江苏省如东高级中学2019-2020学年2019-2020学年高一下学期期末热身练数学试题(答案图片版)
江苏省如东高级中学2019_2020学年高一物理下学期第二次阶段测试试题含解析
C. 小球从A到C与从C到B的过程，速度的变化率相等
D. 小球从A到C与从C到B的过程，损失的机械能相等
【答案】BCD
【解析】
【详解】A．位移是从初位置指向末位置的有向线段故小球从A出发到返回A，位移为0，但整个过程中摩擦力的方向与小球运动的方向始终相反，故整个过程中摩擦力对物体做负功，重力不做功，支持力不做功，所以整个过程中外力对物体做负功，故A错误；
CD．断开开关S，电荷量不变，根据 、 、 可得
将A板向右平移，可知 减小，则电场强度不变，电场力不变，所以 不变，故C正确，D错误；
故选C。
7.如图所示，一足够长的木板在光滑水平面上以速度v向右匀速运动，现将质量为m的物体竖直向下轻轻地放置在木板上的右端，已知物体和木板之间的动摩擦因数为μ。为保持木板的速度不变，须对木板施一水平向右的作用力F。从物体放到木板上到它相对木板静止的过程中，力F做的功为（ ）
C. 从开始运动至落地，重力对两小球做功相同
D. 从开始运动至落地，重力对两小球做功 平均功率相同
【答案】AC
【解析】
【详解】A．在整个过程中，只有重力做功，这两个小球在整个过程中重力做功相等，根据动能定理则有
可知落地时的动能相等，速度大小相等，故A正确；
B．因为两小球落地时的速度大小相等，但方向不同，所以A球落地时竖直方向的速度小于B落地时竖直方向的速度，根据重力的瞬时功率 可知A球落地时重力的瞬时功率小于B落地时重力的瞬时功率，故B错误；
A. 到达M、N时两粒子速率仍相等
B. 到达M、N时两粒子速率vM>vN
C. 到达M、N时两粒子的电势能相等
D. 两个粒子的电势能都是先减小后增大
江苏省如东高级中学2019-2020学年高一英语下学期期中学情检测试题(含解析)-江苏省如东高级
江苏省如东高级中学2019-2020学年高一英语下学期期中学情检测试题（含解析）选择题部分（共85分）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。
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。
1．What time is it in New York?A.It’s5：00 pmB.It’s 10:00 p.m.C.It’s 7:00 p.m.2.What do we know about the man’s ticket?A. It was super expensive.B. He got it at the last minute.C. He bought it a week ago.3.What does the man want to do?A. Buy a book on the Internet.B. Borrow a book.C. Return a book to the library.4.What does the man want?A. A chocolate cake.B. Iced tea.C. A hot drink.5.What does the man think the woman should do?A. Speak out how she feels.B. Cancel her trip to Spain.C. Go to another country.第二节（共15小题；每小题1.5分，满分22.5分）听下面5段对话或独白b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。
听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，各小题将给出5秒钟的作答时间。
江苏省如东高级中学2019-2020学年高一下学期第二次阶段测试数学试题
江苏省如东高级中学第二学期第二次阶段性测试高一数学试题考试时间：120分钟 2020.06一．选择题（共10小题，每小题5分，共50分）1．在空间直角坐标系中，点(2P -，1，4)关于xOy 平面对称点的坐标是 ( ) A ．(2-，1，4)-B ．(2-，1-，4)-C ．(2，1-，4)D ．(2，1，4)-2．圆222210x y x y +--+=的点到直线4x y -=距离的最小值是 ( )A ．1+B ．2C ．1D ．1+3．已知三棱柱111ABC A B C -的体积为120，点P ，Q 分别在侧棱1AA ，1CC 上，且1PA QC =，则三棱锥1B BPQ -的体积为 () A ．20 B ．30 C ．40 D ．604．一车间为规定工时定额，需要确定加工零件所花费的时间，为此进行了4次试验，测得数据如表：根据上表可得回归方程ˆ811y x =+，则实数a 的值为 ( )A ．34B ．35C ．36D ．375．已知圆锥的表面积为3π，它的侧面展开图是一个半圆，则此圆锥的体积为 ( )A B C D6．在四面体ABCD 中，E ，F 分别为棱AC ，BD 的中点,6AD =，4BC =，EF =，则异面直线AD 与BC 所成角的余弦值为 ( ) A ．34B ．56C ．910D ．11127．抛掷一个质地均匀的骰子的试验，事件A 表示“小于5的偶数点出现”，事件B 表示“不小于5的点数出现”，则一次试验中，事件A 或事件B 至少有一个发生的概率为 ( )A ．23 B ．13C ．12D ．568．在平面直角坐标系中，已知圆22:4O x y +=，过点(1,1)P 的直线l 交圆O 于,A B 两点，且2AP PB =，则满足上述条件的所有直线斜率之和为 ( )A ．83-B ． 83C ． 38-D ． 389．在ABC ∆中，内角A 、B 、C 所对边分别为a 、b 、c ，若2cos 3cos 5cos a b cA B C==，则B ∠的大小是 ( ) A ．12π B ．6π C ．4π D ．3π 10．如图所示，三棱锥S ABC -中，ABC ∆与SBC ∆都是边长为1的正三角形，32SA =，若S ，A ，B ，C 四点都在球O 的表面上，则球O 的表面积为( ) A ．73πB ．133π C ．43πD ．3π二．多选题（本题共2小题，每小题5分，共10分，选对得5分，漏选得3分，选错得0分） 11．如图，正方体1111ABCD A B C D -棱长为1，线段11B D 上有两个动点E ，F ，且2EF =，则下列结论正确的是 ( ) A ．AC ⊥平面BEFB ．AE ，BF 始终在同一个平面内C ．//EF 平面ABCDD ．三棱锥A BEF -的体积为定值12．在三角形ABC 中，下列命题正确的有( ) A ．若30A =︒，4b =，5a =，则三角形ABC 有两解B ．若0tan tan 1A B <<，则ABC ∆一定是钝角三角形C ．若cos()cos()cos()1A B B C C A ---=，则ABC ∆一定是等边三角形D ．若cos cos a b c B c A -=-，则ABC ∆的形状是等腰或直角三角形 三．填空题（本题共4小题，每小题5分，共20分）13．若一组数据3，x ，2，4，5的平均数为3，则该组数据的方差是 ． 14．过点(1,1)P 作圆22210x y x ++-=的切线，切点为A ，则||PA = ．15．在四面体ABCD 中，E 、G 分别是CD 、BE 的中点， 若记,,AB a AD b AC c ===，则AG = ．16．从正方体1111ABCD A B C D -上截下一个角，得三棱锥A EFG -．如果该三棱锥的三个侧面面积分别为1，2，4，则该三棱锥的底面EFG 的面积是 ．四．解答题（本大题共6小题，计70分．应写出必要的文字说明、证明过程或演算步骤） 17．（本小题满分10分）锐角ABC ∆中，角A ，B ，C 所对的边分别为a ，b ，c ，若3b =且2cosB(a cosC+ccosA)=b ． （1）求ABC ∆的外接圆直径； （2）求ac +的取值范围． 18．（本小题满分12分）在平面直角坐标系xOy 中，O 为坐标原点，以O 为圆心的圆与直线340x y --=相切．(1)求圆O 的方程．(2)直线:3l y kx =+与圆O 交于A ，B 两点，在圆O 上是否存在一点M ，使得四边形OAMB 为菱形？若存在，求出此时直线l 的斜率；若不存在，说明理由． 19．