山西省太原市第十二中学2018届高三1月月考

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山西省太原市高新经济开发区多校2024-2025学年上学期第一次月考七年级数学试卷

山西省太原市高新经济开发区多校2024-2025学年上学期第一次月考七年级数学试卷一、单选题1．下面的几何体中，属于棱柱的有（）A ．1个B ．2个C ．3个D ．4个2．今年春节电影《热辣滚烫》《飞驰人生2》《熊出没·逆转时空》《第二十条》在网络上持续引发热议，根据国家电影局2月18日发布数据，我国2024年春节档电影票房达80.16亿元，创造了新的春节档票房纪录．其中数据80.16亿用科学记数法表示为（） A ．880.1610⨯B ．98.01610⨯C ．100.801610⨯D ．1080.1610⨯3．下列各组数相等的有（） A ．()22-与22- B ．()31-与()21-- C ．0.3--与0.3D ．a 与a4．两江新区正加快打造智能网联新能源汽车产业集群，集聚了长安、长安福特、赛力斯、吉利、理想等10家整车企业，200余家核心零部件企业．小虎所在的生产车间需要加工标准尺寸为4.5mm 的零部件，其中()4.50.2mm ±范围内的尺寸为合格，则下列尺寸的零部件不合格的是（） A ．4.4mmB ．4.5mmC ．4.6mmD ．4.8mm5．用一个平面去截以下几何体：圆柱，圆锥，球，三棱柱，长方体，七棱柱；能截得三角形截面的几何体有（）个． A ．3B ．4C ．5D ．66．国庆期间，小郑与小州一起去爬山，他们想知道山的高度．小郑提议利用温差测量山峰的高度，小郑在山顶上测得温度是1-℃，小州此时在山脚测得温度为5℃．已知该地区高度每增加100米，气温大约降低0.8℃，则该山的高度大约是（） A ．600米B ．800米C ．650米D ．750米7．有理数a ，b 在数轴上的对应点的位置如图所示，下列结论中正确的是（）A ．2a >-B ．0ab >C ．a b -<D ．a b >8．用小立方块搭成的几何体从正面和上面看的视图如图，这个几何体中小立方块的个数不可以是（）A ．8B ．9C ．10D ．129．几何图形由点、线、面组成，点动成线、线动成面、面动成体．下列现象中能反映“线动成面”的是（） A ．流星划过夜空 B ．笔尖在纸上快速滑动 C ．汽车雨刷的转动D ．旋转门的旋转10．如图，将一刻度尺放在数轴上（数轴的单位长度是1cm ），刻度尺上“0cm ”和“3cm ”分别对应数轴上的3和0，那么刻度尺上“5.6cm ”对应数轴上的数为（）A ． 1.4-B ． 1.6-C ． 2.6-D ．1.6二、填空题11．“霜降”是秋季的最后一个节气，“霜降”之后气温骤降、昼夜温差更大，今年霜降后的某天，本市清徐、阳曲、娄烦、古交四个县市的最低气温分别是：1℃、1-℃、0℃、2-℃，其中最低温度是℃．12．如图是一个正方体的平面展开图，要使展开图折叠成正方体后，相对面上的两个数互为相反数，则图中x y z ++=．13．如图是一张长12cm ，宽10c m 的长方形铁皮，将其剪去两个完全相同的边长为2cm 的正方形和两个完全相同的长方形，剩余部分(阴影部分)可制成有盖的长方体铁盒，这个铁盒的体积是3cm ．14．你喜欢吃拉面吗？拉面馆的师傅，用一根很粗的面条，把两头捏合在一起拉伸，再捏合，再拉伸，反复几次，就把这根很粗的面条拉成了许多细的面条，如下面草图所示．请问这样第次可拉出128根面条．15．一个小正方体的六个面分别标有数字1，2，3，4，5，6．将它按如图所示的方式顺时针滚动，每滚动90°算一次，则滚动第2024次时，小正方体朝下一面标有的数字是三、解答题 16．计算下列各题： (1)(3)15(12)--+-； (2)(3)(2)(16)4-⨯---÷；(3)3135(2)428⎛⎫-⨯-+- ⎪⎝⎭；(4)221510336⎛⎫⎛⎫⎛⎫-÷-+⨯- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭． 17．把下面的有理数填在相应的括号里2+，12-，0.3，7-，79+，132-，0，100，23%(1)非负数集合：{ …} (2)负数集合：{ … } (3)分数集合：{ …} (4)非负整数集合：{ …}18．把下列各数表示在数轴上，并用“>”把它们连接起来． ()2122,,, 4.5,0,323⎛⎫-------+ ⎪⎝⎭19．在一个大正方体的角上切去一个小正方体，剩余的几何体如图所示，其中从正面、左面、上面看这个几何体时，看到的形状图如图①②③所示．(1)从正面看到的形状图是图_______，从左面看到的形状图是图_______，从上面看到的形状图是图_______；（填序号）(2)若大正方体的边长为20cm ，小正方体的边长为10cm ，求这个几何体的表面积与体积． 20．快递员小王在某商业大厦乘坐电梯取送快递，假定乘电梯向上一楼记作1+，向下一楼记作1-，小王从1楼出发，电梯上下楼层依次记录如下（单位：层）：16+，5-，14+，9-，18+，12-，22-．(1)请通过计算说明小王最后是否回到出发点1楼．(2)该大楼每层高3m ，电梯每向上或向下1m 需要耗电0.2kW h g ，根据小王现在所处位置，请你算算，他取送快递时电梯需要耗电多少千瓦时？ 21．如图1至图3是将正方体截去一部分后得到的多面体．(1)根据要求填写表格(2)猜想f v e ，，三个数量间有何关系．(3)一个多面体的面数等于顶点数，且这个多面体有30条棱，求这个多面体的面数． 22．【概念学习】规定：求若干个相同的有理数（均不等于0）的除法运算叫做除方，如()()()()222,3333÷÷-÷-÷-÷-等．类比有理数的乘方，我们把222÷÷记作2③，读作2的圈3次方，()()()()3333-÷-÷-÷-记作()3-④，读作3-的圈4次方．一般地，把n 个a 相除，记作a ⓝ．【初步探究】（1）直接写出计算结果：5=③______；13⎛⎫-= ⎪⎝⎭④______．【深入思考】我们知道，有理数的减法运算可以转化为加法运算，除法运算可以转化为乘法运算，则有理数的除方运算也可以按如图所示的方式转化为乘法运算．【探究应用】（2）试一试：仿照图中算式，将下列运算结果直接写成乘方的形式：()3-=⑤______；5=⑧______；12⎛⎫= ⎪⎝⎭⑩______；（3）算一算：4211195345⎛⎫⎛⎫⎛⎫-÷-⨯---÷ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⑤④④．23．已知在纸面上有一数轴，根据给出的数轴，解答下面的问题：(1)已知A 、B 两点相距3.5个单位长度，请你根据图中A 、B 两点的位置，分别写出它们所表示的有理数．(2)在数轴上标出与点A的距离为2的点（用不同于A、B的字母表示），并写出这些点表示的数．(3)折叠纸面，若数轴上1 对应的点与5对应的点重合，回答以下问题：①10对应的点与_______对应的点重合；②若数轴上M、N两点之间的距离为2024（M在N的左侧），且M、N两点经折叠后重合，求M、N两点表示的数．(4)如图，半径为2的圆上有一点Q落在数轴上A点处，求将圆在数轴上向右滚动（无滑动）一周后点Q在数轴上所表示的数．。

2020年高考数学冲刺复习知识点精讲：与圆有关的最值问题含解析

与圆有关的最值问题一、考情分析通过对近几年的高考试题的分析比较发现,高考对直线与圆的考查,呈现逐年加重的趋势,与圆有关的最值问题,更是高考的热点问题.由于圆既能与平面几何相联系,又能与圆锥曲线相结合,命题方式比较灵活,故与圆相关的最值问题备受命题者的青睐. 二、经验分享1. 与圆有关的最值问题的常见类型及解题策略(1)与圆有关的长度或距离的最值问题的解法．一般根据长度或距离的几何意义，利用圆的几何性质数形结合求解．(2)与圆上点(x ，y )有关代数式的最值的常见类型及解法．①形如u ＝y －bx －a 型的最值问题，可转化为过点(a ，b )和点(x ，y )的直线的斜率的最值问题；②形如t ＝ax ＋by 型的最值问题，可转化为动直线的截距的最值问题；③形如(x －a )2＋(y －b )2型的最值问题，可转化为动点到定点(a ，b )的距离平方的最值问题． 2.与圆有关的最值问题主要表现在求几何图形的长度、面积的最值，求点到直线的距离的最值，求相关参数的最值等方面．解决此类问题的主要思路是利用圆的几何性质将问题转化三、知识拓展1.圆外一点P 到圆C 上点的距离距离的最大值等于，最小值等于PC r -.2.圆C 上的动点P 到直线l 距离的最大值等于点C 到直线l 距离的最大值加上半径，最小值等于点C 到直线l 距离的最小值减去半径.3.设点M 是圆C 内一点，过点M 作圆C 的弦，则弦长的最大值为直径，最小的弦长为.四、题型分析(一) 与圆相关的最值问题的联系点 1.1 与直线的倾斜角或斜率的最值问题利用公式k ＝tan α(α≠90°)将直线的斜率与倾斜角紧密联系到一起,通过正切函数的图象可以解决已知斜率的范围探求倾斜角的最值,或者已经倾斜角的范围探求斜率的最值.处理方法：直线倾斜角的范围是[0,π),而这个区间不是正切函数的单调区间,因此根据斜率求倾斜角的范围时,要分⎣⎢⎡⎭⎪⎫0，π2与⎝ ⎛⎭⎪⎫π2，π两种情况讨论．由正切函数图象可以看出,当α∈⎣⎢⎡⎭⎪⎫0，π2时,斜率k ∈[0,＋∞)；当α＝π2时,斜率不存在；当α∈⎝ ⎛⎭⎪⎫π2，π时,斜率k ∈(－∞,0)．【例1】坐标平面内有相异两点,经过两点的直线的的倾斜角的取值范围是( ).A ．,44ππ⎡⎤-⎢⎥⎣⎦ B ． C ．D ．3,44ππ⎡⎤⎢⎥⎣⎦ 【答案】C 【解析】,且0AB k ≠.设直线的倾斜角为α,当01AB k <≤时,则,所以倾斜角α的范围为04πα≤≤.当时,则,所以倾斜角α的范围为34παπ≤<. 【点评】由斜率取值范围确定直线倾斜角的范围要利用正切函数y ＝tan x 的图象,特别要注意倾斜角取值范围的限制；求解直线的倾斜角与斜率问题要善于利用数形结合的思想,要注意直线的倾斜角由锐角变到直角及由直角变到钝角时,需依据正切函数y ＝tan x 的单调性求k 的范围．【小试牛刀】若过点的直线与圆224x y +=有公共点,则该直线的倾斜角的取值范围是（）A ．0 6π⎛⎫ ⎪⎝⎭，B ．0 3π⎡⎤⎢⎥⎣⎦， C. 0 6π⎡⎤⎢⎥⎣⎦， D ．0 3π⎛⎤ ⎥⎝⎦，【答案】B【解析】当过点的直线与圆224x y += 相切时,设斜率为k ,则此直线方程为,即.由圆心到直线的距离等于半径可得,求得0k =或k =故直线的倾斜角的取值范围是[0,]3π,所以B 选项是正确的.1.2 与距离有关的最值问题在运动变化中,动点到直线、圆的距离会发生变化,在变化过程中,就会出现一些最值问题,如距离最小,最大等.这些问题常常联系到平面几何知识,利用数形结合思想可直接得到相关结论,解题时便可利用这些结论直接确定最值问题.【例2】过点()1,2M 的直线l 与圆C ：交于,A B 两点,C 为圆心,当ACB ∠最小时,直线l 的方程是．答案:解析：要使ACB ∠最小,由余弦定理可知,需弦长AB 最短.要使得弦长最短,借助结论可知当()1,2M 为弦的中点时最短.因圆心和()1,2M 所在直线的,则所求的直线斜率为1-,由点斜式可得.【点评】与圆有关的长度或距离的最值问题的解法．一般根据长度或距离的几何意义,利用圆的几何性质数形结合求解．此题通过两次转化,最终转化为求过定点的弦长最短的问题. 【例3】若圆C ：关于直线对称,则由点(,)a b 向圆C 所作的切线长的最小值是（）A ．2B ．3C ．4D ．6 【答案】C【解析】圆C ：化为（x+1）2+（y-2）2=2,圆的圆心坐标为（-1,2．圆C ：关于直线2ax+by+6=0对称,所以（-1,2）在直线上,可得-2a+2b+6=0,即a=b+3．点（a,b ）与圆心的距离,,所以点（a,b ）向圆C 所作切线长：当且仅当b=-1时弦长最小,为4【点评】与切线长有关的问题及与切线有关的夹角问题,解题时应注意圆心与切点连线与切线垂直,从而得出一个直角三角形．【小试牛刀】【安徽省合肥一中、马鞍山二中等六校教育研究会2019届高三第二次联考】已知抛物线上一点到焦点的距离为，分别为抛物线与圆上的动点，则的最小值为（） A ．B ．C ．D ．【答案】D 【解析】由抛物线焦点在轴上，准线方程，则点到焦点的距离为，则，所以抛物线方程：，设，圆，圆心为，半径为1，则，当时，取得最小值，最小值为，故选D.1.3 与面积相关的最值问题与圆的面积的最值问题,一般转化为寻求圆的半径相关的函数关系或者几何图形的关系,借助函数求最值的方法,如配方法,基本不等式法等求解,有时可以通过转化思想,利用数形结合思想求解.【例4】在平面直角坐标系中,,A B 分别是x 轴和y 轴上的动点,若以AB 为直径的圆C 与直线相切,则圆C 面积的最小值为（）A.45πB.34πC.(6π-D.54π 【答案】A 【解析】设直线l ：.因为,所以圆心C 的轨迹为以O 为焦点,l 为准线的抛物线.圆C 半径最小值为,圆C 面积的最小值为选A.【例5】动圆C 经过点(1,0)F ,并且与直线1x =-相切,若动圆C 与直线总有公共点,则圆C的面积（）A ．有最大值8πB ．有最小值2πC ．有最小值3πD ．有最小值4π 【答案】D【解析】设圆心为(,)a b ,半径为r ,,即,即214a b =,∴圆心为21(,)4b b ,2114r b =+,圆心到直线的距离为,∴或2b ≥,当2b =时,,∴.【小试牛刀】【山东省恒台第一中学2019届高三上学期诊断】已知O 为坐标原点，直线．若直线l 与圆C 交于A ，B 两点，则△OAB 面积的最大值为（）A ．4B ．C ．2D ．【答案】C 【解析】由圆的方程可知圆心坐标，半径为2，又由直线，可知，即点D 为OC 的中点，所以,设，又由，所以，又由当，此时直线，使得的最小角为，即当时，此时的最大值为2，故选C 。

