yx=+.(1)为了评价两种模型的拟合效果,完成以下任务:①完成下表(计算结果精确到)(备注:i i ie y y=-,ie称为相应于点(,)i ix y的残差(也叫随机误差));租用单车数量x (千辆) 2 3 4 5 8 每天一辆车平均成本y (元)2模型甲估计值(1)i y残差(1)ie0 模型乙估计值(2)iy2残差(2)ie0 0②分别计算模型甲与模型乙的残差平方和1Q 及2Q ,并通过比较1Q ,2Q 的大小,判断哪个模型拟合效果更好.(2)这个公司在该城市投放共享单车后,受到广大市民的热烈欢迎,共享单车常常供不应求,于是该公司研究是否增加投放,根据市场调查,这个城市投放8千辆时,该公司平均一辆单车一天能收入10元,6元收入的概率分别为,;投放1万辆时,该公司平均一辆单车一天能收入10元,6元收入的概率分别为,,问该公司应该投放8千辆还是1万辆能获得更多利润?(按(1)中拟合效果较好的模型计算一天中一辆单车的平均成本,利润=收入-成本).20. 如图,设椭圆C :22221(0)x y a b a b +=>>的离心率为12,,A B 分别为椭圆C 的左、右顶点,F 为右焦点,直线6y x =与C 的交点到y 轴的距离为27,过点B 作x 轴的垂线l ,D 为l 上异于点B 的一点,以BD 为直径作圆E .(1)求C 的方程;(2)若直线AD 与C 的另一个交点为P ,证明:直线PF 与圆E 相切.21. 已知函数21()ln 12f x x ax bx =-++的图像在1x =处的切线l 过点11(,)22. (1)若函数()()(1)(0)g x f x a x a =-->,求()g x 的最大值(用a 表示);(2)若4a =-,121212()()32f x f x x x x x ++++=,证明:1212x x +≥. 请考生在22、23两题中任选一题作答,如果多做,则按所做的第一题记分.22.选修4-4:坐标系与参数方程在极坐标系中,曲线 C 的极坐标方程为2cos 2sin (02)ρθθθπ=+≤<,点(1,)2M π,以极点O 为原点,以极轴为x 轴的正半轴建立平面直角坐标系,已知直线22:21x t l y ?==+??(t 为参数)与曲线C 交于,A B 两点,且||||MA MB >.(1)若(,)P ρθ为曲线C 上任意一点,求ρ的最大值,并求此时点P 的极坐标;(2)求||||MA MB . 23.选修4-5:不等式选讲已知函数()|2|f x x =-.(1)求不等式()5|1|f x x ≤--的解集;(2)若函数1()(2)g x f x a x =--的图像在1(,)2+∞上与x 轴有3个不同的交点,求a 的取值范围.试卷答案一、选择题1-5: CBADC 6-10: DCCBA 11、12:BA二、填空题13.2936 14. 593 15. 14- 16. 22三、解答题17.(1)∵3412a a +=,∴112521012a d a +=+=,∴11a =,∴21n a n =-,∴212(21)143n a n n -=--=-,2(143)22 n n nS n n +-==-(2)若25,,m a a a 成等比数列,则225m a a a =,即23(21)9m -=,∴14m = ∵11111()(21)(21)22121n nn a S n n n n +==--+-+,∴141111111114(1)(1)2335272922929m T T ==-+-++-=-=. 18. (1)证明:由已知四边形ABCD 为矩形,得AB BC ⊥,∵PB AB ⊥,PBBC B =,∴AB ⊥平面PBC .又//CD AB ,∴CD ⊥平面PBC .∵CD ?平面PCD ,∴平面PBC ⊥平面PCD .(2)解:以B 为坐标原点,建立如图所示的空间直角坐标系B xyz -.设1PB AB==,(0)BC a a=>,则(0,0,0)B,(0,0,)C a,(1,0,0)P,(0,1,)D a,所以(1,0,)PC a=-,(0,1,)BD a=,则||cos60||||PC BDPC BD=,即22112aa=+,解得1a=(1a=-舍去).设111(,,)n x y z=是平面PBD的法向量,则n BPn BD==,即111xy z=+=,可取(0,1,1)n=-.设222(,,)m x y z=是平面PCD的法向量,则m PDm CD==即2222x y zy-++==,可取(1,0,1)m=,所以1cos,2||||n mn mn m<>==-,由图可知二面角B PD C--为锐角,所以二面角B PD C--的大小为60.19. 解:(1)①经计算,可得下表:②22210.1(0.1)0.10.03Q=+-+=,220.10.01Q==,12Q Q>,故模型乙的拟合效果更好.(2)若投放量为8千辆,则公司获得每辆车一天的收入期望为100.660.48.4+?=,所以一天的总利润为(8.4 1.7)800053600-?=(元)若投放量为1万辆,由(1)可知,每辆车的成本为26.41.6 1.66410+=(元),每辆车一天收入期望为100.460.67.6?+?=,所以一天的总利润为(7.6 1.664)1000059360-?=(元)所以投放1万辆能获得更多利润,应该增加到投放1万辆. 20.(1)解:由题可知,12c a =,∴2a c =,223b c =,设椭圆C 的方程为2222143x y c c+=,由22221436x y c c y x+=?=?,得22||77c x ==,∴1c =,2a =,23b =,故C 的方程为22143x y +=. (2)证明:由(1)可得:(1,0)F ,设圆E 的圆心为(2,)(0)t t ≠,则(2,2)D t ,圆E 的半径为||R t =,直线AD 的方程为(2)2ty x =+. 设过F 与圆E 相切的直线方程为1x ky =+,2||1t k =+,整理得:212t k t-=,由2(2)2112t y x t x y t ?=+-?=+??,得22262363t x t t y t ?-=??+??=?+?,又∵22222626()()33143t t t t -+++=,∴直线PF 与圆E 相切. 21.(1)由'1()f x ax b x=-+,得'(1)1f a b =-+,l 的方程为1(1)(1)(1)2y a b a b x --++=-+-,又l 过点11(,)22,∴111(1)(1)(1)222a b a b --++=-+-,解得0b =. ∵21()()(1)ln (1)12g x f x a x x ax a x =--=-+-+,∴2'1()(1)1(1)1()1(0)a x x ax a x a g x ax a a x x x --+-+-+=-+-==>,当1(0,)x a ∈时,'()0g x >,()g x 单调递增;当1(,)x a∈+∞时,'()0g x <,()g x 单调递减.故2max 111111()()ln ()(1)1ln 22g x g a a a a a a a a==-+-+=-.(2)证明:∵4a =-,∴2212121211221212()()3ln 21ln 213f x f x x x x x x x x x x x x x ++++=++++++++,212121212ln()2()22x x x x x x x x =++++-+=,∴2121212122()ln()x x x x x x x x +++=-令12(0)x x m m =>,()ln m m m ?=-,'1()m m m,令'()0m ?<得01m <<;令'()0m ?>得1m >.∴()m ?在(0,1)上递减,在(1,)+∞上递增,∴()(1)1m ??≥=,∴212122()1x x x x +++≥,120x x +>,解得:1212x x +≥. 22. (1)2cos 2sin 22)4πρθθθ=+=+,02θπ≤<,∴当4πθ=时,ρ取得最大值22P 的极坐标为(22,)4π.(2)由2cos 2sin ρθθ=+,得22cos 2sin ρρθρθ=+,即22220x y x y +--=,故曲线C 的直角坐标方程为22(1)(1)2x y -+-=.将2212x y ?==+??代入22(1)(1)2x y -+-=并整理得:2210t t -=,解得262t =,∵||||MA MB >,∴由t 的几何意义得,62||2MA =,62||2MB =,||6223||62MA MB +==+-. 23.(1)由()5|1|f x x ≤--,得|1||2|5x x -+-≤,∴2235x x >??-≤?或1215x ≤≤??≤?或1325x x <??-≤?,解得14x -≤≤,故不等式()5|1|f x x ≤--的解集为[1,4]-.(2)122,111()(2)|22|1122,12x x xh x f x x x x x x x-+≥??=-=--=??+-<<??,当112x <<时,11()22222222h x x x x x =+-≥?-=-,当且仅当1 2x x =,即22x =时取等号,∴min ()222h x =-,当1x ≥时,1()22h x x x=-+递减,由1()(2)0g x f x a x=--=,得()h x a =,又1()(1)12h h ==,结合()h x 的图像可得(222,1)a ∈-.。