（本小题满分12分）某市场研究人员为了了解产业园引进的甲公司前期的经营状况，对该公司2019年连续六个月(510-月）的利润进行了统计，并根据得到的数据绘制了相应的折线图，如图所示． （1）由折线图可以看出，可用线性回归模型拟合月利润y （单位：百万元）与月份代码x 之间的关系，求y 关于x 的线性回归方程，并据此预测该公司2020年5月份的利润； （2）甲公司新研制了一款产品，需要采购一批新型材料，现有A ，B 两种型号的新型材料可供选择，按规定每种新型材料最多可使用4个月，但新材料的不稳定性会导致材料损坏的年限不同，现对A ，B 两种型号的新型材料对应的产品各100件进行科学模拟测试，得到两种新型材料使用寿命的频数统计表（表)．若从产品使用寿命的角度考虑，甲公司的负责人选择采购哪款新型材料更好？使用寿命 材料类型1个月 2个月 3个月 4个月 总计A 20 35 35 10 100 B10304020100参考数据：196i i y ==∑，1371i i i x y ==∑．参考公式：回归直线方程ˆˆˆybx a =+，其中1122211()()()ˆ()nniii ii i nniii i x x yy x y nxyb xx xnx ====---==--∑∑∑∑，ˆˆˆay bx =-． 20．（本小题满分12分）如图，在直三棱柱111ABC A B C -中，点D 是线段AB 上的动点． （1）线段AB 上是否存在点D ，使得1//AC 平面1B CD ？若存在，请写出ADDB值，并证明此时，1//AC 平面1B CD ；若不存在，请说明理由； （2）已知平面11ABB A ⊥平面1CDB ，求证：CD AB ⊥． 21．（本小题满分12分）为了贯彻落实中央、省、市关于新型冠状病毒肺炎疫情防控工作要求，积极应对新型冠状病毒疫情，切实做好2020年春季开学工作，保障校园安全稳定，普及防控知识，确保师生生命安全和身体健康．某校开学前，组织高三年级800名学生参加了“疫情防控”网络知识竞赛（满分150分）．已知这800名学生的成绩均不低于90分，将这800名学生的成绩分组如下：第一组[90，100)，第二组[100，110)，第三组[110，120)，第四组[120，130)，第五组[130，140)，第六组[140，150]，得到的频率分布直方图如图所示．（1）求a 的值并估计这800名学生的平均成绩（同一组中的数据用该组区间的中点值代表）； （2）该校“群防群控”督查组为更好地督促高三学生的“个人防控”，准备从这800名学生中选取2名学生参与督查工作，其选取办法是：先在第二组、第五组、第六组中用分层抽样的方法抽取6名学生，再从这6名学生中随机抽取2名学生．记这2名学生的竞赛成绩分别为x 、y ．求事件||20x y -的概率．22．（本小题满分12分）平行四边形ABCD 中，AB BD ⊥，2AB =，2BD =，沿BD 将BCD ∆折起，使二面角A BD C --是大小为锐角α的二面角，设C 在平面ABD 上的射影为O ．（1）当α为何值时，三棱锥C-OAD 的体积最大？最大值为多少？ （2）当AD BC ⊥时，求α的大小．江苏省如东高级中学第二学期第二次阶段性测试高一数学参考答案1．A ． 2．C ． 3．C ． 4．C ． 5．A ． 6．D ． 7．A ． 8．A ． 9．D ． 10．A ． 11．ACD ． 12．BCD ． 13．2． 1415．111244a b c =++ 1617．解：（1）因为2cos (cos cos )B a C c A b +=， 由正弦定理可得，2cos (sin cos sin cos )sin B A C C A B +=， 即2cos sin()sin B A C B +=，所以2cos sin sin B B B =， 因sin 0B ≠，故1cos 2B =且(0,)B π∈，故3B π=， ……3分由正弦定理21sin bR B ===，即外接圆直径1， ……5分 （2）由正弦定理可得，2sin sin a cR A C==，1sin sin sin sin()sin sin 32a c A C A A A A A π∴+=+=++=+3sin 2A A =)6A π=+， 7分 由题意可得，022032A A πππ⎧<<⎪⎪⎨⎪<-<⎪⎩，解可得62A ππ<<，所以2(,)633A πππ+∈，sin()6A π∴+∈，∴3(2a c +∈． ……10分18．（1）设圆O 的半径长为r ,因为直线xy -4=0与圆O 相切,所以 r=2.所以圆O 的方程为 x 2+y 2=4. ……5分 （2）假设存在点M ,使得四边形OAMB 为菱形,则OM 与AB 互相垂直且平分,所以原点O 到直线l :y =kx +3的距离d =12|OM|=1.=1,解得k 2=8,即k =±, 经验证满足条件.所以存在点M ,使得四边形OAMB 为菱形，此时直线l 的斜率为±. ……12分 19．解：（1）由折线图可知统计数据(,)x y 共有6组， 即(1,11)，(2,13)，(3,16)，(4,15)，(5,20)，(6,21)．计算可得1(123456) 3.56x =+++++=，1(111315162021)166y =+++++=，616222163716 3.516ˆ2916356i ii ii x yxybxx ==--⨯⨯===-⨯-∑∑， ……2分ˆˆ162 3.59ay bx =-=-⨯=． ……4分 ∴月度利润x 与月份代码x 之间的线性回归方程为ˆ29y x =+， ……5分 当13x =时，ˆ213935y=⨯+=．故预计甲公司2020年5月份的利润为35百万元； ……6分（2）由频率估计概率，A 型材料可使用1个月，2个月，3个月、4个月的概率分别为0.2，0.35，0.35，0.1，A ∴型新材料对应产品的使用寿命的平均数为110.220.3530.3540.1 2.35x =⨯+⨯+⨯+⨯=；……9分B 型材料可使用1个月，2个月，3个月、4个月的概率分别为0.1，0.3，0.4，0.2， B ∴型新材料对应的产品的使用寿命的平均数为210.120.330.440.2 2.7x =⨯+⨯+⨯+⨯=．12x x <，∴应该采购B 新型材料． ……12分20．（1）解：在线段AB 上存在点D ，当1ADDB=时，1//AC 平面1B CD ． 证明如下：连接1BC ，交1B C 于点E ，连接DE ，则点E 是1BC 的中点， 又当1ADDB=，即点D 是AB 的中点，由中位线定理得1//DE AC ， DE ⊂平面1B CD ，1AC ⊂/平面1B CD ， 1//AC ∴平面1B CD ． ……6分（2）证明：过B 作1BP DB ⊥并交1DB 于点P ，又平面11ABB A ⊥平面1CDB ，BP ⊂平面11ABB A ，平面11ABB A ⋂平面11CDB DB =，BP ∴⊥平面1CDB ，又CD ⊂平面1CDB ，CD BP ∴⊥．在直三棱柱111ABC A B C -中，1BB ⊥平面ABC ，CD ⊂平面ABC ， 1CD BB ∴⊥，又1BB ⊂平面11ABB A ，BP ⊂平面11ABB A ，1BB BP B =，CD ∴⊥平面11ABB A ．又AB ⊂平面11ABB A ，CD AB ∴⊥． ……12分21．解：（1）由频率分布直方图可知(0.01020.0250.0150.005)101a ⨯++++⨯=，解得0.035a =． ……2分 这800名学生数学成绩的平均数为：950.010101050.010101150.025101250.035101350.015101450.00510120⨯⨯+⨯⨯+⨯⨯+⨯⨯+⨯⨯+⨯⨯=．……5分（2）由题意可知：第二组抽取2名学生，其成绩记为A ，B ，则100A ，110B <， 第五组抽取3名学生，其成绩记为C ，D ，E ，则130C ，D ，140E <， 第六组抽取1名学生，其成绩记为F ，则140150F ， 现从这6名学生中抽取2名学生的成绩的基本事件为：(,)A B ，(,)A C ，(,)A D ，(,)A E ，(,)A F ，(,)B C ，(,)B D ，(,)B E ， (,)B F ，(,)C D ，(,)C E ，(,)C F ，(,)D E ，(,)D F ，(,)E F 共15个．其中事件||20x y -包含的基本事件为：(,)A B ，(,)C D ，(,)C E ，(,)C F ，(,)D E ，(,)D F ，(,)E F 共7个，记“这2名学生的竞赛成绩分别为x 、y ，其中||20x y -”为事件M ， 则事件||20x y -的概率为7()15P M =．……12分22．(1) 当45α=︒时，三棱锥O ACD -的体积最大，最大值为2;(2) 60α=︒. 【解析】（1）由题意可得BD ⊥OD ，可得1.2AODSOD BD =，OC ⊥平面ABDO ，利用三棱锥的体积计算公式和正弦函数的单调性即可得出；（2）由AD ⊥BO ，即可得出． 解：（1）由题知OD 为CD 在平面ABD 上的射影， ∵BD ⊥CD ，CO ⊥平面ABD ，∴BD ⊥OD ， ∴∠ODC=α，则OC=CDsin α，OD=CDcos α． ∴ -------3分==，当且仅当sin2α=1，即α=45°时取等号，∴当α=45°时，三棱锥O﹣ACD的体积最大，最大值为． --------6分（2）连接OB，分面面面面面9---,OCBDBOCOCBOCADBOCBCCOCBCCOBCADCOADABDADABDCO⊥⇒⎪⎪⎪⎭⎪⎪⎪⎬⎫⊂⊥⇒⎪⎪⎪⎭⎪⎪⎪⎬⎫⊂=⋂⊥⊥⇒⎭⎬⎫⊂⊥，又因为BD⊥OD2,2ππ=∠+∠=∠+∠OBDABHABHDABBDORtABDRt∆∆∴相似于1,222,=∴=∴=ODODABBDBDOD中，的平面角，在为二面角）知由（ODCRtCBDAODC∆--∠121cos==∠CDODODC分12----3πα=∴。