分段函数的性质、图象以及应用 (1)

7
分段函数与周期和最值
分段函数的值域是各段值域的并集,最大值是各段最大值中的最大者是函数的最大值,最小值是各段最小值中的最小者,一般可借助于图像来解决.
例 7【2018届山西省太原十二中高三1月月考】已知8m n -<<,函数()()823log ,8,
{
2,,
x x m f x x x m x n --≤<=-≤≤若()f x 的值域为[]1,3-,则n m -的最大值与最小值之积为( ) A. 4 B. 6 C. 8 D. 10 【答案】B 【解析】()()22
C. D.
或其它参数的值的问题,一般按自变量
的实根个数不可能为(
5
解决由函数零点的存在情况求参数的值或取值范围问题,关键是利用函数方程思想或数形结合思想,
1) B. [0
6 分段函数与解析式
分段函数是定义域中各段的x 与y 的对应法则不同,函数式是分两段或几段给出的.因此求解析式时,也是分段求解析式的.
log ,8{
2,x<=-≤≤,分别作出()2log y x =-和22y x x =-的图像, ()f x 在
,否则
上无意义,最大值为
,所以零点之和为
分段函数的单调性
D. (
数首先是函数,且是一个函数,不是多个函数;分段函数的处理方法
自变量在哪一段,该代哪个解析式,这样就要分段讨论、求解,即要重视分类讨论思想在解题过程中的应
用.
因而分段函数已成为高考命题的一个热点
于分段函数的性质、图象的考查,重点放在函数的奇偶性、周期性以及函数的零点问题与分段函数结合上;
知识是学生掌握比较模糊,看到就头疼的题目
要是学生的作图能力普遍较弱,还有就是没有形成解题的模式和套路,以至于遇到类似的题目便产生畏惧届湖南省长沙市第一中学高考模拟卷一

山西省太原市杏花岭区中学的综合排名

山西省太原市杏花岭区中学的综合排名杏花岭区是中国山西省太原市市辖区，与迎泽区同为太原市的中心城区，总面积170.2平方公里，那么口碑好的中学都有哪些呢?小编为大家准备了相关的资料，接下来就让小编带大家一睹为快!山西省太原市杏花岭区中学排名(本排名榜由家长学生打分产生，仅供参考)太原市第十二中学的简介太原市第十二中学校创建于1956年，是山西省首批重点中学，分初中部和高中部两址。

2006年被评为"山西省示范高中"。

学校占地面积48000平方米，建筑面积39000多平方米，有60个教学班，在校学生3300余人。

校园几经绿化、美化、亮化，现已成为一所环境优雅、设施齐备、功能完善的现代化学校，具有良好的人文地理优势和浓郁的教育文化氛围。

学校师资力量雄厚，拥有一支既有高尚师德和现代教育理念，又有娴熟的教学技能和艺术才能的教师队伍。

现有教职工375人，其中特级教师4人，高级教师104人，国家级、省市学科带头人和骨干教师100多人，省市教学能手30多人，他们为学校的持续发展奠定了良好的基础。

在50多年的发展里程中，学校规模不断扩大，教学质量逐年提高，培养了大批优秀、合格的人才。

学校始终坚持"一切为了学生成长"的办学理念和"厚德、博学、慎思、笃行"的校训，以提高教学质量、深化教育改革和创新为重点，全面实施素质教育，注重培养学生的创新实践能力，重视学生人文精神、科学素养和文化底蕴的培养，使学生达到知识能力、身体素质和健康人格的和谐发展。

1991年，杨景农同学以636分获高考山西省理科状元;2001年杜强同学以687分获高考山西省理科第二名，太原市理科第一名;2004年张婷同学以634分获太原市中考状元。

近几年，教学成绩一年上一个台阶，高考本科一次达线： 2006年274人，2007年313人，2008年433人，2009年498人，2010年513人，2011年687人。

2023-2024学年山西省知名高中高一上学期11月月考(期中)英语试题

2023-2024学年山西省知名高中高一上学期11月月考（期中）英语试题1. What happened to the man?A．He hurt his knee. B．He failed a test. C．He missed a party.2. Where is Larry now?A．At home. B．In the office. C．At school.3. What does Sam prefer to do?A．Run. B．Sing. C．Dance.4. What are the speakers mainly talking about?A．Where to have a trip.B．Where to park the car.C．Where to see a movie.5. What is the probable relationship between the speakers?A．Teacher and student. B．Schoolmates. C．Co-workers.听下面一段较长对话,回答以下小题。

6. How many rooms does the woman want to book?A．One. B．Two. C．Three.7. Why does the woman book a room with no beds?A．For her cats. B．For her bags. C．For her dogs.听下面一段较长对话,回答以下小题。

8. How did the speakers come to see the match?A．By car. B．By taxi. C．By bike.9. Why did the speakers arrive early?A．To get cheap tickets. B．To find a good place. C．To meet their friends. 听下面一段较长对话,回答以下小题。

2023届山西省太原市高三1月第一次联考英语试题

2023届山西省太原市高三1月第一次联考英语试题学校:___________姓名：___________班级：___________考号：___________一、阅读理解Canada is home to many of the world’s top universities. Here are some scholarships in Canada that will help reduce some of the financial burden for students.University de Montreal (UdeM) International Student Scholarship ProgramThe UdeM scholarship program aims to support international candidates in their university studies. They offer graduate scholarships for students. The scholarship is valued at CA $9,420 per year or CA $3,140 per session. Deadline to apply is March 1st, 2023.China Scholarship Council—University of Saskatchewan Joint Funding Program The China Scholarship Council (CSC), partnering with the University of Saskatchewan (USask), is offering outstanding Chinese graduate students the opportunity to pursue PhD and post-doctor studies through the CSC-USask Joint Funding Program. To be qualified, you must be a citizen of the People’s Republic of China. Applicants should not be older than 35 years old at the time of the application. Deadline is February 10th, 2023 (China time).University of Saskatchewan Graduate ScholarshipIf you are not a Chinese citizen but want to pursue your postgraduate studies at the University of Saskatchewan, you can choose this scholarship. Their PhD scholarships are valued at CA $20,000 each; alternatively, their Master’s thesis(论文)is valued at CA$16,000. Deadline to apply is February 22nd, 2023.Canada-ASEAN Scholarships and Educational Exchange for Development (SEED) If you are keen to engage in a short-term exchange opportunity for study or research in Canadian post-secondary(中学后)institutions at the college, undergraduate and graduate levels, this scholarship is for you. Do note that only Canadian institutions can submit applications on behalf of candidates. If you are interested in this scholarship program, contact your institution to make your interest known and to request information about the application process. Deadline is March 4th, 2023.1．When should you apply for the UdeM International Student Scholarship Program? A．Before February 10th, 2023. B．Before March 4th, 2023.C．Before March 1st. 2023. D．Before February 22nd, 2023. 2．Which scholarship program is only available to Chinese citizens?A．Canada-ASEAN Scholarships and Educational Exchange for Development. B．University of Saskatchewan Graduate Scholarship.C．UdeM International Student Scholarship Program.D．CSC-USask Joint Funding Program.3．What can we learn about the feature of the SEED scholarship program? A．Applications can only be submitted by Canadian institutions.B．Each applicant can get CA$16,000 from it.C．There is an age limit for the applicants.D．Students of any level can apply for it.I was a newcomer in a class. So was Alice. That’s where the si milarities ended. I was tall and she was small. My thick, black hair had been recently cut short. Her natural blonde hair flowed to her waist and looked great. I was awkward and shy. She wasn’t. I couldn’t stand her.I considered her my enemy. She liked me. She wanted to be my friend.One day, she invited me over and I said yes—I was too shocked to answer any other way. No one had invited me over to play. But this girl who wore the latest fashions wanted me to go to her home with her after school. I got very surprised when she led me into an apartment building. She lived on the fourth floor in a two-room place with her mother, her stepfather, her two brothers and her sister. When we got to the room she shared with her sister, she took out a big case of Barbies—which was my next surprise. I would have thought she’d grown too mature for them. I had never played with them. But we sat on the floor of a walk-in closet, laughing as we made up crazy stories about the Barbies. That’s when we found out that we both wanted to be writers when we were older and we both had wild imaginations. We had a great day that afternoon. Our jaws ached from smiling so much.She showed me her wardrobe, which had mostly come from a designer clothing(品牌服装)store down the block. The woman who owned it used her as a model sometimes for her newspaper ads and gave her clothes in exchange.Alice had the whole neighborhood charmed. The bookstore owners lent her fashion magazines, the movie theater gave her free passes and the pizza place let her have free slices. Soon I was included in her magic world. We slept over at each other’s houses, and spent every free moment together. My dark hair grew out and I learned to love being tall.Alice, my first real friend since childhood, has taught me an amazing and very surprisingthing about making friends—your worst enemy can turn out to be your best friend.4．What can we infer from the first paragraph?A．The author was jealous of Alice.B．The author looked up to Alice.C．The author wished to be similar to Alice.D．The author attempted to get along well with Alice.5．What did the author think of the Barbies for Alice?A．The Barbies had been kept well by Alice.B．The Barbies were not suitable for Alice.C．Alice must have made them by herself.D．Alice must have spent much money on them.6．How did Alice get the designer clothes?A．By working as a designer.B．By serving a bookstore.C．By acting as a model.D．By advertising for a pizza place.7．What did Alice and the author have in common?A．They were both humorous.B．They both had blonde hair.C．They were both outgoing.D．They both intended to be writers.A study, led by researchers from the University of California, San Diego, followed the same 875 mother-child pairs in Chile for 16 years, assessing them at ages 1, 5, 10 and 16. At each visit, researchers examined the mother for signs of depression and used cognitive (认知的) development tests on the child. They also asked questions to assess the home life, describing the level of connection between the mother and child.Researchers found that signs of depression in moms when the child is one are connected with lower scores on cognitive development tests for the child at age 16.“We found that mothers who were highly depressed didn’t show as much care or provid e as much learning material to support their children, such as toys and books, as mothers who were not depressed. This, in turn, influenced the children’s IQ at ages 1, 5,10 and 16,” Patricia East, researchscientist with the Department of Pediatric at UC San Diego School of Medicine, said in a statement.The authors found the relationship in reverse (反向) to be true as well—lower development scores early in the child’s life resulted in less connection from the mom and that only increased signs of the mother’s depression as the child entered into the age of teenagers.Children who had seriously depressed mothers were found to have an average IQ score of 7.30 compared to a score of 7.78 in children without depressed mothers. “Differences in IQ from 7.78 to 7.30 are highly meaningful when it comes to children’s language-speaking skills and vocabulary,” said East. “Our study results show the long-term results that a child can experience due to the mother’s long-term depression.”However, the authors recognize these families in Chile can be very different from mothers and children of other cultural backgrounds or nationalities. Besides, all the families studied were from a similar cultural background and had a similar level of education. 8．What kind of boy may have a lower IQ score?A．His mother often feels sad and without enthusiasm.B．His mother often feels strong impatience about something.C．His mother often gives little attention and thought to what she is doing.D．His mother often has a feeling of being happy with her own character and abilities. 9．What do we learn about highly-depressed mothers?A．Their IQ scores were relatively low.B．They cared about their children very much.C．They had little influence on their children’s IQ scores.D．They couldn’t supply t heir children with much learning material.10．How did the researchers carry out the study?A．By doing some experiments.B．By referring to some materials.C．By asking questions and testing.D．By analyzing a number of problems.11．What is the main idea of the last paragraph?A．The results might not be universal.B．The results were only accepted in Chile.C．All the families had a similar level of education.D．All the families studied were from a similar cultural background.A concert that features Ancient Tang poems is set to hit the US next month to celebrate the Chinese Lunar New Year and the 50th anniversary of The Philadelphia Orchestra’s historic 1973 tour of China. The program entitled Echoes of Ancient Tang Poems is set to be released on Jan 6 and 7,2023 in Philadelphia and New York. iSING! Suzhou and The Philadelphia Orchestra present the program. The show, led by former Philadelphia Orchestra Assistant Conductor Lio Kuokman, features ancient Chinese lyrical texts(剧本)from young composers who were selected from the 2020 iSING! Composition Competition.The 2020 iSING! Composition Competition is a five-month-long process of selecting winning composers from more than 200 entries from nine countries. The panel of judges included Hao Jiang Tian, well-known bass(男低音歌手), iSING! founder and artistic director. Founded in 2011, iSING! Art Festival is the first international vocal art festival in China. Since 2014, over 380 singers from more than 30 countries have been selected to come to China to participate in the annual iSING! Suzhou Art Festival. Poets of the Tang Dynasty featured in the incoming program Echoes of Ancient Tang Poems include Li Bai, Bai Juyi, Du Fu, Du Mu, Zhang Ji, and Wang Bo.“I was initially worried about being able to connect both historically and culturally,” said Fernando Buide del Real in an official press statement. “But I soon realized Tang poetic sentiments(诗意)are universal and go beyond geographical boundaries. During the COVID lockdown, I found resonance(共鸣)in Wang Bo’s poem about fr iendship, loneliness, and separation. The precision and depth of the Chinese language is amazing. Every character and every phrase is filled with meaning.”January’s concert also aims to celebrate the 50th anniversary of The Philadelphia Orchestra’s 1973 t our of China. Over the past 50 years, the Orchestra has returned to China 12 times, more than any other US orchestra.“I think this event is very important to China and the US, especially in terms of cultural exchanges,” Hao Jiang Tian said. “During the CO VID-19 pandemic, performing arts around the world has been greatly affected and impacted. Our incoming event integrates ancient Tang poems, new forms of music, and singers from 9 different countries—this is the first time in history. In the current situation, such performance is particularly important,” Tian said. 12．What can we learn about iSING! Art Festival?A．It was established in the year of 2014.B．Its founders include Conductor Lio Kuokman.C．It is the first international vocal art festival in China.D．It is held every other year in Suzhou and Philadelphia.13．How did Fernando Buide del Real like Wang Bo’s poem?A．He found it tough to understand.B．He couldn’t think too highly of it.C．He could hardly share Wangbo’s sad emotions.D．He felt its contents were unrealistic and strange.14．Which can replace the underlined word “integrates” in the last paragraph? A．Affects. B．Balances. C．Inspects. D．Combines. 15．What can be a suitable title for the news report?A．Tang Dynasty poems concert to celebrate Chinese New YearB．Poets of Tang Dynasty featured in Echoes of Ancient Tang PoemsC．50th anniversary of The Philadelphia Orchestra’s 1973 tour of ChinaD．Process of selecting winning composers for 2020 iSING! Composition Competition二、七选五Some people say house dust is mostly human skin cells, but it’s only a little bit true.____16____, but there are many other components(成分)on top of your ceiling fans. These include paint, hair, building materials, bacteria, viruses, insect body parts, etc.That list is based on the Canadian House Dust Study, in which researchers collected dust samples from 1,025 Canadian homes. The proportions(比例)of the components vary from household to household. A house near a busy road, for example, is likely to harbor a high level of outdoor pollutants from car exhaust(废气). ____17____.The commonly-cited statement that 70% to 80% of house dust is human skin cells is likely not true for most houses. ____18____, and 40%came from dirt and other materials from outside. That indoor 60%included everything from organic fibers to building materials, not just dead skin cells.People do drop lots of skin cells as they go about their business. ____19____. Not all dead skin flakes(剥落)off onto your home’s floors. Many run down the drain(下水道)in theshower, and others are contained by clothing and end up being washed out in the washing machine.____20____. One study found that higher levels of cholesterol and squalene (oils found in dead skin) in dust were associated with lower levels of ozone(臭氧)indoors. Ozone is a pollutant that can cause lung discomfort. Ozone reacts with these oils, thus reducing indoor ozone by between 2% and 15%.A．But our body needs all kinds of nutrientsB．Skin cells are part of the makeup(组成)of house dustC．The skin helps people maintain the right internal temperatureD．A newly-built house might have much dust from constructionE．The average adult loses about 500 million skin cells each dayF．It may not be a bad thing to have some dead skin in your homeG．According to a study of house dust, 60% of the components of the dust were from inside三、完形填空My husband and I had dreamed about moving to the countryside for years. And finally, we found our own piece of paradise on Tennessee’s Cumberland Plateau. We built a house and ____21____ a big vegetable garden. Life was good.But as most country-dwellers(乡村居民)know, at some point you have to deal with uncooperative(不合作的)____22____. Our mistake was thinking that because we had a dog named Heidi, we were ____23____.Jeff and I ____24____ town one day to find some of our new neighbors standing in our driveway. It seemed that a small group of cows had been in our street lately and none was sure where they came from or who ____25____ them. All they were sure of was that these cows were a grade-A problem. ____26____, our thoughtful neighbors had come to____27____ our property while we were gone, driving them away.But about a week later, we suddenly heard mooing coming from the front yard. Our beautiful vegetable garden, which we had put so much time and money into, had just been____28____ by the mooing cows.We ____29____ finding out who owned the cows and soon learned they were from another town. A field owner was renting some land to a cattle owner, but didn’t keep the fencein good ____30____. The cows’ owner apologized and paid for our damaged ____31____. We thought that was the end of the ____32____.But one day, Melissa, our neighbor from two doors away, ____33____ at my door. She said, “Those ____34____ cows are at my house. They are trying to get into the pool!” But when we got there, it turned out that the cows were just ____35____ and were drinking from it. I helped Melissa drive them off and gave her the owner’s name and phone number.Either the fence was finally fixed or the owner sold them, because that was the last time we saw those cows.21．A．abandoned B．displayed C．reserved D．planted 22．A．farmers B．birds C．animals D．neighbors 23．A．healthy B．calm C．blessed D．safe 24．A．went through B．passed by C．returned from D．left for 25．A．owned B．beat C．caught D．counted 26．A．Formally B．Fortunately C．Optimistically D．Occasionally 27．A．evaluate B．transfer C．hide D．defend 28．A．ruined B．observed C．sought D．threatened 29．A．adapted to B．set about C．put off D．gave up 30．A．repair B．faith C．mood D．time 31．A．clothes B．driveway C．crops D．house 32．A．match B．story C．play D．film 33．A．laughed B．apologized C．negotiated D．appeared 34．A．heavy B．clever C．crazy D．ugly 35．A．hungry B．thirsty C．exhausted D．frightened四、用单词的适当形式完成短文阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。