河北省邯郸市高三英语上学期摸底考试试题
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2018届河北省邯郸市高三上学期摸底考试英语第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
例:How much is the shirt?A. £ 19.15.B. £ 9.18.C. £9.15.答案是C。
1. What will the woman do today?A. Visit a friend.B. Do an experiment.C. Attend a lecture.2. What does the man find it difficult to do?A. Fix a toy train.B. Understand the instructions.C. Put together the folding table.3. Where is the woman’s cell phone?A. In the classroom.B. In her bag.C. In the dining hall.4. When does the woman need the book?A. On April 3rd.B. On April 2nd.C. On April 1st.5. What does the man mean?A. Few people read his article.B. The woman can’t convince him.C Most readers don’t agree with him.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对活或独白。
每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
2018届河北省邯郸市高三上学期摸底考试物理试题及答案 精品
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邯郸市高三摸底考试物理试题2017.9 (考试时间:90分钟满分:100分)第I卷一、选择题:本题共10小题,每小题4分,共40分。
在每小题给出的四个选项中,第1~6题只有一个选项符合题目要求,7~10题有多个选项符合题目要求。
全部选对的得4分,选对但不全的得2分,有选错或不选的得0分。
1.下列关于牛顿运动定律的说法中正确的是A.惯性就是物体保持静止状态的性质B.力的国际制单位“牛顿”是根据牛顿第二定律定义的C.物体运动状态改变的难易程度就是加速度D.一对作用力与反作用力的作用效果总相同2.某大型游乐场内的新型滑梯可以等效为如图所示的物理模型,一个小朋友与斜面在AB段的动摩擦因数μ1<tan θ,BC段的动摩擦因数μ2>tan θ,他从A点开始下滑,滑到C点恰好静止,整个过程中滑梯保持静止状态。
该小朋友从斜面顶端A点滑到底端C点的过程中A.地面对滑梯的摩擦力方向先水平向左,后水平向右B.地面对滑梯始终无摩擦力作用C.地面对滑梯的支持力的大小始终等于小朋友和滑梯的总重力的大小D.地面对滑梯的支持力的大小先大于、后小于小朋友和滑梯的总重力的大小3.如图,这是物体做匀变速曲线运动的轨迹的示意图。
已知物体在B点的加速度方向与速度方向垂直,则下列说法中正确的是A.C点的速率小于B点的速率B.A点的加速度比C点的加速度大C.A点的速率大于B点的速率D.从A点到C点加速度与速度的夹角先增大后减小,速率是先减小后增大4.如图,飞船从轨道1变轨至轨道2。
若飞船在两轨道上都做匀速圆周运动,不考虑质量变化,相对于在轨道1上,飞船在轨道2上的A.动能大 B.向心加速度大C.运行周期短 D.角速度小5.汽车以额定功率在平直公路上匀速行驶,在t1时刻突然使汽车的功率减小一半,并保持该功率继续行驶,到t2时刻汽车又开始做匀速直线运动(设汽车所受阻力不变),则在t1~t2时间内A.汽车的加速度保持不变B.汽车的加速度逐渐减小C.汽车的速度先减小后增大D.汽车的速度先增大后减小6.如图,跳水运动员最后踏板的过程可以简化为下述模型:运动员从高处落到处于自然状态的跳板(A位置)上,随跳板一同向下运动到最低点(B位置)。
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河北省邯郸市2018届高三上学期模拟考试第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)1. What will the woman do today?A. Visit a friend.B. Do an experiment.C. Attend a lecture.2. What does the man find it difficult to do?A. Fix a toy train.B. Understand the instructions.C. Put together the folding table.3. Where is the woman’s cell phone?A. In the classroom.B. In her bag.C. In the dining hall.4. When does the woman need the book?A. On April 3rd.B. On April 2nd.C. On April 1st.5. What does the man mean?A. Few people read his article.B. The woman can’t convince him.C Most readers don’t agree with him.第二节(共15小题;每小题1.5分,满分22.5分)听第6段材料,回答第6、7题。
6. Why is the man unwilling to join the woman?A. He has a pain in his left arm.B. He wants to watch TV.C. He is too lazy.7. What will the woman probably do next?A. Stay at home.B. Take Harry to hospital.C. Play with Helen. 听第7段材料,回答第8、9题。
8. When will the man be home from work?A. At 5:45.B. At 6:15.C. At 6:50.9. Where will the speakers go?A. The Green House Cinema.B. The New State Cinema.C. The UME Cinema. 听第8段材料,回答第10至12题。
10. What kind of room does the man want?A. A business room.B. A single room.C. A double room.11. What is needed on the form?A. An address and the signature.B. The passport number and the signature.C. The passport number and an address.12. How is the man paying for the room?A. In cash.B. By Visa card.C. By check.听第9段材料,回答第13至16题。