2019-2020学年江苏省南通市如东县高一下学期期末数学试卷 (解析版)
2019-2020学年江苏省南通市如东县高一第二学期期末数学试卷一、选择题（共10小题）.1．若直线ax+3y﹣5＝0经过点A（2，1），则实数a的值（）A．1B．2C．3D．42．将一颗质地均匀的骰子（一种各个面上分别标有1，2，3，4，5，6的正方体玩具）先后抛掷两次，则向上的点数之和为4的概率为（）A．B．C．D．3．在△ABC中，已知，则角C等于（）A．B．C．D．4．已知直线l1：mx+y﹣1＝0，l2：（3m﹣2）x+my﹣2＝0，若l1∥l2，则实数m的值为（）A．2B．1C．1或2D．0或5．已知l，m为两条不同直线，α，β为两个不同平面，则下列命题中真命题的是（）A．若l∥m，m⊂α，则l∥αB．若l⊥m，m⊂α，则l⊥αC．若α∥β，m⊂α，则m∥βD．若α⊥β，m⊂α，则m⊥β6．圆x2+y2﹣2x﹣8y+13＝0截直线ax+y﹣1＝0所得的弦长为，则实数a＝（）A．﹣B．﹣C．D．27．在长方体ABCD﹣A1B1C1D1中，AB＝BC＝2，AA1＝1，则直线BC1与平面BB1DD1所成角的正弦值为（）A．B．C．D．8．已知关于某设备的使用年限x（单位：年）和所支出的维修费用y（单位：万元）有如表的统计资料：x23456y 2.2 3.8 5.5 6.57.0由上表可得线性回归方程，若规定当维修费用y＞12时该设备必须报废，据此模型预报该设备使用的年限不超过为（）A．7B．8C．9D．109．已知△ABC的内角A，B，C的对边分别为a，b，c，AD为角A的角平分线，交BC于D，，，BD＝2，则b＝（）A．B．C．D．10．已知锐角三角形△ABC的内角A，B，C的对边分别为a，b，c．且b＝2a sin B，则cos B+sin C 的取值范围为（）A．（0，]B．（1，]C．（，）D．（，）二、多项选择题：本题共2小题，每小题5分，共10分．在每小题给出的四个选项中，有多项符合题目要求．全部选对的得5分，部分选对的得3分，有选错的得0分．11．若圆C1：（x﹣1）2+y2＝1与圆C2：x2+y2﹣8x+8y+m＝0相切，则m的值可以是（）A．16B．7C．﹣4D．﹣712．如图，在三棱锥P﹣ABC中，D、E、F分别为棱PC、AC、AB的中点，PA⊥平面ABC，∠ABC＝90°，AB＝PA＝6，BC＝8，则（）A．三棱锥D﹣BEF的体积为18B．平面DEF截三棱锥P﹣ABC所得的截面面积为12C．点P与点A到平面BDE的距离相等D．直线PB与直线DF垂直三、填空题：本大题共4小题，每小题5分，共20分．不需写出解答过程，请把答案直接填写在答题卡相应位置上．13．若x1，x2，…，x n的方差为2，则2x1+3，2x2+3，…，2x n+3的方差为．14．△ABC中，角A，B，C的对边分别为a，b，c．已知2a＝b+c，sin2A＝sin B sin C，则△ABC一定为．（用“直角三角形”“等边三角形”“等腰直角三角形”填空）15．设长方体的长、宽、高分别为3、2、1，其顶点都在同一个球面上，则该球的半径为．16．已知凸四边形ABCD（指把四边形的任意一条边向两端无限延长成一直线时，其他各边都在此直线的同旁）中，边，对角线AC＝4，且∠ACB＝90°，又顶点D满足AD2+CD2＜16，则凸四边形ABCD的对角线BD长的范围是．四、解答题：本大题共6小题，共70分．请在答题卡指定区域内作答.解答时应写出文字说明、证明过程或演算步骤．17．△ABC的内角A，B，C的对边为a，b，c，a cos C+c cos A＝2b cos A．（1）求A；（2）若B＝45°，a＝2，求b，c．18．在正四棱锥P﹣ABCD中，E，F分别为棱PA，PC的中点．（1）求证：EF∥平面ABCD；（2）求证：EF⊥平面PBD．19．某校疫情期间“停课不停学”，实施网络授课，为检验学生上网课的效果，高三年级进行了一次网络模拟考试．全年级共1500人，现从中抽取了100人的数学成绩，绘制成频率分布直方图（如图所示）．已知这100人中[110，120）分数段的人数比[100，110）分数段的人数多6人．（1）根据频率分布直方图，求a，b的值；并估计抽取的100名同学数学成绩的平均数（假设同一组中的每个数据可用该组区间的中点值代替）；（2）现用分层抽样的方法从分数在[130，140），[140，150]的两组同学中随机抽取6名同学，从这6名同学中再任选2名同学作为“网络课堂学习优秀代表”发言，求这2名同学的分数恰在同一组内的概率．20．如图，在长方体ABCD﹣HKLE中，底面ABCD是边长为3的正方形，对角线AC与BD相交于点O，点F为线段AH上靠近点A的三等分点，BE与底面ABCD所成角为．（1）求证：AC⊥BE；（2）求二面角F﹣BE﹣D的余弦值．21．已知圆C：x2+（y﹣2）2＝r2（r＞0）与直线l：3x+4y+12＝0相切．（1）求圆C的标准方程；（2）若动点M在直线y+6＝0上，过点M引圆C的两条切线MA、MB，切点分别为A，B．①记四边形MACB的面积为S，求S的最小值；②证明直线AB恒过定点．22．已知函数．（1）求函数f（x）的定义域；（2）设g（x）＝f（x）+a，若函数g（x）在（1，2）上有且仅有一个零点，求实数a 的取值范围；（3）设，是否存在正实数m，使得函数y＝h（x）在[1，2]内的最大值为4？若存在，求出m的值；若不存在，请说明理由．参考答案一、单选题：本大题共10小题，每题5分，共50分．在每小题提供的四个选项中，只有一项是符合题目要求的．1．若直线ax+3y﹣5＝0经过点A（2，1），则实数a的值（）A．1B．2C．3D．4【分析】把点A（2，1）代入直线方程即可得出．解：∵直线ax+3y﹣5＝0经过点A（2，1），∴2a+3﹣5＝0，解得a＝1．故选：A．2．将一颗质地均匀的骰子（一种各个面上分别标有1，2，3，4，5，6的正方体玩具）先后抛掷两次，则向上的点数之和为4的概率为（）A．B．C．D．【分析】分别求出基本事件数，“点数和为4”的种数，再根据概率公式解答即可．解：所有的基本事件共6×6＝36个，其中，点数和为4的有（1，3）、（2，2）、（3，1）共3个，∴出现向上的点数之和为4的概率是＝，故选：B．3．在△ABC中，已知，则角C等于（）A．B．C．D．【分析】由已知可得a2+b2﹣c2＝﹣ab，利用余弦定理可求cos C＝﹣，结合范围C∈（0，π），可求C的值．解：∵，即a2+b2﹣c2＝﹣ab，∴cos C＝＝＝﹣，∵C∈（0，π），∴C＝．故选：D．4．已知直线l1：mx+y﹣1＝0，l2：（3m﹣2）x+my﹣2＝0，若l1∥l2，则实数m的值为（）A．2B．1C．1或2D．0或【分析】根据直线方程中一次项系数之比相等，但不等于常数项之比，求出m的值．解：当m＝0时，直线l1：y﹣1＝0，l2：x＝1，不满足l1∥l2．当m≠0时，∵直线l1：mx+y﹣1＝0，l2：（3m﹣2）x+my﹣2＝0，若l1∥l2，则＝≠，则实数m＝1，故选：B．5．已知l，m为两条不同直线，α，β为两个不同平面，则下列命题中真命题的是（）A．若l∥m，m⊂α，则l∥αB．若l⊥m，m⊂α，则l⊥αC．若α∥β，m⊂α，则m∥βD．若α⊥β，m⊂α，则m⊥β【分析】由直线与平面的位置关系判断选项A；由线面垂直的判定定理判断选项B；由面面平行的定义判断选项C；由面面垂直的性质定理判断选项D．