【同步课堂】 8.2.3 倍角公式练习(1)(解析版)

8.2.3 倍角公式【基础练习】一、单选题1．在平面直角坐标系xOy 中，角θ的顶点在原点，始边与x 轴的非负半轴重合，终边经过点()3,1-，则cos2θ=（）A ．35 B ．35C ．45-D ．45【答案】D 【解析】解：∵角θ的顶点在原点，始边与x 轴的非负半轴重合，终边经过点()3,1-，∴OP =∴sin 10θ=．则224cos212sin 125θθ=-=-⨯=⎝⎭. 故选D ．2．函数()212sin f x x =-是（） A ．偶函数且最小正周期为2πB ．奇函数且最小正周期为2π C ．偶函数且最小正周期为π D ．奇函数且最小正周期为π【答案】C 【解析】解：∵()212sin cos 2f x x x =-=， ∴()()cos 2cos2f x x x -=-=，∴函数()f x 是偶函数且最小正周期22T ππ==，故选：C ．3．已知tan α=tan2α的值为（）A ．B C ．D【答案】B 【解析】22tan tan 21tan ααα===-故选：B4．已知1sin 23πα⎛⎫+=- ⎪⎝⎭，则cos2=α（）A ．79- B ．79C ．89-D ．89【答案】A 【解析】由1sin 23πα⎛⎫+=- ⎪⎝⎭，得1cos 3α=-，则cos2=α2272cos 1199α-=-=-.故选：A. 5．13cos80cos10-的值为（） A ．2 B ．4C ．6D ．8【答案】B 【解析】13cos80-13sin10=-cos103sin10sin10cos10-=()2sin 3010sin10cos10-=2sin 2041sin 202==.故选：B二、填空题 6．已知1cos 23α=，则()22cos 2cos 2παπα⎛⎫+-- ⎪⎝⎭的值为_____. 【答案】1- 【解析】解：由1cos 23α=，得212cos 13α-=，即22cos 3α=. 所以()222222cos 2cos sin 2cos 13cos 13123παπαααα⎛⎫+--=-=-=-⨯=- ⎪⎝⎭故答案为：1-.7．已知tan α和tan β是方程2260x x +-=的两个根，则()tan 2αβ+⎡⎤⎣⎦=____________ 【答案】1663- 【解析】由题得1tan tan ,tan tan 32αβαβ+=-⨯=-.所以1tan tan 12tan()1tan tan 138αβαβαβ-++===--+.所以()212tan()164tan 2=11tan ()63164αβαβαβ-++==-⎡⎤⎣⎦-+-. 故答案为：1663-. 8．已知tan 2α=，则cos2sin cos ααα-=__________．【答案】1- 【解析】依题意，222222cos sin sin cos 1tan tan 1421cos sin 1tan 14ααααααααα------===-+++.三、解答题9．已知函数()()2cos sin cos f x x x x =⋅+，求54f π⎛⎫⎪⎝⎭的值. 【答案】2 【解析】解:方法一：55552cos sin cos 4444f ππππ⎛⎫⎛⎫=+ ⎪⎪⎝⎭⎝⎭2cos sin cos 2444πππ⎛⎫=---= ⎪⎝⎭.方法二：因为()22sin cos 2cos sin 2cos21f x x x x x x =+=++214x π⎛⎫=++ ⎪⎝⎭，所以511112444f πππ⎛⎫=+=+=⎪⎝⎭. 10．如图，扇形AOB ，圆心角AOB 等于60°，半径为2，在弧AB 上有一动点P ，过P 引平行于OB 的直线和OA 交于点C ，设∠AOP ＝θ，求△POC 面积的最大值及此时θ的值.【答案】3. 【解析】因为CP ∥OB ，所以∠CPO ＝∠POB ＝60°﹣θ，∴∠OCP ＝120°．在△POC 中，由正弦定理得OP CPsin PCO sin θ=∠，∴2120CPsin sin θ=︒，所以CP =又()260120OC sin sin θ=︒-︒，∴OC =（60°﹣θ）．因此△POC 的面积为S （θ）12=CP •OCsin120°12=••（60°﹣θ）==12-sinθ）=2sinθcosθ12-sin 2θ）=12+cos2θ12-）=（2θ﹣60°）12-]，θ∈（0°，60°）．所以当θ＝30°时，S【提升练习】一、单选题1．若1tan 3θ= ，则cos2θ=( )A ．45-B ．15-C ．15D ．45【答案】D 【解析】222222cos cos2cos cos sin sin sin θθθθθθθ-=-=+. 分子分母同时除以2cos θ，即得：2211149cos211519tan tan θθθ--===++. 故选D.2．下列四个等式：①tan 25tan 3525n 3ta 5︒︒︒+︒=+ ②2tan 22.511tan 22.5︒=-︒； ③221cossin 882ππ-=；④14sin10cos10-=︒︒．其中正确的等式个数是（） A ．1 B ．2 C ．3 D ．4【答案】B 【解析】①因为()tan 25t 6an 35tan tan 25351tan 25tan 350+-︒==︒︒+︒，所以tan 25tan 35tan 25tan 35++-︒︒︒︒)，所以tan 25tan 3525n 3ta 5︒︒︒+︒=+ ②22tan 22.512tan 22.511tan 451tan 22.521tan 22.522︒︒===-︒-︒，故错误；③22cos sin cos8842πππ-==，故错误；④1cos10sin10cos10sin10cos10︒︒-=︒︒︒︒，12(cos10)2sin 20224112sin10cos10sin 2022︒︒===⨯︒︒，故正确.故选：B3．0>ω函数()sinsin22xxf x ωπω+=在[]43ππ-,上单调递增，则ω的范围是A ．20,3⎛⎤ ⎥⎝⎦B ．30,2⎛⎤ ⎥⎝⎦C ．(]0,2D ．[)2,+∞【答案】B 【解析】由题得111()=sin cos sin x 222f x wx wx w =，所以函数的最小正周期为2T wπ=, 因为函数()sinsin22xxf x ωπω+=在[]43ππ-,上单调递增，所以24w 324w4ππππ⎧≥⎪⎪⎨⎪-≤-⎪⎩，又w ＞0，所以302w <≤. 故选B4．已知α为第三象限角，且sin cos 2m αα+=，2sin 2m α=，则m 的值为（）A ．13-B．2-C．D．3-【答案】C 【解析】sin cos 2m αα+=，则()222sin cos 1sin 214m m ααα+=+=+=，解得m =，α为第三象限角，则sin cos 20m αα+=<，故m =故选：C.5．17世纪德国著名的天文学家开普勒曾经这样说过：“几何学里有两件宝，一个是勾股定理，另一个是黄金分割.如果把勾股定理比作黄金矿的话，那么可以把黄金分割比作钻石矿.”黄金三角形有两种，其中底与腰之比为黄金分割比的黄金三角形被认为是最美的三角形，它是一个顶角为36︒的等腰三角形(另一种是顶角为108°的等腰三角形).例如，五角星由五个黄金三角形与一个正五边形组成，如图所示，在其中一个黄金ABC中，BC AC =.根据这些信息，可得sin 234︒=（）A．14- B．38+-C．14-D．48+-【答案】C 【解析】由题意可得：72ACB ∠=，且112cos 4BCACB AC ∠==，所以251cos1442cos 7214=-=-，所以()51sin 234sin 14490cos144+=+==-，故选：C二、填空题6．设α为锐角，若4cos()65πα+=，则sin(2)12πα+的值为______.【答案】50【解析】247cos(2)213525πα⎛⎫+=⋅-=⎪⎝⎭，24sin(2)325πα+=，所以sin(2)sin(2)1234πππαα+=+-2472525⎫=-=⎪⎝⎭7．()sin 501︒︒ 的值__________. 【答案】1 【解析】解： ()cos10sin501sin50cos10︒+︒︒+︒=︒⨯︒()2sin50cos30sin10sin 30cos102sin50sin 402sin50cos50cos10cos10cos10︒︒︒+︒︒︒︒︒︒===︒︒︒()sin 10902sin50cos50sin100cos101cos10cos10cos10cos10︒+︒︒︒︒︒====︒︒︒︒.故答案为:1.8．（山西省太原十二中2018届高三上学期1月月考数学（理））《九章算术》是我国古代内容极为丰富的数学名著，书中由一道著名的“引葭赴氨”问题：“今有池方一丈，葭生其中央，出水一尺.引葭赴岸，适与岸齐.问水深、葭长各几何？”其意思为：“今有水池1丈见方（即10CD =尺），芦苇生长在水的中央，长处水面的部分为1尺.将芦苇向池岸牵引，恰巧与水岸齐接（如图所示），问水深、芦苇的长度各是多少？”现假设BAC θ=∠，则tan 24θπ⎛⎫+=⎪⎝⎭__________．【答案】5 【解析】设BC x =，则1AC x =+ ()2225,51,12AB x x x ∴=∴+=+∴=22tan1222tan ,tan 5231tan 2θθθθ==∴=-（负根舍去） tan 524θπ⎛⎫+= ⎪⎝⎭即答案为5三、解答题9．已知函数()2sin (sin cos )f x x x x =+．（1）求函数()f x 的最小正周期和最大值；（2）画出函数()y f x =在,22ππ⎡⎤-⎢⎥⎣⎦上的图像．【答案】（1）π；1（2）图像见解析【解析】（1）2()2sin (sin cos )2sin 2sin cos 1cos 2sin 2f x x x x x x x x x =+=+=-+1sin 2cos cos 2sin 12444x x x πππ⎫⎛⎫=-=+-⎪ ⎪⎭⎝⎭，所以函数()f x 的最小正周期为22T ππ==，当sin 214x π⎛⎫-= ⎪⎝⎭时，函数()f x的最大值为1+（2）列表故函数()y f x =在,22⎡⎤-⎢⎥⎣⎦上的图像如下图所示：10．已知函数2()sin cos cos (0)f x a x x x b a =⋅-++> (1)写出函数的最小正周期；(2)设[0]2x ，π∈，()f x 的最小值是2-，求实数,a b 的值．【答案】（1）π；（2）2,2a b ==-+. 【解析】（1）1cos 2()sin 2222a x f x x ab +=-++1sin 22sin 2223a x x b a x b π⎛⎫⎛⎫=-+=-+ ⎪ ⎪ ⎪⎝⎭⎝⎭，故()f x 的最小正周期为22T ππ==. （2）当0,2x π⎡⎤∈⎢⎥⎣⎦时，22333x πππ-≤-≤，故sin 213x π⎛⎫≤-≤ ⎪⎝⎭，又0a >，故sin 223a b a x b a b π⎛⎫-+≤-+≤+ ⎪⎝⎭，高一数学同步课堂提升坚持就是胜利！所以2a b b ⎧+=⎪⎨+=-⎪⎩，故22a b =⎧⎪⎨=-+⎪⎩。