13. When does the microbiology lecture on Tuesday start?A. At 9 am.B. At 11 am.C. At 2 pm.14. What lesson does the man have at 12 o’clock on Wednesday?A. Math.B. Physics.C. Chemistry.15. Whi ch day is the man’s busiest day?A. Tuesday.B. Wednesday.C. Thursday.16. What has the man got on Friday’s schedule?A. One lecture.B. Two lectures.C. Nothing.听第10段材料,回答第17至20题。
17 What do restaurants in America seldom accept?A. Cash.B. Checks.C. Credit cards.18. What do we know about popular restaurants?A. They have many bookings every day.B. They serve alcohol to guests under 20.C. They accept bookings for large parties.19. How much should you tip the server for superior service in a restaurant?A. About 15% of the total bill.B. About 20% of the total bill.C. About 25% of the total bill.20. What can we learn from the speech?A. Americans smoke less than Europeans.B. Asians smoke less than Europeans.C. The legal smoking age in America is 21.第二部分阅读理解(共两节,满分40分)AThe islands of Malta and Gozo are brilliant for a family holiday, packed with fun places to visit whatever your children’s ages. The islands’ small size means everywhere is within easy reach.Sandy beaches and swimming spotsOne of Malta’s best beaches for families is soft-sanded, sheltered Golden Bay. Older children can try activities such as stand-up paddleboarding, sailing or windsurfing, while the gentle slope of the beach makes it easy for younger kids to safely paddle in the sea.Eating outChildren are welcome at most restaurants, though more upmarket places often only accept older kids. There are often kids’ menus that tend to offer pizza, but you can always ask for a half portion (一份) of a starter dish (portions are huge in Maltese restaurants). With a wide range of cuisines on offer, children are bound to find something they’ll like, Maltese food is strongly influenced by Italian cuisine, so there’s pizza and pasta galore, and some kids will love the national dish-fried rabbitAccommodationTry to schedule at least a few days on Gozo as there are lots of self-catering farmhouses with pools to rent and it’s even easier to get around than Malta. Smaller hotels in Valletta usually only accept older children.Transport and other tipsThe easiest way to get around Malta and Gozo is to drive, but the local bus service is reliable, easy to use, and fairly inexpensive. Buses are frequent between major towns, but only roughly hourly to and from smaller places. Ferries run between Malta and Gozo, and you can take tourist boats over to Comino.2l. What can younger kids do on the islands of Malta and Gozo?A. Try windsurfing.B. Experience paddling.C. Have dinner at all restaurants.D. Live in smaller hotels in Valletta.22. What do we know about Maltese food?A. It has much to do with Italian cuisine.B. Children can always eat up a portion of dish.C. Fried rabbit is the most popular among visitors.D. Pizza is only offered to kids in Maltese restaurants.23. What transportation may be the best choice for visitors in Malta and Gozo?A. Tourist boats.B. Ferries.C. Sailing boats.D. Motor vehicles.I got a call that an opossum (负鼠) was in trouble on the Wantagh State Parkway. The initial report had come around noon on a recent Saturday via a call to the hotline of Volunteers for Wildlife, an organization where I am a board member. I got to the spot first and,no surprise, no opossum. People unaccustomed to seeing nocturnal animals (夜行动物) during the day often think that the animals are in trouble. Although the callers are well-intentioned, sometimes the animal is neither in trouble nor a threat.When DiGiacomo, a trooper, drove up, he said the opossum had been spotted farther