解：对于A，若l∥m，m⊂α，则l∥α或者l⊂α，错误；对于B，l⊥m，m⊂α，l只与平面α内的一条直线垂直，根据线面垂直的判定定理，不能得到l垂直平面α，错误；对于C，若α∥β，m⊂α，则m∥β，正确；对于D，α⊥β，m⊂α，根据面面垂直的性质定理，若m不垂直于平面α和β的交线，则不能得到m垂直平面β，错误；故选：C．6．圆x2+y2﹣2x﹣8y+13＝0截直线ax+y﹣1＝0所得的弦长为，则实数a＝（）A．﹣B．﹣C．D．2【分析】由圆的方程，得到圆心与半径，再求得圆心到直线的距离，利用勾股定理解．解：圆的方程可化为（x﹣1）2+（y﹣4）2＝4，所以圆心坐标为（1，4），由点到直线的距离公式得：d＝＝1，解得a＝﹣，故选：A．7．在长方体ABCD﹣A1B1C1D1中，AB＝BC＝2，AA1＝1，则直线BC1与平面BB1DD1所成角的正弦值为（）A．B．C．D．【分析】以D为原点，DA为x轴，DC为y轴，DD1为z轴，建立空间直角坐标系，利用向量法能求出直线BC1与平面BB1DD1所成角的正弦值．解：在长方体ABCD﹣A1B1C1D1中，AB＝BC＝2，AA1＝1，以D为原点，DA为x轴，DC为y轴，DD1为z轴，建立空间直角坐标系，B（2，2，0），C1（0，2，1），D（0，0，0），D1（0，0，1），＝（﹣2，0，1），＝（2，2，0），＝（0，0，1），设平面BB1DD1的法向量＝（x，y，z），则，取x＝1，得＝（1，﹣1，0），设直线BC1与平面BB1DD1所成角为θ，则直线BC1与平面BB1DD1所成角的正弦值为：sinθ＝＝＝．故选：D．8．已知关于某设备的使用年限x（单位：年）和所支出的维修费用y（单位：万元）有如表的统计资料：x23456y 2.2 3.8 5.5 6.57.0由上表可得线性回归方程，若规定当维修费用y＞12时该设备必须报废，据此模型预报该设备使用的年限不超过为（）A．7B．8C．9D．10【分析】由已知求得样本点的中心的坐标，代入线性回归方程求得，再由＞12求得x 的范围得答案．解：，．则样本点的中心的坐标为（4，5.0），代入，得5.0＝4+0.08，即．∴线性回归方程为．由1.23x+0.08＝12，解得x＝9.69．∴据此模型预报该设备使用的年限不超过9年．故选：C．9．已知△ABC的内角A，B，C的对边分别为a，b，c，AD为角A的角平分线，交BC于D，，，BD＝2，则b＝（）A．B．C．D．【分析】利用正弦定理以及三角形的内角和结合三角形的形状忙着求解即可．解：因为，BD＝2，，，所以，，所以，，又因为，所以△ADC为等腰三角形，所以．故选：A．10．已知锐角三角形△ABC的内角A，B，C的对边分别为a，b，c．且b＝2a sin B，则cos B+sin C 的取值范围为（）A．（0，]B．（1，]C．（，）D．（，）【分析】由已知结合正弦定理进行化简可求sin A，进而可求A，结合锐角三角的条件可求B的范围，然后结合和差角公式及辅助角公式进行化简后结合正弦函数的性质即可求解．解：因为b＝2a sin B，由正弦定理可得，sin B＝2sin A sin B，因为sin B≠0，故sin A＝，因为A为锐角，故A＝，由题意可得，，解可得，，则cos B+sin C＝cos B+sin（），＝cos B+cos B+，＝＝sin（B+）．故选：C．二、多项选择题：本题共2小题，每小题5分，共10分．在每小题给出的四个选项中，有多项符合题目要求．全部选对的得5分，部分选对的得3分，有选错的得0分．11．若圆C1：（x﹣1）2+y2＝1与圆C2：x2+y2﹣8x+8y+m＝0相切，则m的值可以是（）A．16B．7C．﹣4D．﹣7【分析】化圆C2为标准方程，求得圆心坐标与半径，再由圆心距与半径的关系列式求解m值．解：圆C2可化简为（x﹣4）2+（y+4）2＝32﹣m（m＜32）．由两圆相切，可得，解得m＝16或﹣4．故选：AC．12．如图，在三棱锥P﹣ABC中，D、E、F分别为棱PC、AC、AB的中点，PA⊥平面ABC，∠ABC＝90°，AB＝PA＝6，BC＝8，则（）A．三棱锥D﹣BEF的体积为18B．平面DEF截三棱锥P﹣ABC所得的截面面积为12C．点P与点A到平面BDE的距离相等D．直线PB与直线DF垂直【分析】A，由S△BEF＝．即可判定B，取PB中点M，连接DM、FM，可得平面DEF截三棱锥P﹣ABC所得的截面为矩形MFED，求得面积即可．C，由PA面DBE，即可判断，D，假设直线PB与直线DF垂直，就可得PB⊥面BDE，由AB⊥面BDE，根据过平面外一点只能作一条直线垂直某平面即可判断．解：对于A，可得DE＝，S△BEF＝．∴三棱锥D﹣BEF的体积为＝＝6，故错．对于B，如图，取PB中点M，连接DM、FM，可得平面DEF截三棱锥P﹣ABC所得的截面为矩形MFED，面积为MF•FE＝12，故正确．对于C，由已知可得PA面DBE，故点P与点A到平面BDE的距离相等，故正确，对于D，易得EF⊥PB，假设直线PB与直线DF垂直，就可得PB⊥面BDE，由AB⊥面BDE，根据过平面外一点只能作一条直线垂直某平面，故错．故选：BC．三、填空题：本大题共4小题，每小题5分，共20分．不需写出解答过程，请把答案直接填写在答题卡相应位置上．13．若x1，x2，…，x n的方差为2，则2x1+3，2x2+3，…，2x n+3的方差为8．【分析】根据数据mx1+a，mx2+a，…，mx n+a的方差为数据x1，x2，…，x n的方差的m2倍，代入数值即可求得结论．解：∵x1，x2，…，x n的方差为2，∴2x1+3，2x2+3，…，2x n+3的方差为x1，x2，…，x n的方差的22倍，即22×2＝8．故答案为：814．△ABC中，角A，B，C的对边分别为a，b，c．已知2a＝b+c，sin2A＝sin B sin C，则△ABC一定为等边三角形．（用“直角三角形”“等边三角形”“等腰直角三角形”填空）【分析】利用正弦定理化简sin2A＝sin B sin C，得到a2＝bc，与2a＝b+c联立得到a＝b ＝c，可得出三角形ABC为等边三角形．解：由正弦定理化简sin2A＝sin B sin C，得到a2＝bc，又2a＝b+c，即a＝，∴a2＝＝bc，即（b+c）2＝4bc，∴（b﹣c）2＝0，即b＝c，∴2a＝b+c＝b+b＝2b，即a＝b，∴a＝b＝c，则△ABC为等边三角形．故答案为：等边三角形15．设长方体的长、宽、高分别为3、2、1，其顶点都在同一个球面上，则该球的半径为．【分析】由长方体的对角线与其外接球的直径之间的关系，求出外接球的半径．解：由长方体的对角线等于其外接球的直径2R可得：（2R）2＝32+22+12，解得：R＝，故答案为：．16．已知凸四边形ABCD（指把四边形的任意一条边向两端无限延长成一直线时，其他各边都在此直线的同旁）中，边，对角线AC＝4，且∠ACB＝90°，又顶点D满足AD2+CD2＜16，则凸四边形ABCD的对角线BD长的范围是（）．【分析】将该四边形置于平面直角坐标系内，然后由AD2+CD2＜16可知，D点在以AC 直径的圆的下半圆内，数形结合即可解决问题．解：因为，对角线AC＝4，且∠ACB＝90°，如图建立平面直角坐标系．则C（0，0），A（4，0），B（0，），以AC为直径的圆M的圆心M（2，0），半径r＝2，故．因为AD2+CD2＜16＝AC2∴，∴∠ADC是钝角，所以D在半圆M内，可见：当D点与C重合时，BD的最小值为；当BD过圆心M，且D与Q重合时，BD取最大值BM+2＝6．因D在半圆M内，故．故答案为：（）．四、解答题：本大题共6小题，共70分．请在答题卡指定区域内作答.解答时应写出文字说明、证明过程或演算步骤．17．△ABC的内角A，B，C的对边为a，b，c，a cos C+c cos A＝2b cos A．（1）求A；（2）若B＝45°，a＝2，求b，c．【分析】（1）先利用正弦定理边化角，然后化简成关于A的一个角的三角函数形式，解方程即可；（2）结合（1）的结论，先求出C，然后利用正弦定理求另两边．解：（1）因为：a cos C+c cos A＝2b cos A由正弦定理得sin A cos C+sin C cos A＝2sin B cos A．即sin（A+C）＝2sin B cos A＝sin B，显然sin B≠0．故cos A＝，∵A∈（0，π），∴．（2）结合（1）知A＝60°，B＝45°，∴C＝180°﹣60°﹣45°＝75°，∴sin75°＝sin（30°+45°）＝sin30°cos45°+cos30°sin45°＝．∴，解得．18．在正四棱锥P﹣ABCD中，E，F分别为棱PA，PC的中点．