太原重点中学

山大附中山西大学附中是山西省教育厅直属的省级示范性高中，正式成立于1955年。

她的前身是太行军区和太岳军区干部子弟学校迁到太原合并而成的山西省干部子弟学校，1963年被确定为省重点中学。

学校现有面积8万平米，位处山西省高校、科研单位最集中的太原市坞城地区，人文地理环境优越。

充满绿色和人文关怀的校园，优美宜人；网络和多媒体环境下的数字化校园，开放现代。

在长期的办学实践中，山西大学附中逐步形成了“以人为本、持续发展”的办学思想和“科学与人文交融、厚德与博学并举”的育人理念，构建了富有特色的校本课程，形成了极具个性的学校文化及办学特色，创建了学生“自主发展、全面发展、个性发展、特长发展”的育人模式，为师生全面、和谐、持续发展提供了优良的环境和广阔的空间，为社会提供了优质教育，培养了一批批国家级和省级拔尖学生。

在全国名牌大学保送生选拔考试中和高考中，山大附中每年有40余人被北大清华录取，80%以上的学生升入全国重点大学，高考一本、二本达线率稳居全省最前列。

太原育英中学太原育英中学，原太原26中。

太原市育英中学是一所完全中学，是市属重点中学，现有初中18个班级，高中22个班级，在校学生2200余名，教职工164人。

学校位于杏花岭区城坊东街中段，占地41亩，有教学楼、图书楼、实验楼、办公楼各一栋，并开辟有学生宿舍，校园环境优美怡人，学校语音室、电教室、计算机室、多媒体视听教室、多功能综合电教室等设备齐全，现代化教学设施初具规模。

学校注重教风、学风和校风建设，追求文化品味与格调，讲求教书育人、管理育人，注重学生参与社会实践活动，注重德育教育、礼仪教育和文明习惯的养成，全面贯彻党的教育方针，初步形成了“和谐发展”的办学特色。

育英中学的电化教学工作自70年代起，一直走在太原市的前列，多次获省市电教先进集体奖。

在推进多媒体辅助教学的进程中，学校投入大量资金，积极引进设备，多次组织教师培训，组建了多媒体创作工作组。

在大面积推广和普及多媒体辅助教学的活动中，该校把重点放在应用教学软件，提高课堂效率上。

山西省太原市第十二中学2024学年高三英语第一学期期末学业质量监测模拟试题含解析

山西省太原市第十二中学2024学年高三英语第一学期期末学业质量监测模拟试题考生须知：1．全卷分选择题和非选择题两部分，全部在答题纸上作答。

选择题必须用2B铅笔填涂；非选择题的答案必须用黑色字迹的钢笔或答字笔写在“答题纸”相应位置上。

2．请用黑色字迹的钢笔或答字笔在“答题纸”上先填写姓名和准考证号。

3．保持卡面清洁，不要折叠，不要弄破、弄皱，在草稿纸、试题卷上答题无效。

第一部分（共20小题，每小题1.5分，满分30分）1．Jess was sad and her friend helped her ___ the first awful weeks after her husband Bill died.A．break through B．break downC．get through D．get rid of2．At that time, my mind was a complete _______ ; I couldn't think of a single answer. A．blank B．bonusC．blow D．bottom3．He gets up at six o’cl ock, runs for half an hour and then has a meal; that is his morning ______.A．route B．routineC．behavior D．tendency4．We offered to pay our half of the cost that was needed to restore the shared doorway but Charles would have of it.A．nothing B．anythingC．none D．any5．Decades ago, scientists believed that how the brain develops when you are a kid______ determines your brain structure for the rest of your life.A．sooner or later B．more or less C．to and from D．up and down6．— Sorry, sir. I ______ follow you. Would you please speak slower?— Ok.A．can’t B．mustn’t C．shouldn’t D．needn’t7．（2018·海淀二模）—Excuse me, sir. Can you spare me a dollar ________ I can buy this book?—Sure, no problem.A．for B．soC．but D．or8．When I got to his house, I found that the walls _____ .A．are being painted B．are paintingC．were being painted D．were painting9．Julie is one of those women who always the latest fashion.A．put up with B．keep up with C．come up with D．get on with10．I hope when you come tomorrow, you _____ the reading and have something to share.A．did B．are doingC．will be doing D．will have done11．The auto factory ______ new profit records through technical innovation — 10% growth rate in the last two years and hopefully 15% this year.A．set B．has setC．is setting D．has been setting12．—You shouldn’t have treated me that way.My heart is broken.—I’m sorry,Paul.I didn’t mean you.A．hurting B．to hurt C．hurt D．having hurt13．— How can I wake up so early?—Set the alarm at 5:00 am., you will make it.A．and B．butC．or D．so14．—I heard Mr. Morgan would be here at 4:00 pm. next Thursday.—No, he _____ at that time.A．was boarding B．would be boardingC．will be boarding D．is boarding15．－I hear Iron Man III is on recently. Let’s set off for the cinema to appreciate it now.－________ It’s about 10 PM. I’m so tired that I must go to bed.A．Let’s find some of the action. B．How do you find it?C．It’s entirely up to you. D．You can’t be serious.16．All of a sudden, the thief walking behind a young and pretty lady _______ her purse, ______into the crowd.A．seizing; rushed B．seized; rushed C．seizing; rushing D．seized; rushing 17．A teacher’s job is not to tell the students what to believe or value, but to ________ them to develop a worldview for themselves.A．urge B．equipC．persuade D．rank18．Criticized as online games are, they never fail to ______ a large number of teenagers．A．appeal to B．object to C．refer to D．turn to19．We completed one third of the project, and the loan _______ in place, we had to delay the rest till the next month.A．not arranged B．was not arrangedC．not arranging D．had not been arranged20．We’d better discuss everything ______before we work out the plan.A．in detail B．in general C．on purpose D．on time第二部分阅读理解（满分40分）阅读下列短文，从每题所给的A、B、C、D四个选项中，选出最佳选项。

山西省太原市第十二中学校2020-2021学年八年级下学期3月月考物理试题

太原十二中学2020~2021学年初二年级第二学期3月月考试题物理一、单选题（本大题共20小题，每题3分，共60分）1.下列关于力的说法正确的是A.只有相互接触的物体才会产生力B.只要有物体，就一定会产生力的作用C.力不能脱离物体而存在D.孤立的一个物体也能产生力的作用2.“足球进校园”是国家教育部增强青少年体质的一项重大举措，下列足球运动中不属于力．．．．改变物体运动状态的是（）A.用力把足球踢出去B.被踩在脚下的足球变形了C.守门员抱住飞来的足球D.草地上滚动的足球最终停下来3.下列体育项目中的一些现象，不能用“力的作用是相互的”来解释的是（）A.跳水运动员踩踏跳板。

身体向上跳起B.铅球运动员投出铅球后，身体随着向前倾倒C.滑冰运动员用力推墙身体离墙而去D.游泳运动员向后划水，身体前进4.在航空领域，常发生小鸟撞毁飞机事件，关于小鸟和飞机相撞时的说法正确的是（）A.小鸟和飞机的受力一样大B.飞机受到的力大C.小鸟受到的力大D.主动撞击的一方产生的力大5.有关弹簧测力计的使用方法，以下说法错误的是（）A.测量前要使指针对准零刻度线B.测量时被测力的大小应在量程之内C.测量时，应使测力计的弹簧伸长方向与所测力的方向在同一直线上D.测量重力时测力计不能拉着物体沿竖直方向匀速直线运动6.关于弹力的叙述中，正确的是()A.只有弹簧、橡皮筋这类物体才有弹力B.只要物体发生形变就会产生弹力C.任何物体的弹性都有一定的限度，因而弹力不可能无限大D.弹力的大小只与物体形变的程度有关7.在图中，A、B两球相互间一定有弹力作用的图是（）A. B.C. D.8.在实验时，小明将一个正常的铁质外壳测力计的挂钩挂在铁架台上，静止时有如图所示的示数。

接着，他把这个测力计像右图中乙那样，上下各挂一个100g的钩码，并挂到甲测力计下，g取10N/kg，则甲乙两测力计的示数分别是（）A.1.0N和2.0NB.3.0N和1.0NC.3.0N和2.0ND.2.0N和3.0N9.过春节时贴年画时就用到我们的物理知识。

独家解析：山西省太原市2024年高三年级模拟考试英语试题(一) (含听力)

独家解析：山西省太原市2024年高三年级模拟考试英语试题（一） (含听力)一、听力选择题1. Why does the woman have to change her flight?A．Her father is ill.B．Her flight has been canceled.C．She suffers from a heart attack.2. What are the two speakers talking about?A．When the woman can get her book back.B．Where the man borrowed the book.C．What they’ll do this weekend.3. Where does the man most likely live?A．In Canada.B．In New York.C．In California.4.A．Read a book.B．Write a composition.C．Talk about a problem.D．Listen to the radio.5. Why is the woman worried?A．Kate is going to drive a long way.B．Kate plans to go to Colorado where there may be a heavy snow.C．Kate will spend her holiday in Colorado.二、听力选择题6. 听下面一段较长对话，回答以下小题。

1. What does the woman think of the coat she has been given?A．It’s too big.B．It is out of fashion.C．It isn’t worth the money.2. What does the man mean in the end?A．The woman has to pay the price difference.B．He doesn’t want to sell the coat to the woman.C．He refuses to exchange the coat for the woman.7. 听下面一段较长对话，回答以下小题。

2020年高考数学(理)二轮复习几何体与球切、接的问题

热点七几何体与球切、接的问题纵观近几年高考对于组合体的考查，与球相关的外接与内切问题是高考命题的热点之一.高考命题小题综合化倾向尤为明显，要求学生有较强的空间想象能力和准确的计算能力，才能顺利解答.从实际教学来看，这部分知识学生掌握较为薄弱、认识较为模糊，看到就头疼的题目.分析原因，除了这类题目的入手确实不易之外，主要是学生没有形成解题的模式和套路，以至于遇到类似的题目便产生畏惧心理. 下面结合近几年高考题对球与几何体的切接问题作深入的探究,以便更好地把握高考命题的趋势和高考的命题思路,力争在这部分内容不失分.从近几年全国高考命题来看,这部分内容以选择题、填空题为主，大题很少见.首先明确定义1：若一个多面体的各顶点都在一个球的球面上，则称这个多面体是这个球的内接多面体，这个球是这个多面体的外接球。