south and we proceeded south on the grass while steady traffic flew by in the other direction!Then I finally saw the grayish opossum sitting precariously at the top of the highway divider, obviously frightened and only a few feet from certain death on each side. The animal was frozen in place, shaking. Its dark eyes looked straight down.DiGiacomo turned on his car’s flashing lights, drove south a few hundred yards farther on the grass-and very carefully turned into the oncoming traffic, which came to a stop behind him. Then he drove slowly ahead, with the cars at a crawl behind him.I grabbed my net and a towel, put on my Kevlar gloves and slowly approached my frightened creature. I dropped the net over him, threw a towel over that to make sure the animal stayed on my side of the highway divider, and then grabbed it.The opossum protested with a few attempts to bite my hand, but I imagined it was relieved to get off that divider. I placed it on the road to check for any injury, but as soon as I lifted the net, the opossum was off like a shot toward the roadside, heading for the safety of the trees.It was a shared moment of joy. I followed the opossum as it disappeared into the woods and smiled.24. What can we learn about the calls to the hotline?A. They usually offer no help.B. They usually cause trouble.C. They often give volunteers a surprise.D. They sometimes give false information.25. What situation was the opossum in when the author spotted it?A. It was extremely weak.B. It was in great danger.C. It got injured in the eye.D. It was shaking with anger.26. What does the underlined word “that” in Paragraph 5 refer to?A. The net.B. The gloves.C. The opossum.D. The divider.27. Where did the opossum run finally?A. On the grass.B. Onto the divider.C. Into the woods.D. Toward the roadside.One night in 1966, Michael Chapman rushed into a folk-music club in Cornwall, England to get out of the rain, and soon found himself onstage playing the guitar.“They offered me a job to play for the rest of the summer. And I’ve been playing ever since,” Chapman says. “I’ve told that story so many times, but it’s absolutely true. If it hadn’t been raining that night, I wouldn’t be talking to you.”The British guitar player has spent the 50 years since then on the road. He released some praised albums in the 1970s, but his health went downhill in the 80s and 90s. Like a lot of British artists in the 1960s, Chapman taught himself to play the guitar by listening to American jazz and blues records.Chapman actually taught photography, but quit when he was 26 to play music full time. When he was starting out, he was often compared to other British acoustic (原声的) guitar heroes of the time, like John Martyn and Bert Jansch.“The fascinating thing about Michael Chapman is that he doesn’t fit in with those guys of his generation,” says Andrew Male, a music journalist. “He’s always been an outsider. He never moved to London.”And while he never became part of the London music scen e, Chapman was noticed by those musicians. “The 70s caught up with him in the 80s,” Male says. “He had a huge heart attack at the end of the 80s that nearly finished him off. When he made his way back into the world in the early 90s, people had forgotten a bout him.”But not everyone. A younger generation of musicians in America, including Thurston Moore of Sonic Youth and a younger singer and guitarist named Steve Gunn, recently rediscovered Chapman.“We wanted to capture Michael’s voice and character where he is now. You know, it’s a little road-weary, but he still has it,” Gunn says. “For me, it was really important to get that right.”