（1）求证：EF∥平面ABCD；（2）求证：EF⊥平面PBD．【分析】（1）由已知利用中位线的性质可证EF∥AC，进而根据线面平行的判定即可证明EF∥平面ABCD．（2）连结AC，BD交于点O，连结PO，由已知可证PO⊥平面ABCD．利用线面垂直的性质可证PO⊥AC，由BD⊥AC，EF∥AC可证EF⊥PO，结合EF⊥BD利用线面垂直的判定即可证明EF⊥平面PBD．【解答】证明：（1）因为E，F分别为棱PA，PC的中点，所以EF∥AC，……………………又因为EF⊄平面ABCD，AC⊂平面ABCD，所以EF∥平面ABCD．……………………（2）连结AC，BD交于点O，连结PO．因为P﹣ABCD为正四棱锥，所以PO⊥平面ABCD．又AC⊂平面ABCD，所以PO⊥AC．又因为BD⊥AC，EF∥AC，所以EF⊥PO，在正方形ABCD中AC⊥BD，EF∥AC，EF⊥BD．……………………又PO，BD⊂平面PBD，PO∩BD＝O，所以EF⊥平面PBD，……………………19．某校疫情期间“停课不停学”，实施网络授课，为检验学生上网课的效果，高三年级进行了一次网络模拟考试．全年级共1500人，现从中抽取了100人的数学成绩，绘制成频率分布直方图（如图所示）．已知这100人中[110，120）分数段的人数比[100，110）分数段的人数多6人．（1）根据频率分布直方图，求a，b的值；并估计抽取的100名同学数学成绩的平均数（假设同一组中的每个数据可用该组区间的中点值代替）；（2）现用分层抽样的方法从分数在[130，140），[140，150]的两组同学中随机抽取6名同学，从这6名同学中再任选2名同学作为“网络课堂学习优秀代表”发言，求这2名同学的分数恰在同一组内的概率．【分析】（1）由频率分布直方图能求出a，b的值，并估计抽取的100名同学数学成绩的平均数．（2）设“抽取的2名同学的分数恰在同一组内”为事件A，在分数为[130，140）的同学中抽取4人，分别用a1，a2，a3，a4表示，在分数为[140，150]的同学中抽取2人，分别用b1，b2表示，从这6名同学中抽取2人，利用列举法能求出抽取的2名同学的分数恰在同一组内的概率．解：（1）依题意a+b＝0.046，1000（b﹣a）＝6，解得a＝0.020，b＝0.026，平均数为：75×0.02+85×0.08+95×0.14+105×0.2+115×0.26+125×0.15+135×0.1+145×0.05＝112．（2）设“抽取的2名同学的分数恰在同一组内”为事件A，由题意，在分数为[130，140）的同学中抽取4人，分别用a1，a2，a3，a4表示，在分数为[140，150]的同学中抽取2人，分别用b1，b2表示，从这6名同学中抽取2人所有可能出现的结果有：（a1，a2），（a1，a3），（a1，a4），（a1，b1），（a1，b2），（a2，a3），（a2，a4），（a2，b1），（a2，b2），（a3，a4），（a3，b1），（a3，b2），（a4，b1），（a4，b2），（b1，b2）共15种，抽取的2名同学的分数恰在同一组内的结果有：（a1，a2），（a1，a3），（a1，a4），a2，a3），（a2，a4），（a3，a4），（b1，b2），共7种，所以抽取的2名同学的分数恰在同一组内的概率为．20．如图，在长方体ABCD﹣HKLE中，底面ABCD是边长为3的正方形，对角线AC与BD相交于点O，点F为线段AH上靠近点A的三等分点，BE与底面ABCD所成角为．（1）求证：AC⊥BE；（2）求二面角F﹣BE﹣D的余弦值．【分析】（1）证明DE⊥AC，AC⊥BD，然后证明AC⊥平面BDE．推出AC⊥BE；（2）以点D为坐标原点，DA、DC、DE所在直线分别为x、y、z轴建立空间直角坐标系D﹣xyz，∠DBE为直线BE与平面ABCD所成的角，求出平面BEF的法向量，利用空间向量的数量积求解二面角F﹣BE﹣D的余弦值．【解答】（1）证明：因为在长方体ABCD﹣HKLE中，有DE⊥平面ABCD，因为AC⊂平面ABCD所以DE⊥AC，因为四边形ABCD是正方形，所以AC⊥BD，又BD∩DE＝D，从而AC⊥平面BDE．而BE⊂平面BDE，所以AC⊥BE；（2）解：因为在长方体ABCD﹣HKLE中，有DA、DC、DE两两垂直，以点D为坐标原点，DA、DC、DE所在直线分别为x、y、z轴建立如下图所示的空间直角坐标系D﹣xyz，由（1）知∠DBE为直线BE与平面ABCD所成的角，又因为BE与平面ABCD所成角为，所以，所以．由AD＝3，得，可知，所以，因为，故，则A（3，0，0），，，B（3，3，0），C（0，3，0），所以，，设平面BEF的法向量为，则，即，令，可得，因为AC⊥平面BDE，所以为平面BDE的法向量，即，所以．根据法向量方向与二面角半平面的位置关系，可知二面角F﹣BE﹣D为锐二面角，所以二面角F﹣BE﹣D的余弦值为．21．已知圆C：x2+（y﹣2）2＝r2（r＞0）与直线l：3x+4y+12＝0相切．（1）求圆C的标准方程；（2）若动点M在直线y+6＝0上，过点M引圆C的两条切线MA、MB，切点分别为A，B．①记四边形MACB的面积为S，求S的最小值；②证明直线AB恒过定点．【分析】（1）根据题意，求出圆心C到直线l的距离，由直线与圆相切的性质可得r＝d，由圆的标准方程的形式分析可得答案；（2）①根据题意，先求出S的表达式，分析可得要使S最小，只要|MC|最小即可；由直线与圆的位置关系可得|MC|min＝8，计算即可得答案；②根据题意，分析可得M、A、C、B四点共圆．且以MC为直径，进而可得该圆的方程，与C的方程联立，计算可得直线AB的方程，进而分析可得答案．解：（1）由题意知，圆心C：（0，2）到直线l：3x+4y+12＝0的距离，圆C：x2+（y﹣2）2＝r2（r＞0）与直线l：3x+4y+12＝0相切，则r＝d＝4，故圆C的标准方程为x2+（y﹣2）2＝16；（2）①圆C的圆心为（0，2），半径r＝4，因为MA、MB是⊙C的两条切线，所以CA⊥MA，CB⊥MB，故又因为，根据平面几何知识，要使S最小，只要|MC|最小即可；易知，当点M坐标为（0，﹣6）时，|MC|min＝8．此时，②设点M的坐标为（a，﹣6），因为∠MAC＝∠MBC＝90°，所以M、A、C、B四点共圆．且以MC为直径．该圆方程为：x（x﹣a）+（y+6）（y﹣2）＝0又圆C的方程为x2+（y﹣2）2＝16两圆方程相减得：﹣ax+8y＝0即直线AB的方程为﹣ax+8y＝0，所以直线AB恒过定点（0，0）．22．已知函数．（1）求函数f（x）的定义域；（2）设g（x）＝f（x）+a，若函数g（x）在（1，2）上有且仅有一个零点，求实数a 的取值范围；（3）设，是否存在正实数m，使得函数y＝h（x）在[1，2]内的最大值为4？若存在，求出m的值；若不存在，请说明理由．【分析】（1）根据对数函数定义可求得其定义域；（2）由（1）可得g（x），则由条件可知其单调递增，则g（1）g（2）＜0，解得a的取值范围即可；（3）换元得，易得h（t）的单调区间，分类讨论得到m的值．解：（1）由题意，函数有意义，则满足3x﹣1＞0，解得x＞0，即函数f（x）的定义域为{x|x＞0}；（2）由g（x）＝f（x）+a，且，可得，且g（x）为单调递增连续函数，又函数g（x）在（1，2）上有且仅有一个零点，所以g（1）•g（2）＜0，即（a+1）•（a+3）＜0，解得﹣3＜a＜﹣1，所以实数a的取值范围是（﹣3，﹣1）；（3）由，设t＝f（x），x∈[1，2]，则，易证h（t）在为单调减函数，在为单调增函数，当时，函数h（t）在[1，3]上为增函数，所以最大值为，解得m＝3，不符合题意，舍去；当时，函数h（t）在[1，3]上为减函数，所以最大值为，解得m＝3，不符合题意，舍去；当时，函数h（t）在上增函数，在上为减函数，所以最大值为h（1）＝4或h（3）＝4，解得m＝3，符合题意，综上可得，存在m＝3使得函数y＝h（x）的最大值为4．。
江苏省南通市如东县2019-2020学年高一数学下学期期末考试试题(含解析)
【点睛】本题考查平面上两点间距离，考查余弦定理，解题关键是确定 点所在平面区域，然后由点到圆的位置关系求得结论．
四、解答题：本大题共6小题，共70分.请在答题卡指定区域内作答.解答时应写出文字说明、证明过程或演算步骤.
17.△ABC的内角A，B，C的对边为 ，
（1）求A；
（2）若B=45°，a=2，求b，c.
【详解】由题意，圆 可化简为 ，
所以，圆 的圆心坐标 ，半径 ，
圆 的圆心坐标 ，半径 ，
所以， ，
所以， 或 ，解得 或 .
故选：AC.
【点睛】本题考查两圆 位置关系的判定，考查分类讨论的数学思想方法，属于基础题.