定义2：若一个多面体的各面都与一个球的球面相切，则称这个多面体是这个球的外切多面体，这个球是这个多面体的内切球.1 球与柱体的切接规则的柱体，如正方体、长方体、正棱柱等能够和球进行充分的组合，以外接和内切两种形态进行结合，通过球的半径和棱柱的棱产生联系，然后考查几何体的体积或者表面积等相关问题.1.1 球与正方体如图所示，正方体1111ABCD A B C D -，设正方体的棱长为a ，,,,E F H G 为棱的中点，O 为球的球心.常见组合方式有三类：一是球为正方体的内切球，截面图为正方形EFGH 和其内切圆，则2a OJ r ==；二是与正方体各棱相切的球，截面图为正方形EFGH和其外接圆，则2GO R a ==；三是球为正方体的外接球，截面图为长方形11ACA C和其外接圆，则1A O R '==.通过这三种类型可以发现，解决正方体与球的组合问题，常用工具是截面图，即根据组合的形式找到两个几何体的轴截面，通过两个截面图的位置关系，确定好正方体的棱与球的半径的关系，进而将空间问题转化为平面问题.（1）正方体的内切球，如图1. 位置关系：正方体的六个面都与一个球都相切，正方体中心与球心重合；=.数据关系：设正方体的棱长为a，球的半径为r，这时有2r a（2）正方体的外接球，如图2. 位置关系：正方体的八个顶点在同一个球面上；正方体中心与球心重合；数据关系：设正方体的棱长为a，球的半径为r，这时有2r=.（3）正方体的棱切球，如图3. 位置关系：正方体的十二条棱与球面相切，正方体中心与球心重合；数据关系：设正方体的棱长为a，球的半径为r，这时有2r=.例1【2018届福建省三明市A片区高中联盟校高三上学期期末】某几何体的三视图如图所示，则该几何体的外接球的表面积为（）A. 4πB. 8πC. 10πD. 12π【答案】D【针对练习】1.如图，虚线小方格是边长为1的正方形，粗实(虚)线画出的是某几何体的三视图，则该几何体外接球的表面积为A ．36πB ． 32πC ．9πD ．8π 答案及解析： 1.B 几何体的直观图如图所示为三棱锥，三棱锥中，，所以外接球的直径为，则半径，所以外接球的表面积,故选B.1.2 球与长方体例 2 自半径为R 的球面上一点M ，引球的三条两两垂直的弦MC MB MA ,,，求222MC MB MA ++的值．【答案】24R .【解析】以MC MB MA ,,为从一个顶点出发的三条棱，将三棱锥ABC M -补成一个长方体，则另外四个顶点必在球面上，故长方体是球的内接长方体，则长方体的对角线长是球的直径． ∴222MC MB MA ++=224)2(R R =．例 3【2018届二轮复习专题】《九章算术》中，将底面为长方形且有一条侧棱与底面垂直的四棱锥称之为阳马；将四个面都为直角三角形的三棱锥称之为鳖臑．若三棱锥P-A BC 为鳖臑，PA ⊥平面ABC ，PA ＝AB ＝2，AC ＝4，三棱锥P-ABC 的四个顶点都在球O 的球面上，则球O 的表面积为( )A. 8πB. 12πC. 20πD. 24π【答案】C【针对练习】1.已知边长为2的等边三角形ABC，D为BC的中点，以AD为折痕，将△ABC折成直二面角，则过A，B，C，D四点的球的表面积为A. 2πB. 3πC. 4πD. 5π答案及解析：1.D5π.故选D.2 球与锥体的切接规则的锥体，如正四面体、正棱锥、特殊的一些棱锥等能够和球进行充分的组合，以外接和内切两种形态进行结合，通过球的半径和棱锥的棱和高产生联系，然后考查几何体的体积或者表面积等相关问题.2.1正四面体与球的切接问题（1）正四面体的内切球，如图4. 位置关系：正四面体的四个面都与一个球相切，正四面体的中心与球心重合；==；（可以利用体积桥证数据关系：设正四面体的棱长为a，高为h；球的半径为R，这时有4R h a明）（2）正四面体的外接球，如图5. 位置关系：正四面体的四个顶点都在一个球面上，正四面体的中心与球心重合；数据关系：设正四面体的棱长为a ，高为h ；球的半径为R ，这时有43R h ==；（可用正四面体高h 减去内切球的半径得到）（3）正四面体的棱切球，如图6. 位置关系：正四面体的六条棱与球面相切，正四面体的中心与球心重合；数据关系：设正四面体的棱长为a ，高为h ；球的半径为R ，这时有4,.R h ===例 4【2018届广西防城港市高三1月模拟】各面均为等边三角形的四面体ABCD 的外接球的表面积为3π，过棱AB 作球的截面，则截面面积的最小值为__________．【答案】2π 【解析】将四面体放回一个正方体中,使正四面体的棱都是正方体的面对角线,那么正四面体和正方体的外接球是同一个球,当AB 是截面圆的直径时,截面面积最小.因外接球的表面积为3π,,棱长为1,截面圆面积最小值为222ππ⎛⨯= ⎝⎭.点评：与球有关的组合体问题，一种是内切，一种是外接．解题时要认真分析图形，明确切点和接点的位置，确定有关元素间的数量关系，并作出合适的截面图.【针对练习】1.设A ，B ，C ，D 是同一个半径为4的球的球面上四点，△ABC 为等边三角形且其面积为，则三棱锥D -ABC 体积的最大值为（）A. B. C. D. 答案及解析：1.C2.在四面体ABCD 中，1AD DB AC CB ====，则四面体体积最大时，它的外接球半径R = ．答案及解析：2.6如图，取AB中点E，连接CE，DE，设AB=2x（0＜x＜1），则CE=DE=，∴当平面ABC⊥平面ABD时，四面体体积最大，为V===．V′=，当x∈（0，）时，V为增函数，当x∈（，1）时，V为减函数，则当x=时，V有最大值．设△ABD的外心为G，△ABC的外心为H，分别过G、H作平面ABD、平面ABC的垂线交于O，则O为四面体ABCD的外接球的球心．在△ABD中，有sin，则cos，∴sin=．设△ABD的外接圆的半径为r，则，即DG=r=．又DE=，∴OG=HE=GE=．∴它的外接球半径R=OD=．2.2其它棱锥与球的切接问题球与正棱锥的组合，常见的有两类，一是球为三棱锥的外接球，此时三棱锥的各个顶点在球面上，根据截面图的特点，可以构造直角三角形进行求解.二是球为正棱锥的内切球，例如正三棱锥的内切球，球与正三棱锥四个面相切，球心到四个面的距离相等，都为球半径R ．这样求球的半径可转化为球球心到三棱锥面的距离，故可采用等体积法解决，即四个小三棱锥的体积和为正三棱锥的体积.球与一些特殊的棱锥进行组合，一定要抓住棱锥的几何性质，可综合利用截面法、补形法等进行求解.例如，四个面都是直角三角形的三棱锥，可利用直角三角形斜边中点几何特征，巧定球心位置.例5【湖南省长沙市长郡中学2017届高三摸底】已知边长为ABCD 中，60BAD ∠=，沿对角线BD 折成二面角A BD C --为120的四面体ABCD ，则四面体的外接球的表面积为（）A ．25πB ．26πC ．27πD ．28π【答案】D图2图1G O EAC ED C BA例6【江西省新余市第一中学2017届高三上学期调研考试（一）】某几何体的正视图和侧视图如图（1）所示, 它的府视图的直观图是'''A B C ,如图（2）所示,其中0''''2,''A O B O C ===则该几何体的外接球的表面积为．【答案】1123π 【解析】由斜二测画法易知，该几何体的俯视图是一个边长为4的等边三角形,再结合正视图和侧视图可知，该几何体是如下图所示的高为4的三棱锥D －ABC ，将其补形为三棱柱ABC-EDF,设球心为O ，EDF ∆的中心为1O ，则124sin 6033O EDE ==,所以该几何体的外接球的半径R OE ====其表面积为211243S R ππ==.EDB A例7【2018届山西省太原十二中高三上学期1月】在四棱锥P ABCD -中， PC ⊥底面ABCD ，底面为正方形， //QA PC ， PBC AQB ∠=∠= 60，记四棱锥P ABCD -的外接球与三棱锥B ACQ -的外接球的表面积分别为12,S S ，则21S S =___．【答案】157【解析】设正方形的边长为a ，设2O 为CQ 的中点，因为PC ⊥平面ABCD ，而,CD CB ⊂平面ABCD ，所以,PC CD PC CB ⊥⊥，又//AQ PC ，故,AQ CD AQ CB ⊥⊥，又CD CB C ⋂=，故AQ ⊥平面ABCD ，AC ⊂平面ABCD ，所以AQ AC ⊥，故QAC ∆为直角三角形， CQ 为斜边，所以222QO CQ AQ ==．同理QAC ∆也为直角三角形，结合60AQB ∠=︒ ，所以AQ =，又CB BA ⊥， AQ AB A ⋂=，所以CB ⊥平面AQB ， QB ⊂平面AQB ，所以CB QB ⊥， QBC ∆为直角三角形，所以22BO QO =， 2O 为三棱锥B AQC - 外接球的球心，且半径212R QC ===．同理设1O 为AP 的中点，则1O 为四棱锥P ABCD -外接球的球心，且半径112R AP ===，所以1252115::4367S S ==．填157．点睛：球的半径的计算，关键在球心位置的确定，三棱锥B AQC -中,QAC QBC ∆∆均为直角三角形，因此外接球的球心就是QC 的中点，因为它到四个顶点的距离是相等的．同理四棱锥P ABCD -外接球的球心就是AP 的中点．【针对练习】 1.已知在三棱锥P -ABC 中，1PA PB BC ===，AB =，AB BC ⊥，平面P AB ⊥平面ABC ，若三棱锥的顶点在同一个球面上，则该球的表面积为（）A B ．3π C ．3D．2π 答案及解析：1.B 试题分析：如下图所示，设球心为O ，则可知球心O 在面ABC 的投影在ABC ∆外心，即AC 中点E 处，取AB中点F ，连PF ，EF ，OE ，OP ，由题意得，PF ⊥面ABC ，∴在四边形POEF 中，设OE h =，∴半径0r h ===，2r =，即球心即为AC 中点，∴表面积243S r ππ==，故选B.2.在四面体S ﹣ABC 中，SA ⊥平面ABC ，∠BAC =120°，SA =AC =2，AB =1，则该四面体的外接球的表面积为 A ．11πB ．328π C ．310πD ．340π答案及解析：2.D ∵AC=2，AB=1，∠BAC=120°， ∴BC=，∴三角形ABC 的外接圆半径为r ，2r= ，r= ，∵SA ⊥平面ABC ，SA=2，由于三角形OSA 为等腰三角形，O 是外接球的球心．则有该三棱锥的外接球的半径R=， ∴该三棱锥的外接球的表面积为S=4πR 2=．选D.3.在四面体ABCD 中，△BCD 与△ACD 均是边长为4的等边三角形，二面角A -CD -B 的大小为60°，则四面体ABCD 外接球的表面积为（）A ．2089πB ．529πC ．643πD ．523π答案及解析：3.A根据题意得到这个模型是两个全等的三角形，二面角大小为，取CD 的中点记为O ，连结OB ，OA ，根据题意需要找到外接球的球心，选择OA的离O点近的3等分店记为E，同理去OB上一点记为F，自这两点分别做两个面的垂线，交于点P，则点P就是球心。

2023届山西省太原市高三上学期1月第一次联考数学试题(解析版)