“He’s the right old musician for present times,” Male says. “I think the music he makes seems quite in tune with where we’re at.”28. Which can best describe Michael Chapman’s first involvement in music?A. Untrue.B. Accidental.C. Unfortunate.D. Reasonable.29. What can we learn about the British guitarist John Martyn from the text?A. He used to teach photography full time.B. He can not get along well with other musicians.C. He may be very familiar with American jazz and blues.D. He may be one of the best guitar players in the early 90s.30. Why did Michael Chapman quit his music career once?A. He lost his special voice.B. He suffered from a severe heart attack.C. He got tired of competing with others.D. He was almost forgotten by everyone.31. Which of the following can be the best title for the text?A. The Rise and Fall of Guitar MusicB. The Sad Story of a Guitar MusicianC. A New Generation Helps an Old Guitar MusicianD. A Guitar Musician’s Back with the Help of a New GenerationIf you’re DIY-minded, you can build your own solar power system. In some cases, you can even build your own solar panels (太阳能电池板), although the amount that you can effectively DIY home solar depends on how much you want to power. Making your own solar panel is a time-consuming process and requires some electrical skills. However, it can also be very rewarding—learning to build your own panel is a great way to understand how solar electricity is generated (产生).Before you can build your own solar panels, you first need to understand how solar cells generate electricity. Once you’ve bought individual solar cells (they can be purchased online), the basic process for building your own solar panel goes like this: Prepare the backing (底板) for your panel. Many DIY solar panel builders prefer to use a wooden board as the base for their solar cell. You’ll need to drill holes in the board so that the wires for each cell can pass through. Wire your solar cells together. This requires some experience with electrical work. Use a soldering iron (烙铁) to attach wire to the solar cells and then link each of the cells together. Attach cells to your backing. If possible, attach each solar cell to the backing individually. This makes it easier to replace a single cell in the event that becomes damaged or is not operating properly.At this point you have a functional solar panel that can produce electricity when the sun shines. However, a solar panel by itself is not useful. If you are trying to generate electricity to power devices in your home, you need to pair your panel with an inverter (换流器) that will turn direct current (DC) power from the sun into the alternating current (AC) power used in most modem electronic devices.32. What does the author think of building DIY solar panel in the first paragraph?A. It is easy.B. It is necessary.C. It is common.D. It is worthwhile.33. What’s the first step of building your own solar panels?A. Master the advanced knowledge of management.B. Buy individual solar cells.C. Prepare the backing for your panel.D. Drill holes in the wooden board.34. Why are solar cells connected to the wooden board separately?A. To produce more power.B. To substitute damaged cells easily.C. To attach cells to the backing tightly.D. To pass through the wooden board easily.35. What can be inferred from the text?A. People with electrical skills have advantages over those without in DIY solar panels,B. A functional solar panel can give power to your home devices directly.C. How to pair an inverter depends on how much you want to power.D. A wooden board is a must for your own solar panels.第二节(共5小题;每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。