12.如图，在三棱锥 中， 、 、 分别为棱 、 、 的中点， 平面 ， ， ， ，则（ ）
因此其面积为 ，故B正确；
因为 ， 平面 ， 平面 ，所以 平面 ，
因此点 与点 到平面 的距离相等，故C正确；
取 中点为 ，连接 ， ，
则 ， ，且 ，
即等于直线 与直线 所成的角，
又 ， ，
因此 ，
所以直线 与直线 不垂直，即D错；
故选：BC.
【点睛】本题主要考查立体几何的综合，涉及棱锥体积，线面垂直，线面平行，异面直线所成角的求法等，属于常考题型.
A. 7B. 8C. 9D. 10
【答案】C
【解析】
试题分析：由已知表格得： ， ，
由于线性回归直线恒过样本中心点 ，所以有： ，解得： ，
所以线性回归方程 ，
由 得： 解得： ，
由于 ，
所以据此模型预报该设备使用年限的最大值为9.
故选C.
考点：线性回归．
9.已知△ABC的内角A，B，C的对边分别为a，b，c，AD为角A的角平分线，交BC于D， ， ，BD＝2，则b＝（ ）
2019-2020学年江苏如东高级中学高三英语期末试卷及参考答案
2019-2020学年江苏如东高级中学高三英语期末试卷及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30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AOver the years, I’ve been guilty of hastily shutting the front door to many strangers when they came knocking with the intention of selling things. But earlier this year, around Easter time, a dear friend of mine had an experience that changed my mind and perception towards these “intruders”.Linda, who recently moved to a new neighbour hood, had been housebound all week suffering from a severe case of flu when, early one morning, there was the terrible knock on her front door. Peering out the window, she saw two young ladies demurely (端庄地) holding leaflets and a TV set model. She knew of a large TV set establishment nearby and assumed a message was about to be gently delivered. “This is the last thing I need today,” she muttered to herself and hesitantly opened the door.Coughing, she poked her head out and impatiently informed them she wasn’t the slightest bit interested in any TV whatsoever. Furthermore, she added, she was feeling quite unwell and abruptly shut the door. The ladies politely turned and left in silence.A few hours later, another knock. Linda glanced out the window and to her surprise, there were the same two ladies, back again. Really annoyed this time, she opened the door ready togive them a piece of her mind. Before she could speak, with concerned looks on their faces, these women handed Linda what looked to be a dish, saying, “We’re so sorry we disturbed you earlier. We thought you may like this homemade chicken soup. Hopefully it might make you feel a little better.”Linda was taken aback by this lovely gesture. Being new to the area, she didn’t know many neighbours or people who could assist her with shopping or errands (跑腿) so the gesture was all the more appreciated. She could only smile sheepishly and sincerely thanked them. With that, the ladies left.After she related this story to me, I thought about how touching this deed really was. Especially in today’s world where sadly, kindness and thoughtfulness seem so rare. I also realized that thinking about others and showing kindness is the real essence of love.1. What does the expression “give them a piece of her mind”in Paragraph 4 mean?A. Scold them severely.B. Drive them away.C. Refuse them directly.D. Speak out her idea.2. What can be learned about Linda from the passage?A. Linda always hastily shut the front door to many people.B. Linda was not a little interested in any TV whatsoever.C. Linda was very grateful to the ladies for their chicken soup.D. Linda would buy their TV set in reward for the ladies’ chicken soup.3. What’s the best title for the passage?A. A Chicken SoupB. Soup for the soulC. The Real Essence of LoveD. A Friend’s experienceB36-year-old Juan Dual likes to joke that he’s empty inside. Juan’s story began when he was only 13. It was then that he was diagnosed with a terrible disease, which left him with a 99.8% chance of developing cancer of the digestive system. At age 19, right after finishing high-school, Juan underwent a tough operation to take away his colon and rectum. Sadly, it was only the beginning. By age 28, Juan’s disease had affected his stomach and gallbladder so he had to go under the knife again.Having just recovered from several serious surgeries, Juan Dual decided to accept the invitation of some friends of his parents and travel to Japan. It was there that things started to change for the better. He didn’t speak a word of Japanese, so he spent most of his time walking his dog. One day, the dog pulled harder, and Juan realized that he was still able to jog, and he started to do just that.Months later, he found himself working in a small, peaceful town in England. There was little in terms of entertainment, but the town was surrounded by hills, so he devoted even more of his time to running. He befriended some like-minded folks and told them what he’d been through, and they seemed amazed at the fact that he was still alive, let alone that he was pushing himself to exercise. That’s when the idea of focusing on motivating others took root in his mind.With the help of Pepa, a nutritionist, Juan Dual slowly relearned how to eat to keep his energy level high enough to sustain him during physical activity. Eight months after his last operation, he finished the Barcelona half marathon in two hours. He then started training for mountain running and ultra-marathons.4. Why does Juan Dual say he is empty inside?A. Because he has no desire for anything.B. Because he doesn’t have much knowledge.C. Because he always suffers from great hunger.D. Because many of his organs have been removed.5. What made Juan Dual aware that he could still run?A. His parents’ support.B. A walk with his dog.C. The idea of challenging himself.D. His quick recovery from surgeries.6. When did Juan Dual decide to inspire others with his story?A. After finishing the Barcelona half marathon.B After being introduced to a nutritionist named Pepa.C. After sharing it with his friends in an English town.D. After making friends with people with similar sufferings.7. Which of the following words can best describe Juan Dual?A. Ambitious and intelligent.B. Inspiring and responsible.C. Unfortunate but determined.D. Confident but stubborn.CShanghairesidents passing through the city’s eastern Huangpu district in Octobermight have astonished at an unusual sight: a “walking” building. An 85-year-old primary school has been lifted off the ground in its entirety and relocated using new technology named the “walking” machine.In the city’s latest effort to preserve historic structures, engineers used nearly 200 mobile supports under the five-story building. The supports act like robotic legs. They’re split into two groups which in turns rise up and down, imitating the human step. Attached sensors help control how the building moves forward.TheLagenaPrimary School, which weighs 7,600 tons, faced a new challenge — it’s T-shaped, while previously relocated structures were square or rectangular. Experts and technicians met to discuss possibilities and test a number of different technologies before deciding on the “walking machine”.Over the course of 18 days, the building was rotated 21degrees and moved 62 meters away to its new location. The old school building is set to become a center for heritage protection and cultural protection. The project marks the first time this “walking machine” method has been used inShanghaito relocate a historical building.In recent years,China’s rapid modernization has seen many historic buildingsrazedto clear land for skyscrapers and office buildings. But there has been growing concern about the architectural heritage loss as a result of destruction across the country.Shanghaihas beenChina’s most progressive city when it comes to heritage preservation. The survival of a number of 1930s buildings in the famous Bund district and 19th-century “Shikumen” houses in the repairedXintiandi neighborhood has offered examples of how to give old buildings new life. The city also has a track record of relocating old buildings. In 2018, the city relocated a 90-year-old building in Hongkou district, which was then considered to beShanghai’s most complex relocation project to date.8. How did the primary school get moved?A. By reducing the weight of it.B. By using movable supports.C. By dividing it into several parts.D. By using robotic legs.9. What does the underlined word “razed” probably mean in Paragraph 5?A. Replaced.B. Burnt.C. Protected.D. Destroyed.10. What can we infer about the heritage preservation inChina?A. The use of advanced technology leads to growing concern.B. Shanghai is the pioneer in preserving architectural heritage.C.A number of old buildings have been given new life.D. Many historic buildings will be relocated.11. What is the passage mainly about?A. New preservation campaigns are launched inChina.B. New technology gives new life to historic buildings.C. A building inShanghai“walks” to a new location.D. “Walking machine” makes heritage protection simpler.DWhen Rich Jean wanted to help his daughter, Abigail, learn to read, he took her to the library near their home in Brooklyn, N. Y. That's where they met Hasina Islam, who Jean says arose her interest in reading and the library.