2023届山西省太原市高三上学期1月第一次联考数学试题一、单选题1．已知集合{}28xA x =<，集合{}B x x a =>，若A B ⋂=∅，则实数a 的取值范围为（）A ．(,2)-∞B ．(2,)+∞C ．(,3]-∞D ．[3,)+∞【答案】D【分析】先求出集合A ，B ，再由A B ⋂=∅求出实数a 的取值范围.【详解】{}{}{}{}328223,x x A x x x x B x x a =<=<=<=>.又A B ⋂=∅，所以a 的取值范围为[3,)+∞. 故选：D2．若复数z 满足()()112i,z z +-=+i 是虚数单位,则z =（）A ．1 BC D ．2【答案】C【分析】先设复数i z a b =+,代入()()112i,z z +-=+即可得关于,a b 的等式,进而可求z . 【详解】解:由题知不妨设()i ,z a b a b =+∈R , 因为()()112i,z z +-=+所以()()()22i 1i 11i a b a b a b ++--=-+ 2212i a b b =+--2i =+,所以2212a b +-=，21b -=, 故223a b +=,12b =-,所以z =故选:C3．抛物线24y x =的焦点为F ,抛物线上一点P 在其对称轴的上方,若3PF =,则点P 的坐标是（）A ．()4,4B ．(3,C ．(2,D ．()1,2【答案】C【分析】先设出P 点坐标,根据抛物线定义列出等式,即可得点P 坐标.【详解】解:由题设点P 的坐标为(),x y , 根据抛物线的定义知13PF x =+=, 所以2,x =代入抛物线中可得22y =, 故点P 的坐标为()2,22. 故选:C4．地震的震级越大,以地震波的形式从震源释放出的能量就越大,震级M 与所释放的能量E 的关系如下: 4.81.510M E +=（焦耳）()10 3.16≈,那么6级地震释放的能量是4级地震释放的能量的（）A ．3.16倍B ．31.6倍C ．100倍D ．1000倍【答案】D【分析】分别设出4级地震释放的能量和6级地震释放的能量,列出各自等式,将两等式相除进行化简,即可得出结果.【详解】解:由题设4级地震释放的能量为1,6E 级地震释放的能量为2E , 所以 4.8 1.5410.8 4.8 1.5613.8121010,1010E E +⨯+⨯====,所以13.83210.811010100010E E ===, 即6级地震释放的能量是4级地震释放的能量的1000倍. 故选:D5．某同学在参加《通用技术》实践课时，制作了一个工艺品，如图所示，该工艺品可以看成是一个球被一个棱长为4的正方体的六个面所截后剩余的部分（球心与正方体的中心重合），若其中一个截面圆的周长为2π，则该球的表面积为（）A ．20πB ．16πC ．12πD ．8π【答案】A【分析】设截面圆半径为r ，球的半径为R ，根据截面圆的周长求得1r =，再利用2222R r 求解.【详解】设截面圆半径为r ，球的半径为R ，则球心到某一截面的距离为正方体棱长的一半即2，根据截面圆的周长可得22r ππ=，则1r =，由题意知2222R r ，即222125R =+=，∴该球的表面积为2420R ππ=．故选：A6．2020年春节联欢晚会以“共圆小康梦、欢乐过大年”为主题，突出时代性、人民性、创新性，节目内容丰富多彩，呈现形式新颖多样.某小区的5个家庭买了8张连号的门票，其中甲家庭需要3张连号的门票，乙家庭需要2张连号的门票，剩余的3张随机分到剩余的3个家庭即可，则这8张门票不同的分配方法的种数为（） A ．48 B ．72 C ．120 D ．240 【答案】C【解析】根据甲、乙2个家庭的5张票是否连号分类计算.【详解】若甲、乙2个家庭的5张票连号，则有142448A A ⋅=种不同的分配方法，若甲、乙2个家庭的5张票不连号，则有323472A A ⋅=种不同的分配方法，综上，这8张门票共有4872120+=种不同的分配方法，故选：C.【点睛】(1)解排列组合问题要遵循两个原则：一是按元素(或位置)的性质进行分类；二是按事情发生的过程进行分步．具体地说，解排列组合问题常以元素(或位置)为主体，即先满足特殊元素(或位置)，再考虑其他元素(或位置)．(2)不同元素的分配问题，往往是先分组再分配．在分组时，通常有三种类型：①不均匀分组；②均匀分组；③部分均匀分组，注意各种分组类型中，不同分组方法的求法．7．在矩形ABCD 中,2AB AD ==,点E 满足23DE DC =,则AE BD ⋅=（）A ．14-B ．14C ．16-D ．-【答案】A【分析】根据题意建立合适的平面直角坐标系,找到各个点的坐标,根据23DE DC =,求出E 点坐标,代入AE BD ⋅中即可得出结果.【详解】解:由题不妨以A 为坐标原点,,AB AD 方向分别为,x y 轴建立如图所示直角坐标系,则()()()()0,0,23,0,23,2,0,2,A B C D 所以()23,0DC =,()23,2BD =-, 因为23,DE DC = 设(),E x y ,所以()(),2223,03x y -=, 解得()33,2E , 所以()33,2AE =,所以()()33,223,214AE BD =⋅-⋅=-. 故选:A8．已知1a b >>,若1e e e a a b b a a ++=+,则（） A ．()ln 1a b +> B ．()ln 0a b -< C ．333a b -+<D ．133a b -<【答案】A【分析】化简1e e eaab b a a ++=+为1e e 111a b a b b +-=++,构造函数()e (1)xf x x x=>,求导求单调性,即可得12a b >+>,即13e a b b b +>++>>,两边取对数即可判断选项A 正误;根据12a b >+>,可得1a b ->即可得选项B 正误;根据12a b >+>,可得233323,a b -+>>即可判断选项C 正误;根据1a b ->,即可得选项D 正误.【详解】解:由题知1e e e a a b b a a ++=+,()()11e e 1a b b a ++=+,即1e e 11++=+a b a b ,即1e e 111a b a b b +-=++, 令()e (1)xf x x x=>,所以()()21e 0x x f x x -'=>, 故()f x 在()1,+∞上单调递增, 因为1a b >>, 所以101b >+, 由1e e 1011a b a b b +-=>++可知:()()1f a f b >+, 根据单调性可得12a b >+>, 所以1a b ->, 故()ln ln10,a b ->= 故选项B 错误;因为13e a b b b +>++>>. 所以()ln 1,a b +> 故选项A 正确; 因为12a b >+>,所以2333a b -+>> 故选项C 错误; 因为1a b ->, 所以133,a b -> 故选项D 错误. 故选:A二、多选题9．2022年北京冬奥会给中国冰雪产业带来快速发展，冰雪运动人数快速上升，冰雪运动市场需求得到释放，并引领相关户外用品行业市场增长.下面是2013年至2020年中国雪场滑雪人次（万人次）与同比增长率（与上一年相比）的统计情况，则下面结论中正确的是（）A ．2013年至2020年，2019年中国雪场滑雪人次最多B ．2013年至2020年，2015年中国雪场滑雪人次的同比增长率最高C ．2013年到2020年，中国雪场滑雪人次在2020年首次出现负增长D ．2013年至2020年，中国雪场滑雪人次的年增加量相近【答案】ABC【分析】根据折线图以及条形图，结合选项即可逐一求解.【详解】由条形图知，2013年至2020年，2019年中国雪场滑雪人次达2000万人次，故滑雪人次最多，故A 正确，根据同比增长率折线图可知2015年中国雪场滑雪人次的同比增长率最高，中国雪场滑雪人次在2020年首次出现负增长，故B ，C 正确，中国雪场滑雪人次的年增加量不相近，故D 错误. 故选：ABC10．将函数π()2sin(2)2cos26=+-f x x x 的图象向左平移π6个单位长度，得到()y g x =的图象,则下列说法正确的是（）A ．函数()g x 的最小正周期为πB ．函数()g x 的最小值为1-C ．函数()g x 的图象关于直线π6x =对称 D ．函数()g x 在2π[,π]3上单调递减【答案】AC【分析】根据两角和的正弦以及辅助角公式，将()f x 化为正弦型函数，再由图象平移关系求出()g x 的解析式，结合正弦函数性质，逐项验证，即可得出结论.【详解】π()2sin(2)2cos23sin2cos22sin(2)66f x x x x x x π=+--=-，()()2sin[2()]2sin(2)6666g x f x x x ππππ=+=+-=+，()g x 的周期为π，选项A 正确；()g x 的最小值为2-，选项B 错误；()2sin(2)2666g πππ=⨯+=为()g x 的最大值，所以直线π6x =是()g x 的一条对称轴，选项C 正确； 2π37[,π],2[,],()3626x x g x πππ∈+∈单调递增，选项D 错误. 故选：AC.【点睛】本题考查三角恒等变换化简、三角函数平移变换求解析式，以及三角函数图象性质，属于基础题.11．直线240x y --=分别与x 轴，y 轴交于,A B 两点，点P 在圆22(2)5x y +-=上，则PAB 面积的可能值是（） A ．2 B ．4 C ．8 D ．16【答案】ABC【分析】先确定弦长||AB 的长，再利用点到直线距离公式求出圆心到直线240x y --=的距离0d ，则点P 到直线240x y --=的距离的最大值为圆心到直线240x y --=的距离加上一个半径，最小值为圆心到直线240x y --=的距离减去一个半径，再结合三角形面积公式即可求出ABP 面积的取值范围，结合选项得答案．【详解】解：由题意可知：()()2,0,0,4A B -，则AB圆22(2)5x y +-=的圆心为()0,2，半径r =则圆心()0,2到直线240x y --=的距离为0d = 设点P 到直线240x y --=的距离为d ，则min0max 0d d r d d r =-===+=，即d ∈⎣⎦，又12ABPSAB d =⨯⨯=，[]1,11ABPS ∴∈结合选项可知，PAB 面积的可能取值是2或4或8. 故选：ABC.12．已知直三棱柱111ABC A B C 中，1,2AB BC AB BC BB ⊥===，点P 是线段1BC 上一点（包含端点），则下列说法正确的是（）A ．点1B 到平面11ACC A 2B ．异面直线1BB 与1A P 所成角的余弦值为定值C ．若P 是1BC 中点，则直线1A P 与平面111A B C 25D ．ACP △的面积S 的取值范围为2622⎡⎢⎣ 【答案】AD【分析】建立空间直角坐标系，利用空间向量进行求解.【详解】如图，1,D D 分别为11,AC A C 中点，易证1,,BD CD DD 两两垂直，建立如图所示直角坐标系，则))12,0,0,2,0,2BB ，()()()()112,2,0,2,0,0,2,2,2,0C A A C -，()112,2,0A B =，由题意可知平面11ACC A 的一个法向量为()1,0,0m =，所以点1B 到平面11ACC A 的距离为112A B m d m⋅==A 正确；设()101BP BC λλ=≤≤，则))()11112121,22A A BP BC P A B B λλλλ==-++=+-，()10,0,2BB =，1111211cos ,21BB A P BB A P BB A Pλλ⋅==-+B 不正确；若P 是1BC 中点，则22P ⎫⎪⎪⎝⎭，123212A P ⎛⎫=- ⎪ ⎪⎝⎭, 易知平面111A B C 的一个法向量为()0,0,1n =，设直线1A P 与平面111A B C 所成的角为θ，则1116sin 6A P n A P nθ⋅===，所以5tan θ=，C 不正确；))()()2121,2,0,1,0AC AP ACλλλμ=-+==，点P 到直线AC 的距离2221423()6,2333d AP AP μλ⎡⎤⎛⎫=-⋅=-+∈⎢⎥ ⎪⎝⎭⎣⎦，所以ACP △的面积S 的取值范围为26,223⎡⎤⎢⎥⎣⎦，D 正确. 故选：AD.三、填空题13．若函数()ln f x x ax =-的图象在()()1,1f 处的切线斜率为12，则实数=a __________.【答案】12##0.5【分析】求出函数()f x 的导数，再利用导数的几何意义及直线斜率的定义可求【详解】因为()ln f x x ax =-，所以()1f x a x'=-，所以()f x 在1x =处的切线斜率()1112k f a ==-='，解得12a =．故答案为：12．14．写出一个最小正周期为3的偶函数__________. 【答案】()2πcos3f x x =（答案不唯一）【分析】通过题意可联想到余弦型函数()()cos 0f x A x A ω=≠，根据周期求出对应参数即可【详解】由最小正周期为3，可考虑三角函数中的余弦型函数()()cos 0f x A x A ω=≠，满足()()cos f x A x f x ω-==，即是偶函数；根据最小正周期2π3T ω==，可得2π3ω=. 故令1A =，()2πcos3f x x =，故答案为：()2πcos3f x x =（答案不唯一） 15．高斯是德国著名的数学家，有“数学王子”之称，以其名字命名的成果有110个.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，若用{}[]x x x =-表示x的非负纯小数，如1=，已知数列{}n a满足[]{}111n n n a a a a +==+，则2021a =__________.【答案】30303030【分析】根据题意求出12345,,,,a a a a a ，找出规律，即可求出2021a 的值.【详解】13a =，2312a ∴==3622a ==+4942a ==51252a ==+，由此可得到规律：当n为奇数时，132n n a -=⨯，202120211330302a -∴=⨯=故答案为：20213030a =16．已知12,F F 为椭圆22164x y +=的左､右焦点，若动直线l 垂直于y 轴，交此椭圆于,A B 两点，P 为l上满足3PA PB ⋅=的点，则点P 的轨迹方程为__________.【答案】221(22)32x y y +=-<<或221(22)96x y y +=-<<【分析】设点P 的坐标为(),x y ，依题意得()()00,,,A x y B x y -，由3PA PB ⋅=，求出点P 的坐标关系式，代入椭圆方程求出P 点的轨迹方程.【详解】设点P 的坐标为(),x y ，依题意得()()00,,,A x y B x y -，因为PA PB ⋅=3，所以2222000033x x x x x x x x -⋅+=-=⇒=±，代入椭圆的方程得223164x y ±+=，即22132x y +=与221(22)96x y y +=-<<. 故答案为：221(22)32x y y +=-<<或221(22)96x y y +=-<<.四、解答题17．已知数列{}n a 的前n 项和为n S ,且满足132,15n n a a S +-==. (1)求数列{}n a 的通项公式; (2)令1111n n n b a a =⋅+-,求数列{}n b 的前n 项和n T . 【答案】(1)21n a n =+ (2)44n nT n =+【分析】(1)根据12n n a a +-=可得{}n a 是等差数列,且公差为2,代入315S =中,即可得首项,即可得数列{}n a 的通项公式;(2)根据(1)中结果,写出{}n b 通项公式,再利用裂项相消法即可得出结果. 【详解】（1）解:由题知,因为12n n a a +-=, 所以数列{}n a 是公差2d =的等差数列, 因为315S =, 所以13315a d +=, 解得13a =,所以数列{}n a 的通项公式是()1121n a a n d n =+-=+; （2）由(1)得21,n a n =+ 所以()11114141n b n n n n ⎛⎫==- ⎪++⎝⎭,故12n n T b b b =+++111111142231n n ⎛⎫=-+-++- ⎪+⎝⎭()1114141n n n ⎛⎫=-=⎪++⎝⎭, 所以数列{}n b 的前n 项和44n nT n =+.18．如图，在ABC 中，已知π3,6,,,3AB AC BAC BC AC ==∠=边上的两条中线,AM BN 相交于点P .(1)求BC 的长度； (2)求MPN ∠的余弦值. 【答案】(1)33BC = (2)714【分析】（1）根据余弦定理求解即可；（2）根据ABC 的三边关系可得π2ABC ∠=，结合直角三角形与三角形重心结论，求解,AP BP 的长，再利用余弦定理即可得MPN ∠的余弦值.【详解】（1）解：在ABC 中，由余弦定理知2222cos BC AB AC AB AC BAC =+-⋅⋅⋅∠，整理得21936236272BC =+-⨯⨯⨯=，解得33BC =. （2）解：因为22292736AB BC AC +=+==，则AB BC ⊥，所以π2ABC ∠=. 所以221373,22BN AC AM AB BM ===+=，由于,BC AC 边上的两条中线,AM BN 相交于点P ，则点P 为ABC 的重心，所以227,233AP AM BP BN ====，由余弦定理得2227cos 214AP BP AB APB AP BP +-∠==⋅. 所以MPN ∠的余弦值为714. 19．如图，已知四边形ABCD 为菱形，对角线AC 与BD 相交于O ，60BAD ∠=︒，平面ADEF 平面BCEF =直线EF ，FO ⊥平面ABCD ，22BC CE DE EF ====（1）求证：//EF DA ；（2）求二面角A EF B --的余弦值．【答案】（1）证明见解析；（2）35．【分析】（1）根据四边形ABCD 为菱形，得到//AD BC ，利用线面平行的判定定理得到//AD 平面BCEF ，然后利用线面平行的性质定理证明.（2）以O 为坐标原点、OA ，OB ，OF 为x ，y ，z 轴建立空间直角坐标系，取CD 中点M ，连EM ，OM ，分别求得平面ADEF 一个法向量为(,,)m x y z =，平面BCEF 一个法向量为(,,)n x y z =，然后由cos ,|||,|m nm n m n ⋅<>=求解.【详解】（1）因为四边形ABCD 为菱形，所以//AD BC ，AD ⊄平面BCEF ，BC ⊂平面BCEF //AD ∴平面BCEF ，因为平面ADEF 平面BCEF =直线,EF AD ⊂平面ADEF ，所以//EF AD ；（2）因为四边形ABCD 为菱形，所以AC BD ⊥，因为OF ⊥平面ABCD ，所以以O 为坐标原点、OA ，OB ，OF 为x ，y ，z 轴建立空间直角坐标系，取CD 中点M ，连EM ，OM ，60BAD ︒∠=，23,1BC OA OC OB OD =∴====，2BC CD CE DE CDE ====∴为正三角形，3EM11//,=,//,=22OM BC OM BC EF BC EF BC ，//,=//,=EF OM EF OM OF EM OF EM ∴∴，从而31(3,0,0),(0,1,0),(3,0,0),(0,1,0),(3)2A B C D E ---，设平面ADEF 一个法向量为(,,)m x y z =，则00m DA m DE ⎧⋅=⎨⋅=⎩，即0102y x y ⎧+=⎪⎨+=⎪⎩，令11,(1,x y z m =∴===-，设平面BCEF 一个法向量为(,,)n x y z =，则00n BC n EC ⎧⋅=⎨⋅=⎩，即0102y y ⎧-=⎪⎨+=⎪⎩，令11,(1,3,1)x y z n =∴==-=--，3cos ,5|||,|m n m n m n ⋅∴<>==，因此二面角A EF B --的余弦值为35.【点睛】方法点睛：求二面角最常用的方法就是分别求出二面角的两个面所在平面的法向量，然后通过两个平面的法向量的夹角得到二面角的大小，但要注意结合实际图形判断所求角是锐角还是钝角．20．某品牌汽车4S 店对2020年该市前几个月的汽车成交量进行统计，用y 表示2020年第x 月份该店汽车成交量，得到统计表格如下：（1）求出y 关于x 的线性回归方程ˆˆˆy bx a =+，并预测该店9月份的成交量；（ˆa，ˆb 精确到整数）（2）该店为增加业绩，决定针对汽车成交客户开展抽奖活动，若抽中“一等奖”获5千元奖金；抽中“二等奖”获2千元奖金；抽中“祝您平安”则没有奖金．已知一次抽奖活动中获得“二等奖”的概率为13，没有获得奖金的概率为16．现有甲、乙两个客户参与抽奖活动，假设他们是否中奖相互独立，求此二人所获奖金总额X （千元）的分布列及数学期望．参考数据及公式：81850i i i x y ==∑，821204ii x ==∑，()()()121ˆniii nii x x y y bx x ==--=-∑∑，ˆ=-ay bx ．【答案】（1）ˆ212y x =+；预计9月份的成交量为30辆；（2）分布列见解析；期望为193．【分析】(1)先分别求出i x ，i y 的平均数,x y ，再利用最小二乘法计算即可得回归直线方程，取x =9可得成交量的预测值；(2)写出随机变量X 的所有可能值，再计算出X 取各个值时的概率，列出分布列即可得解. 【详解】（1）由题意得：1+2+3+4+5+6+7+89=82x =，14+12+20+20+22+24+30+26=218y =，∴81822219885082194229422048()82i ii i i x y x yb x x==-⋅-⨯⨯===≈-⨯-∑∑，∴12a y b x =-⋅= 所以，回归直线方程为ˆ212yx =+， ∴当9x =时，ˆ2912=30y=⨯+，即预计9月份的成交量为30辆；（2）由题意得：获得“一等奖”的概率为12，所以X 的可能取值为0，2，4，5，7，10，∴()11106636P X ==⨯=，()11111236639P X ==⨯+⨯=，()1114339P X ==⨯=，()11111526626P X ==⨯+⨯=，11111723323P X，()11110224P X ==⨯=，所以X 的分布列为：∴()11111119024571036996343E X =⋅+⋅+⋅+⋅+⋅+⋅=. 21．已知双曲线C ：()222210,0x y a b a b-=>>与双曲线22123x y -=有相同的焦点；且C 的一条渐近线与直线220x y 平行. (1)求双曲线C 的方程；(2)若直线l 与双曲线C 右支相切（切点不为右顶点），且l 分别交双曲线C 的两条渐近线于,A B 两点，O 为坐标原点，试判断AOB 的面积是否为定值，若是，请求出；若不是，请说明理由.【答案】(1)2214x y -=(2)是，2【分析】（1）根据题意列式求解,,a b c ，即可得方程；（2）设直线:l y kx m =+，联立方程由Δ0=可得2214k m -=-，根据题意求,A B 的坐标，即可求AOB 的面积，化简整理即可. 【详解】（1）设双曲线C 的焦距为()20c c >，由题意可得：22212c c a b b a ⎧⎪=⎪=+⎨⎪⎪=⎩，则21a b c ⎧=⎪=⎨⎪=⎩，则双曲线C 的方程为2214x y -=.（2）由于直线l 与双曲线C 右支相切（切点不为右顶点），则直线l 的斜率存在，设直线l 的方程为y kx m =+，则2214y kx m x y =+⎧⎪⎨-=⎪⎩，消y 得：()222418440k x kmx m -+++=，则()()2222Δ64441440k m k m =--+=，可得：2214k m -=-①设l 与x 轴交点为,0m D k ⎛⎫- ⎪⎝⎭，则11222OAB AOD BOD A B A BA B m mS S S OD y y k x x x x k --=+=⨯-=⋅⋅-=⋅-△△△， ∵双曲线两条渐近线方程为：12y x =±，联立12y kx m y x =+⎧⎪⎨=⎪⎩，解得21212m x km y k ⎧=⎪⎪-⎨⎪=⎪-⎩，即2,1212m m A k k ⎛⎫ ⎪--⎝⎭，同理可得：2,2121mm B k k -⎛⎫ ⎪++⎝⎭，则22224422212122142AOB A B m m m m m m mm S x x k k k m ----=⋅-=⋅+=⋅=⋅=-+--△（定值）. 22．已知函数()2e sin x f x x ax =-．（e 是自然对数的底数） (1)若0a =，求()f x 的单调区间；(2)若06a <<，试讨论()f x 在(0,)π上的零点个数．（参考数据：2e 4.8π≈）【答案】(1)单调递增区间为32,2()44k k k ππππ⎛⎫-++∈ ⎪⎝⎭Z ，单调递减区间为372,2()44k k k ππππ⎛⎫++∈⎪⎝⎭Z (2)答案见解析【分析】（1）求出导函数()f x '，令()0f x '>可得增区间，()0f x '<可得减区间；（2）利用导数判断()f x '在0,2π⎛⎫ ⎪⎝⎭上单调递增，在,2ππ⎛⎫ ⎪⎝⎭上单调递减，又(0)2f a '=-，02f π⎛⎫'> ⎪⎝⎭，()0f π'<，从而分02a <和26a <<两种情况讨论，根据函数零点存在定理及函数()f x '的单调性，求出()f x 的单调区间，从而即可求解.【详解】（1）解：0a =，则()2e sin x f x x =，定义域为R ，()2e (sin cos )sin 4x xf x x x x π⎛⎫'=+=+ ⎪⎝⎭，由()0f x '>，解得sin 04x π⎛⎫+> ⎪⎝⎭，可得22()4k x k k ππππ<+<+∈Z ，解得322()44k x k k ππππ-<<+∈Z ，由()0f x '<，解得sin 04x π⎛⎫+< ⎪⎝⎭，可得222()4k x k k πππππ+<+<+∈Z ，解得3722()44k x k k ππππ+<<+∈Z ， ()f x ∴的单调递增区间为32,2()44k k k ππππ⎛⎫-++∈ ⎪⎝⎭Z ，单调递减区间为372,2()44k k k ππππ⎛⎫++∈⎪⎝⎭Z ；（2）解：由已知()2e sin x f x x ax =-，()2e (sin cos )x f x x x a '∴=+-，令()()h x f x '=，则()4e cos x h x x '=．(0,)x π∈，∴当0,2x π⎛⎫∈ ⎪⎝⎭时，()0h x '>；当,2x ππ⎛⎫∈ ⎪⎝⎭时，()0h x '<，()h x ∴在0,2π⎛⎫ ⎪⎝⎭上单调递增，在,2ππ⎛⎫ ⎪⎝⎭上单调递减，即()f x '在0,2π⎛⎫ ⎪⎝⎭上单调递增，在,2ππ⎛⎫⎪⎝⎭上单调递减．(0)2f a '=-，22e 02f a ππ⎛⎫'=-> ⎪⎝⎭，()2e 0f a ππ'=--<．①当20a -时，即02a <时，(0)0f '， 0,2x ππ⎛⎫∴∃∈ ⎪⎝⎭，使得()00f x '=，∴当()00,x x ∈时，()0f x '>；当()0,x x π∈时，()0f x '<， ()f x ∴在()00,x 上单调递增，()0,x π上单调递减．(0)0f =，()00f x ∴>，又()0f a ππ=-<，∴由函数零点存在性定理可得，此时()f x 在(0,)π上仅有一个零点； ②若26a <<时，(0)20f a =-<'，又()f x '在0,2π⎛⎫ ⎪⎝⎭上单调递增，在,2ππ⎛⎫ ⎪⎝⎭上单调递减，而22e 02f a ππ⎛⎫'=-> ⎪⎝⎭，10,2x π⎫⎛∴∃∈ ⎪⎝⎭，2,2x ππ⎛⎫∈ ⎪⎝⎭，使得()10f x '=，()20f x '=，且当()10,x x ∈、()2,x x π∈时，()0f x '<；当()12,x x x ∈时，()0f x '>．()f x ∴在()10,x 和()2,x π上单调递减，在()12,x x 上单调递增．(0)0f =，()10∴<f x ，222e 2e 3022f a πππππ⎛⎫=->-> ⎪⎝⎭，()20f x ∴>，又()0f a ππ=-<，∴由零点存在性定理可得，()f x 在()12,x x 和()2,x π内各有一个零点，即此时()f x 在(0,)π上有两个零点．综上所述，当02a <时，()f x 在(0,)π上仅有一个零点；当26a <<时，()f x 在(0,)π上有两个零点．【点睛】关键点点睛：本题（2）问的解题关键是根据函数零点存在定理及()f x '的单调性，求得函数()f x 的单调区间.。