“You see what you started? You see that spark that you put in this child?” Jean told Hasina Islam at aStoryCorpsconversation in 2016. At the time, Abigail was 7 and Islam was 27. Their friendship began when Abigail was 3. Through the years, Islam has offered book suggestions that Abigail has read with great enthusiasm. “What's cool is that Hasina has recommended a lot of books that I, at the time, thought might be a little too advanced foryou," Jean told Abigail. “Like Charlie and the Chocolate Factory." Abigail said.Islam's own love of the library was sparked when she was in the third grade. She lived near the main branch of the Queens Public Library in New York City, and she went there to research Henry Hudson, an English explorer, for a school project."The librarian made me feel so special. She remembered my name, and my favorite thing was that she gave me book recommendations," she said." When I was graduating from college, I thought about how I was going to make a difference in the world. And I remembered my librarian，and I remembered that feeling that she gave me every single time I went to the library. ”12. When might Abigail and Hasina Islam first meet?A. In 2012.B. In 2016.C. In 2018.D. In 2020.13. What do we know about Charlie and the Chocolate Factory from the text?A. It was Islam's favorite book.B. It might be hard for Abigail.C. It was a best seller at that time.D. It was important for Abigail.14. How did Hasina Islam help Abigail?A. By offering books to her.B. By reading together with her.C. By giving advice on books.D. By introducing great libraries.15. What is paragraph 3 mainly about?A. Islam's special college life.B. Islam's working experiences.C Islam's living conditions. D. Islam's reasons for loving library.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
江苏省如东高级中学2019_2020学年高一生物下学期第二次阶段测试试题
江苏省如东高级中学2019-2020学年高一生物下学期第二次阶段测试试题第Ⅰ卷（选择题，共45分）一、单项选择题（本部分包括15小题，每题2分，共30分。
每题只有一个正确选项。
）1. 下列关于高等动物减数第一次分裂主要特征的叙述,不正确的是( )A.细胞中同源染色体会出现两两配对的现象B.染色体复制后每条染色体上的着丝粒分裂C.四分体中的非姐妹染色单体发生交叉互换D.同源染色体分离后分别移向细胞两极2. 下图为某生物一个细胞的分裂图像,着丝点均在染色体端部,图中①②③④各表示一条染色体,下列表述正确的是( )A.图中细胞处于减数第二次分裂前期B.图中细胞的每条染色体上只有一个DNA分子C.染色体①和③可能会出现在同一个子细胞中D.染色体①和②在后续的分裂过程中会移向同一极3．图1为某二倍体生物(AaBb)细胞不同分裂时期每条染色体上的DNA含量变化,图2表示其中某一时期的细胞图像。
下列有关叙述正确的是( )图1 图2A.图1若为减数分裂,则A与a的分离和A与B的组合发生在cd段B.图1若为有丝分裂,则ef段的细胞都含有两个染色体组C.图2细胞可能是次级精母细胞或次级卵母细胞或极体D.图2细胞中①与②、③与④为同源染色体4．下图表示同一个初级卵母细胞形成的一个卵细胞和三个极体以及受精作用(图中省略了减数分裂中表现正常的其他型号的染色体)。
下列有关叙述正确的是( )A.卵细胞继承了初级卵母细胞1/4的细胞质B.图中卵细胞形成过程中,在减数第二次分裂发生异常C.图示形成的受精卵发育成的个体患有先天智力障碍D.图中受精作用过程中会发生基因重组5．孟德尔一对相对性状的杂交实验中,实现3∶1的分离比必须同时满足的条件是( )①观察的子代样本数目足够多②F1形成的雌雄配子数目相等且生活力相同③雌雄配子结合的机会相等④F2不同基因型的个体存活率相等⑤等位基因间的显隐性关系是完全的A.①②⑤B.①③④⑤C.①②③④⑤D.①②③④6．某昆虫常染色体上存在灰身(B)和黑身(b)基因,现查明雌性含B基因的卵细胞有50%没有活性。
江苏省如东高级中学2019_2020学年高一化学下学期第二次阶段测试试题含解析
江苏省如东高级中学2019-2020学年高一化学下学期第二次阶段测试试题（含解析）选择题单项选择题：本题包括10小题，每小题3分，共计30分。
每小题只有一个．．．．选项符合题意。
1.下列是同周期元素基态原子的最外层电子排布式，所表示的原子最容易得到电子的是A. ns2B. ns2np1C. ns2np4D. ns2np5【答案】D【解析】【详解】最外层电子数≥4时容易得到电子，并且最外层电子数越多越容易得电子，D表示最外层电子数为7，最容易得电子，故选D。
2.下列表示氮原子结构的化学用语中,对核外电子运动状态描述正确且能据此确定电子能级的是（）A. B. C. 1s22s22p3 D.【答案】C【解析】【详解】A. 是N原子的原子结构示意图，只能看出在原子核外各个电子层上含有的电子数的多少，不能描述核外电子运动状态，故A错误；B. 是N原子的电子式，可以得到原子的最外电子层上有5个电子，不能描述核外电子运动状态，故B错误；C. 是N原子的核外电子排布式，不仅知道原子核外有几个电子层，还知道各个电子层上有几个电子亚层，及核外电子运动状态，故C正确；D. D是N原子的轨道表示式，但原子核外的电子总是尽可能的成单排列，即在2p的三个轨道上各有一个电子存在，且自旋方向相同，这样的排布使原子的能量最低，故D错误。
故选C。
【点睛】核外电子是分层排布的，每层可以分为不同的能级，核外电子排布应符合构造原理，泡利原理，洪特规则。
3.某物质可溶于水、乙醇，熔点为209.5℃，其分子结构简式如图，下列说法不正确的是（）A. 该分子与水分子可以形成氢键B. 该分子中原子最外层均达到8电子稳定结构的为C、NC. 1mol该物质中σ键和π键的个数比为5：3D. 该分子中C原子的杂化方式有2种【答案】C【解析】【详解】A．该分子中含有原子半径小、电负性大的N原子以及N—H键，它与水分子间可形成氢键，A正确；B．该分子中，C原子和N原子的最外层均达到8电子稳定结构，B正确；C．单键为σ键，双键中含1个σ键和1个π键，三键中含1个σ键和两个π键，1mol 该物质中含有9molσ键和3molπ键，二者的个数比为9：3，C不正确；D．该分子中C原子的杂化方式分别为sp（碳氮三键中C）、sp2（碳氮双键中C），共有2种杂化方式，D正确；故选C。
江苏省如东高级中学2019-2020高一英语下学期第二次阶段测试试题(含解析)
3. Why is the man so happy？
A。 He finished writing his opera.
B。 He got a ticket to an Italian opera。
Children’s Museum of Houston
This museum lies in Texas。 Termed as the best playground for mind, this museum offers scores of activities for kids of every age. It even offers free family adventure events where families with their kids are encouraged to work on math activities and games。 Children also learn about broadcasting activities and preserving ecosystems。
江苏省如东高级中学2019—2020学年高一英语下学期第二次阶段测试试题(含解析)
时间:120分钟 总分：150分
注意事项:
1。答卷前，考生务必将自己的姓名、考生号等填写在答题卡和试卷指定位置上。
2.回答选择题时,选出每小题答案后， 用铅笔把答题卡上对应题目的答案标号涂黑.如需改动,用橡皮擦干后，再选涂其他答案标号。回答非选择题时, 将答案写在答题卡上,写在本试卷上无效。
第一节 (共15题，每小题2分,满分30分)
阅读下列短文，从每题所给的四个选项(A、B、C和D）中，选出最佳选项,并在答题卡上将该项涂黑。
2019-2020学年江苏省如东高级中学高一下学期期末热身练英语试题
江苏省如东高级中学2019-2020学年高一下学期期末热身练英语试题第一部分听力第一节(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『答案』。
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个小题。
每段对话仅读一遍。
1. What did the man take out of the cupboard?A. A cake.B. Bowls.C. Some sugar.2. What are the speakers talking about?A. Sightseeing.B. Weather.C. Fishing.3. What is the relationship between the speakers?A. Boss and secretary.B. Salesgirl and customer.C. Doctor and patient.4. What does the man imply?A. He is stressed.B. He works too hard.C. He needs some excitement.5. How long will the boy be at summer camp?A. For one week.B. For two weeks.C. For three weeks.第二节(共15小题；每小题1.5分，满分22.5分)听下面5段对话或独白。
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。
听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，每个小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料，回答第6至7题。
6. How does the man feel?A. Elegant.B. Painful.C. Focused.7. What advice does the woman give the man?A. Go to the opera house.B. Take a short break.C. Relax and keep going.听第7段材料，回答第8、9题。
江苏省如东高级中学2019-2020学年高一语文下学期第二次阶段测试试题不分版本
江苏省如东高级中学2019-2020学年高一语文下学期第二次阶段测试试题不分版本江苏省如东高级中学2019-2020学年高一语文下学期第二次阶段测试试题一、现代文阅读(35分)(一)现代文阅读Ⅰ(此题共5小题，19分)材料一：“新基建〞，主要包括5G基站、特高压、城际高速铁路和城际轨道交通、新能源汽车充电桩、大数据中心、人工智能和工业互联网等领域。
“新基建〞投资不是简单的以铁路公路建设为主的“铁公基〞(铁路、公路、机场、水利等重大根底设施建设)形式的投资，而是着眼于前沿科技开展、经济高质量开展、提升人们生活幸福感的新型基建投资。
比方，5C的意义不仅仅是网速更快，而是通过毫秒级时延和超高密度连接有效支持海量物联网设备接入，实现机器间大规模的相互通信，为研究成果创造更多走向现实应用的可能。
“新基建〞着眼长远，但当前的疫情确实赋予其更多责任。
要想尽快降低疫情对经济的负面影响，推动经济恢复正常轨道，需要通过狠抓以“新基建〞为代表的领域，给予总需求一个回弹力，进而通过需求端的率先复苏，带动生产端走出困局。
随着疫情全球蔓延，今年我国面临的外部环境压力有可能进一步加大，外需增长有可能遇到更多困难。
这样的背景下，更要依靠内需发力，推动经济平稳开展。
一定意义上，“新基建〞是为疫情后中国经济恢复打下的一针强心剂，更是未雨绸缪为未来开展铺下的一条路。
(摘编自邹蕴涵《加快“新基建〞对中国经济有长远意义》) 材料二：要用好“新基建〞，使其成为化解疫情影响，撬动经济的有力杠杆，关键还要把握好这个“新〞字。
“新基建〞之“新〞，首先新在开展理念。
不可否认，加快新型根底设施建设，会拉动相当可观的投资。
有机构曾测算，到2025年我国5G网络建设投资累计将到达1.2万亿元，带动产业链上下游以及各行业应用投资超过3.5万亿元。
但与拉动投资相比，“新基建〞更多是通过提升经济运行效率来带动社会开展。
在线办公助力恢复生产运行，远程教育实现“停课不停学〞，智能制造对冲了负面影响……疫情发生以来，经济社会没有因物理“隔离〞而“停摆〞，离不开“新基建〞的支撑。
2019-2020学年江苏如东高级中学高三英语下学期期末考试试题及参考答案
2019-2020学年江苏如东高级中学高三英语下学期期末考试试题及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30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AArtificial intelligence (Al) is practically everywhere today. There are so many products out there which use Al. Some are being developed, some are already in use, and some failed and are being improved, so it’s very difficult to name a few of them and regard them as the best.ViIt is an Al personal trainer which is mainly concerned with fitness and coaching. It, however, requires the use of bio-sensing earphones and other fitness tracking equipment! It can play your favourite music while you work out and all you have to worry about is the exercise you're doing.Deep TextDo you ever wonder how an ad appears suddenly just when you are looking for something similar? This is because of Deep Text. It uses real-time consumer information to produce data which in turn is used to target consumers. Thus, if you search online for flight tickets fromBangaloretoDelhi, it is very likely that an ad relating to hotels inDelhiwill soon follow.Hello EggIf you live alone and miss your mother because you always miss your breakfast or don’t know what to eat for dinner, then Hello Egg is exactly what you are looking for. A very healthy choice of the 2-minute noodles and oats, Hello Egg provides you with a detailed weekly meal plan about the needs of your body. It is truly a modern AI-powered home cooking tool for the young.WordsmithYou can put Mr. Smith