2019届高三数学备考冲刺140分问题32与圆有关的最值问题含答案解析

问题32 与圆有关的最值问题一、考情分析通过对近几年的高考试题的分析比较发现,高考对直线与圆的考查,呈现逐年加重的趋势,与圆有关的最值问题,更是高考的热点问题.由于圆既能与平面几何相联系,又能与圆锥曲线相结合,命题方式比较灵活,故与圆相关的最值问题备受命题者的青睐. 二、经验分享1. 与圆有关的最值问题的常见类型及解题策略(1)与圆有关的长度或距离的最值问题的解法．一般根据长度或距离的几何意义，利用圆的几何性质数形结合求解．(2)与圆上点(x ，y )有关代数式的最值的常见类型及解法．①形如u ＝y －bx －a 型的最值问题，可转化为过点(a ，b )和点(x ，y )的直线的斜率的最值问题；②形如t ＝ax ＋by 型的最值问题，可转化为动直线的截距的最值问题；③形如(x －a )2＋(y －b )2型的最值问题，可转化为动点到定点(a ，b )的距离平方的最值问题．2.与圆有关的最值问题主要表现在求几何图形的长度、面积的最值，求点到直线的距离的最值，求相关参数的最值等方面．解决此类问题的主要思路是利用圆的几何性质将问题转化三、知识拓展1.圆外一点P 到圆C 上点的距离距离的最大值等于，最小值等于PC r -.2.圆C 上的动点P 到直线l 距离的最大值等于点C 到直线l 距离的最大值加上半径，最小值等于点C 到直线l 距离的最小值减去半径.3.设点M 是圆C 内一点，过点M 作圆C 的弦，则弦长的最大值为直径，最小的弦长为.四、题型分析(一) 与圆相关的最值问题的联系点 1.1 与直线的倾斜角或斜率的最值问题利用公式k ＝tan α(α≠90°)将直线的斜率与倾斜角紧密联系到一起,通过正切函数的图象可以解决已知斜率的范围探求倾斜角的最值,或者已经倾斜角的范围探求斜率的最值.处理方法：直线倾斜角的范围是[0,π),而这个区间不是正切函数的单调区间,因此根据斜率求倾斜角的范围时,要分⎣⎢⎡⎭⎪⎫0，π2与⎝ ⎛⎭⎪⎫π2，π两种情况讨论．由正切函数图象可以看出,当α∈⎣⎢⎡⎭⎪⎫0，π2时,斜率k ∈[0,＋∞)；当α＝π2时,斜率不存在；当α∈⎝ ⎛⎭⎪⎫π2，π时,斜率k ∈(－∞,0)．【例1】坐标平面内有相异两点,经过两点的直线的的倾斜角的取值范围是( ). A ．,44ππ⎡⎤-⎢⎥⎣⎦ B ． C ．D ．3,44ππ⎡⎤⎢⎥⎣⎦ 【答案】C 【解析】,且0AB k ≠.设直线的倾斜角为α,当01AB k <≤时,则,所以倾斜角α的范围为04πα≤≤.当时,则,所以倾斜角α的范围为34παπ≤<. 【点评】由斜率取值范围确定直线倾斜角的范围要利用正切函数y ＝tan x 的图象,特别要注意倾斜角取值范围的限制；求解直线的倾斜角与斜率问题要善于利用数形结合的思想,要注意直线的倾斜角由锐角变到直角及由直角变到钝角时,需依据正切函数y ＝tan x 的单调性求k 的范围．【小试牛刀】若过点的直线与圆224x y +=有公共点,则该直线的倾斜角的取值范围是（）A ．0 6π⎛⎫ ⎪⎝⎭，B ．0 3π⎡⎤⎢⎥⎣⎦， C. 0 6π⎡⎤⎢⎥⎣⎦， D ．0 3π⎛⎤ ⎥⎝⎦，【答案】B【解析】当过点的直线与圆224x y += 相切时,设斜率为k ,则此直线方程为,即.由圆心到直线的距离等于半径可得,求得0k =或3k =故直线的倾斜角的取值范围是[0,]3π,所以B 选项是正确的.1.2 与距离有关的最值问题在运动变化中,动点到直线、圆的距离会发生变化,在变化过程中,就会出现一些最值问题,如距离最小,最大等.这些问题常常联系到平面几何知识,利用数形结合思想可直接得到相关结论,解题时便可利用这些结论直接确定最值问题. 【例2】过点()1,2M 的直线l 与圆C ：交于,A B 两点,C 为圆心,当ACB∠最小时,直线l 的方程是．答案:解析：要使ACB ∠最小,由余弦定理可知,需弦长AB 最短.要使得弦长最短,借助结论可知当()1,2M 为弦的中点时最短.因圆心和()1,2M 所在直线的,则所求的直线斜率为1-,由点斜式可得.【点评】与圆有关的长度或距离的最值问题的解法．一般根据长度或距离的几何意义,利用圆的几何性质数形结合求解．此题通过两次转化,最终转化为求过定点的弦长最短的问题. 【例3】若圆C ：关于直线对称,则由点(,)a b 向圆C 所作的切线长的最小值是（）A ．2B ．3C ．4D ．6 【答案】C 【解析】圆C ：化为（x+1）2+（y-2）2=2,圆的圆心坐标为（-1,2）半径．圆C ：关于直线2ax+by+6=0对称,所以（-1,2）在直线上,可得-2a+2b+6=0,即a=b+3．点（a,b ）与圆心的距离,,所以点（a,b ）向圆C 所作切线长：当且仅当b=-1时弦长最小,为4【点评】与切线长有关的问题及与切线有关的夹角问题,解题时应注意圆心与切点连线与切线垂直,从而得出一个直角三角形．【小试牛刀】【安徽省合肥一中、马鞍山二中等六校教育研究会2019届高三第二次联考】已知抛物线上一点到焦点的距离为，分别为抛物线与圆上的动点，则的最小值为（）A ．B ．C ．D ．【答案】D 【解析】由抛物线焦点在轴上，准线方程，则点到焦点的距离为，则，所以抛物线方程：，设，圆，圆心为，半径为1，则，当时，取得最小值，最小值为，故选D.1.3 与面积相关的最值问题与圆的面积的最值问题,一般转化为寻求圆的半径相关的函数关系或者几何图形的关系,借助函数求最值的方法,如配方法,基本不等式法等求解,有时可以通过转化思想,利用数形结合思想求解. 【例4】在平面直角坐标系中,,A B 分别是x 轴和y 轴上的动点,若以AB 为直径的圆C 与直线相切,则圆C 面积的最小值为（）A.45πB.34πC.(65)π-D.54π 【答案】A 【解析】设直线l ：.因为,所以圆心C 的轨迹为以O 为焦点,l 为准线的抛物线.圆C 半径最小值为,圆C 面积的最小值为选A. 【例5】动圆C 经过点(1,0)F ,并且与直线1x =-相切,若动圆C 与直线总有公共点,则圆C 的面积（）A ．有最大值8πB ．有最小值2πC ．有最小值3πD ．有最小值4π 【答案】D【解析】设圆心为(,)a b ,半径为r ,,即,即214a b =,∴圆心为21(,)4b b ,2114r b =+,圆心到直线的距离为,∴或2b ≥,当2b =时,,∴.【小试牛刀】【山东省恒台第一中学2019届高三上学期诊断】已知O 为坐标原点，直线．若直线l 与圆C 交于A ，B 两点，则△OAB 面积的最大值为（） A ．4 B ． C ．2 D ．【答案】C 【解析】由圆的方程可知圆心坐标，半径为2，又由直线，可知，即点D 为OC 的中点，所以,设，又由，所以，又由当，此时直线，使得的最小角为，即当时，此时的最大值为2，故选C 。

2025届山西省太原市第十二中学数学高一上期末质量检测试题含解析

【点睛】本题主要考查两条直线平行的充要条件，意在考查对基础知识的理解与应用，属于基础题.
10、B
【解析】作函数 f x 与 y a 的图象，从而可得函数 F x f x a 有 5 个零点，设 5 个零点分别为
b c d e f ，从而结合图象解得
【详解】解：作函数 f x 与 y a 的图象如下，
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5．已知扇形的圆心角为 2 弧度，其所对的弦长为 2，则扇形的弧长等于 ( )
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度，m（单位：kg）是火箭（除推进剂外）的质量，M（单位：kg）是推进剂与火箭质量的总和， M 称为“总质比”， m
已知 A 型火箭的喷流相对速度为 2000m/s
参考数据： ln 230 5.4，1.648 e0.5 1.649
（1）当总质比为 230 时，利用给出的参考数据求 A 型火箭的最大速度；
取值范围是______.
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人教版必修一英语Unit 5-3

Unit 5Using LanguageⅠ．单词拼写导学号2405065 1．He had finished the study of four years in Peking University, so he was better e __ducated__ than other young people in his village.解析：译文：他结束了北大的四年学习，比村里的其他年轻人受到了更好的教育。

2．Can you give us your o __pinion__ about this matter?解析：译文：你能不能就此事发表一下你的看法。

3．It was so cold inside the house that she had to wrap herself in a b __lanket__ to keep warm.解析：译文：屋子里很冷，她只好披上毯子取暖。

4．He managed to e __scape__ from the big fire.解析：译文：他成功从大火中逃了出来。

5．A $1,500 __reward__ (奖金)has been offered for the return of the necklace.解析：译文：拾到项链归还者获得了1500美元的奖赏。

Ⅱ．单句语法填空导学号2405065 1．You can try __a__ second time if you fail __the__ first time.解析：句意：如果你第一次失败了，你可以再尝试一次。

序数词和不定冠词a连用，表示“再一次，又一次”。

the first time表示“第一次”。

2．A new school was set __up__ in the village last year.解析：句意：去年这个村庄建起了一所新学校。

set up表示“建立；设立；设置”。

山西太原市第十二中学校高一物理第一学期10月月月考考试卷( Word版含答案)

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2023届山西省高三1月适应性调研考试(一模)英语试题(含答案解析)

2023届山西省高三1月适应性调研考试（一模）英语试题学校:___________姓名：___________班级：___________考号：___________一、阅读理解With the cost of living in the United States soaring,more and more people are setting their sights on an international destination.EcuadorFor real bargain hunters,Ecuador is cheap and safe with a very comfortable year-round climate.The cost of living for one person in the country is$750.Ecuador is geographically diverse with miles of endless beaches,hills and rainforests.Housing is economical with the average rent for one person costing around$307per month.Costa RicaCosta Rica is the perfect place to live abroad without depleting(耗尽)your savings.The cost of living is around$893and rent for one person runs approximately$406per month.The tropical destination has cost-effective healthcare.The weather is gorgeous,the water is warm and the housing is highly affordable.UruguayThe country is exquisite(精美的),offering nearly400miles of coast,rolling mountains and four distinct seasons.The cost of living is nearly half of what it is in the states at$1,090 and you can find a nice apartment at the reasonable price of$500per month.PanamaPanama could be a good option for anyone to live cheaply.The cost of living is under half of what it would cost in the U.S.at$1,040for one person.Rent runs approximately$530 per month and food takes up another$353,but outside of that costs are minimal.The country has a mild climate and offers world-class beaches.1．Which country has economical healthcare?A．Ecuador.B．Costa Rica.C．Uruguay.D．Panama. 2．What makes Uruguay special?A．Its reasonable rent.B．Its good bargains.C．Its diverse climates.D．Its cheap apartments.3．How much is the cost of living in the United States?A．Close to$1,090.B．Around$750.C．Under$893.D．Above$2,080.The idea of people taking photographs in front of Van Gogh’s Sun flowers or Botticelli’s The Birth of Venus was once considered so unsatisfactory that galleries banned visitors from using selfie(自拍)sticks.However,the disgust at social media platforms such as Instagram has turned into a bit of a love affair—with one gallery creating a session to help visitors take better photos and videos to share with their followers.The Old Royal Naval College,in Greenwich,southeast London,is reducing visitor numbers on Mondays for its Museum of the Moon artwork“for those looking for the perfect Insta shot”.It will sell400tickets a session in its Painted Hall rather than the usual800so influencers can show their best sides without the crowds.The college,which is displaying a seven-metre model of the moon,created by the artist Luke Jerram,tells ticket holders:“Posing,influencer photo shoots and selfies are encouraged.”Sarah Codrington,head of marketing at the college,said“Social Media Mondays”were partly about appealing to influencers but also freeing up the rest of the week for visitors who might be inconvenienced by people dressing themselves up for their Instagram feed.Jerram,whose Museum of the Moon has been displayed in numerous venues,including Gloucester Cathedral and the Natural History Museum,said:“I had not been consulted about Social Media Mondays but approved.People sometimes complain that there are so many people there that they weren’t having the best experience.People were lying on the floor and doing selfies.Part of the fun of the artwork is watching others interact with it.”He said he welcomed selfie-takers because they usually put away their cameras after ten minutes and engaged with the work.“People are going to have their close encounter with the moon,not necessarily to see an artwork,”he said.Brian Sewell,the late art critic,once was disappointed that it was impossible to see paintings“because people are too busy taking photos”.But Jerram said galleries should embrace it.“Now everyone is their own media and broadcasting company,”he said.“They are broadcasting their experience of the world.”4．What does the gallery mentioned in paragraph1do?A．It forces visitors into using Instagram.B．It assists visitors in taking photographs. C．It forbids visitors from using selfie sticks.D．It charges visitors for sharing photographs.5．What’s the purpose of“Social Media Mondays”?A．To attract influencers of media.B．To sell more tickets than usual. C．To show the model of the moon.D．To meet different needs of visitors. 6．What can we learn from Jerram’s words in the last paragraph?A．He is proud of Museum of the Moon.B．He objects to Social Media Mondays. C．Selfie-takers prefer to enjoy artworks.D．Everyone can broadcast themselves. 7．What does the passage mainly talk about?A．Useful tips on taking photographs.B．Galleries’support for taking selfies. C．The importance of using social media.D．A guide to visiting artworks in galleries.Suppliers of artificial grass have been told to stop describing it as“environmentally friendly”and to expect enforcement(强制执行)if they do not obey.The Advertising Standards Authority,a watchdog,will punish severely firms that make the claim after criticizing a leading supplier for saying that one of its products was“eco”with “no evidence”to back it up.The decision by the watchdog followed a complaint by the campaign group Plastics Rebellion,which argued that Perfectly Green’s“Soul”eco-grass was actually harmful to the planet.In response,Perfectly Green said it believed that the product was environmentally friendly because it did not need to be watered,did not require maintaining with chemical-based products such as pesticides(杀虫剂).However,the Advertising Standards Authority said that to make the claim,Perfectly Green would have to prove that it was not harmful to the environment in any way throughout its full life cycle.It added that no such evidence had been provided and the claim was misleading.The watchdog made it clear that other suppliers should review their use of the term:“This official decision acts as setting an example and we expect advertisers for artificial grass to take note of our decision and act accordingly.”At present dozens of suppliers are claiming that their products are recyclable and environmentally friendly.The Times understands that the watchdog will undertake a “compliance sweep”(合规清查)in the near future,focusing on those companies.Janet Storey,of Plastics Rebellion,described Perfectly Green’s claims as “greenwashing”.She said:“Just9percent of plastic waste has been recycled,with an estimated79percent of all plastic waste ever created either in the open environment or in landfill.Tonnes of plastic grass are burnt in the UK as there are no recycling facilities here.”She added that it carried ecological,microplastic and climate change risks and“causes an unhealthy disconnect from nature”.The group is campaigning for a ban on the sale of plastic grass,which has so far been rejected by the government.Perfectly Green agreed to obey the official decision and remove the words“eco-grass”from its product’s name. 8．Why was the Perfectly Green blamed?A．Its products set a bad example.B．It launched a campaign of protest. C．It wasn’t qualified as a leading supplier.D．Its advertisement didn’t match its products.9．What function is expected of the watchdog?A．To direct suppliers’products.B．To speak for leading companies. C．To decide the companies’future.D．To promote environmental protection. 10．What’s Janet Storey’s attitude to the artificial grass?A．Favorable.B．Indifferent.C．Disapproving.D．Objective. 11．Which of the following is a suitable title for the text?A．Eco-friendly Fake Grass Cut Down to Size B．A Ban on the Sale of Artificial Grass C．Eco-grass Does Harm to the Planet D．The Recycling of Plastic WasteAn academic paper published in Nature Human Behavior on July1suggested smart technology is not making people more stupid but is,in fact,freeing up brain space to allow people to learn more.The paper said that despite the fierce attack of smart technology that has influenced people’s daily lives,it actually assists people to become more excellent rather than less intelligent.For example,if a person needs directions for how to get to a nearby stadium to attend a game,without smart technology,he would have to fish out a map,figure out where the stadium is on that map,and then determine a reasonable route from his current location to his destination.However,with smart technology,the person could simply type in the location and have directions,find parking and even pick a place to eat afterward in mere minutes,freeing up room in their brains to take in other information.The“popular description”surrounding smart technology is that it blocks a person’s cognitive(认知的)abilities such as memory,the paper continued.“And when participants expected information to later be accessible on the computer,they were less likely to remember what its content was,but more likely to remember where it could be found.Theseeffects were temporary and they emerged when access to digital technology or information stored there was available,not when access was unavailable,”the paper continued.The researchers did recognize that long-term use of smart technology did diminish people’s abilities to perform cognitive tasks such as“associations of media multitasking with task switching,episodic(片段的)memory and attention missing.”However,these impacts were all“relevant,not causal.”12．What did the paper find about smart technology?A．It blocks memory abilities.B．It gives memory more space.C．It allows people to study less.D．It makes people more brilliant. 13．Why does the author give an example in paragraph2?A．To illustrate the importance of maps.B．To demonstrate the function of brains. C．To show the ways of choosing directions.D．To stress the strengths of smart technology.14．What does the underlined word“diminish”probably mean?A．Weaken.B．Maintain.C．Increase.D．Display. 15．In which section of a newspaper may this text appear?A．Culture.B．Health.C．Science.D．Education.二、七选五You’ve probably heard of a hybrid(杂交物种)in films or books.The centaurs in the Harry Potter films are hybrids of human and horse.____16____Hybridization occurs in both plants and animals.____17____What’s more,many plants that people eat,from strawberries to carrots,corn and potatoes,have been created by hybridization.Cross-breeding(杂交)sometimes occurs when animals’normal territory overlaps(部分重叠).____18____The grolar bear is a hybrid that happens when grizzly and polar bears mate. As the environment warms,grizzlies in Alaska and Canada wander further north,bringing them into contact with polar bears.Artificial mixing often happens in zoos,where different species that don’t normally meet in the wild can be introduced to each other.Ligers(lion and tiger hybrids)are unlikely to occur in nature,because the only place that their habitats overlap is the Gir Forest in India.____19____Mixed species are more likely to be born with health issues than other animals. Most hybrids born in zoos are infertile,which means they are unable to have babies.However, many cross-breeds born in the wild,like the grolar bear,can have their own babies.____20____“It’s part of the evolutionary process,”he says.Fertile hybrids can lead to the creation of a new species and increase the variety of living things.A．Wildlife conservation is of significance.B．Did you know hybridization is a biological process?C．That’s due to habitat destruction or changing climate.D．Did you know that hybrids appear in the natural world,too?E．Jimmy believes species hybridization in nature is a good thing.F．Many conservationists think that creating hybrids is a bad thing.G．Living things,especially plants,will naturally hybridize in the wild.三、完形填空For LaFont,a bike was always a vehicle of opportunity.Growing up in a poor family,he and his brother often____21____to and from school.The passion continued into his college days.He would also spend hours____22____his bike in his front lawn.In2010,____23____by local kids who gathered in his____24____for tips,tools,and bike parts,LaFont____25____“Front Yard Bike Shop”.Then LaFont became a full-time middle school history teacher,and every year,he saw the number of____26____in his bike program increasing.As a teacher,he also saw how eagerly young people in his classroom needed a safe after-school space before they become ____27____or lazybones instead.In2015,LaFont turned Front Yard Bikes into a____28____operation and left his teaching job to____29____all his time to it.“That was quite a(n)_____30_____,”he said.“But I thought it was_____31_____.”Today,Front Yard Bikes is a place where students_____32_____and learn to saw,drill, measure,cut and where they learn to paint,design,and plan.LaFont recently opened a bike repair shop where older students can get certified in mechanics,receive hands-on training and gain employment at the store to help_____33_____their resume(简历)for future work.Front Yard Bikes_____34_____nearly400young people a year.To date,50students have been certified in mechanics,and2,000kids have_____35_____from the program. 21．A．drove B．walked C．biked D．skated 22．A．repairing B．pushing C．selling D．cleaning 23．A．guided B．inspired C．ordered D．touched 24．A．school B．store C．yard D．garden 25．A．contacted B．imagined C．joined D．formed 26．A．users B．employees C．participants D．workers 27．A．fundraisers B．troublemakers C．bookworms D．shopkeepers 28．A．full-time B．non-profit C．trading D．public 29．A．adjust B．devote C．spare D．limit 30．A．blessing B．luck C．opportunity D．risk 31．A．creative B．available C．secure D．meaningful 32．A．show up B．drop by C．hang about D．check out 33．A．build B．submit C．send D．update 34．A．unite B．employ C．serve D．select 35．A．learned B．benefited C．recovered D．graduated四、用单词的适当形式完成短文阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。