into your Microsoft Excel using their free API, and let it write up detailed analysis of the stories behind your numbers. It can produce detailed reports on thousands of pages of spreadsheets in seconds.1. What can we learn about Vi from the text?A. It is an AI music player.B. It is a bio-sensing earphone.C. It doesn't work without bio-sensing earphones.D. It can make you more energetic while you work out.2. Which can help you improve cooking skill?A. Deep Text.B. Vi.C. Wordsmith.D. Hello Egg.3. What can Wordsmith do for us?A. Produce a detailed report.B. Provide us with a detailed meal plan.C. Book a ticket ahead of time.D. Offer us information on hotels for traveling.BIvy League schools are considered to be the most prestigious of all colleges in the United States. These schools are primarily located in the Northeastern part of the country. There are eight total colleges that are considered to be Ivy League. These schools are Brown, Harvard, Cornell, Princeton, Dartmouth, Yale, and Columbia universities and the University of Pennsylvania. Of all institutions of higher learning, these elite schools are considered to be the most outstanding and the most sought-after in terms of acceptance and graduation.The term “Ivy League” came about in 1954 when the NCAA athletic conference for Division I was formed. At the time, the elitism of these schools was really due to their prestige in the realm of sports like basketball. Although the term “Ivy League” was not created until the 1950s, many of these schools were in existence as far back as 1636, when John Harvard became the first benefactor of Harvard University.Although this group of elite schools is considered to be part of one big league of the elite, there have been plenty of internal rivalries over the years. The sports that these colleges play were so popular that some teams began playing games in New York City so spectators could come from far away and watch the games. The popularity of both the athletes who played and the college team rivalries brought in a good deal of attention to the schools as well as revenue from ticket sales. There have also been academic rivalries between schools. Mostly, these rivalries are a matter of opinion in terms of which school has the most honor graduates, which schools offer the most prestigious scholarships, and what famous graduates have come from each school.Each Ivy League college has its own unique accomplishments that make it important. All carry a certain reputation with them, and each school has programs that excel primarily in the medical and law fields, making them some of the most sought-after schools in the world. Their admission process is very selective, which helps the schools ensure that they only accept the best and brightest. Many famous people have graduated from IvyLeague schools, including recent presidents George W. Bush, Bill Clinton, and Barack Obama. This prestige leads many to believe that these colleges are only for the wealthy and elite. Often, companies look for Ivy League graduates as potential employees, usually preferred by law firms, medical facilities, and large corporations. It has long beencovetedto have earned a degree from an Ivy League school. Today, the Ivy League schools are still excellent in both academia and in sports, and they have left a legacy of higher education with an exceptional track record and reputation to go along with them.4. Which of the following statements is true according to the passage?A. Ivy League schools were initially famous for their reputation in sports.B. Ivy League schools didn’t come into existence until the 1950s.C. Ivy League schools do not compete with each other within the league.D. Ivy League schools are most popular for their excellence in the medical and law fields.5. Which of the following aspects is NOT mentioned in the passage?A. The history of the Ivy LeagueB. The rivalry of the Ivy LeagueC. The accomplishments and cultural impacts of the Ivy LeagueD. The future development of the Ivy League6. Why do many people believe that Ivy League colleges are only for the wealthy and elite?A. Because they are the most sought-after schools in the world.B. Because they have selective admission process to help ensure the quality of their students.C. Because many famous people have graduated from Ivy League schools.D. Because manycompanies look for Ivy League graduates as potential employees.7. What does the underlined word “coveted” in the last paragraph probably mean?A. advocatedB.DesiredC. restrictedD. sponsoredCImaginary friends in childhood refer to the invisible beings that a child gives a personality to and plays with for over three months.Crabbycrab（蟹）appeared on a holiday in Norway by running out of my four-year-old son Fisher's ear after a night of tears from an earache. Like other childhood imaginary friends, Crabby should be a sign thatFisher's mind is growing and developing positively. Indeed, research shows that imaginary friends can help develop children'ssocial skills.Research has shown that the positive effects of having imaginary friends as a child continue into adulthood. Adolescents who remember their imaginary playmates have been found to use more activecoping（应对）styles, such as seeking advice from loved ones rather than bottle things up inside. Even adolescents with behavioral problems who had imaginary friends as children have been found to have better coping skills through the teenage years.Scientists thinkthis could be because these teens have been able to adjust themselves to the social world with imagination rather than choose to be involved in relationships with more difficult classmates. It could also be because the imaginary friends help to reduce these adolescents,loneliness.These teens are also more likely to seek out social connections -they tend to turn to others for advice. Current research by Tori Watson is taking this evidence and looking at how adolescents who have imaginary friends as children deal withbullying（欺凌）at school. It is found that teens who remember their imaginary friends are better at dealing with bullying.While we know a lot about childhood imaginary friends such as Crabby Crab and the positive effects they can have, there is still a lot to learn about imaginary friends.8. What is Crabby crab?A. It is a crab Fisher caught inNorway.B. It is Fisher's imaginary friend.C. It is a toy Fisher like much.D. It is a cause of earache.9. Why do children with imaginary friends have better coping skills?A. Imaginary friends help improve their adjustment.B. Having imaginary friends makes them smarter.C. They have rich imagination.D. They are no longer alone.10. What will a child with imaginary friends probably do if he is bullied?A. Escape from the bully.B. Fight with the bully bravely.C. Keep silent about being bullied.D. Ask a parent or a teacher for help.11. What is the author's attitude towards the effect of imaginary friends?A. Concerned.B. Doubtful.C. Optimistic.D. Indifferent.DBritish anthropologists (人类学家) Russell Hill and Robert Barton of the University of Durham, after studying the results of one-on-one boxing, tae kwon do, Greco-Roman wrestling and freestyle wrestling matches at the Olympic Games, conclude that when two competitors are equally matched in fitness and skill, the athlete wearing red is more likely to win.Hill and Barton report that when one competitor is much better than the other, colour has no effect on the result. However, when there is only a small difference between them, the effect of colour is enoughtipthe balance. The anthropologists say that the number of times red wins is not simply by chance, but that these results are statistically significant.Joanna Setchell, a primate (灵长目动物) researcher at the University of Cambridge, has found similar results in nature. She studies the large African monkeys known as mandrills. Mandrills have bright red noses that stand out against their white faces. Setchell’s work shows that the powerful males — the ones who are more successful with females — have a brighter red nose than other males.As well as the studies on primates by Setchell, another study shows the effect of red among birds. In an experiment, scientists put red plastic rings on the legs of male zebra finches and this increased the birds’ success with female zebra finches, Zebra finches already have bright red beaks (鸟喙), so this study suggests that, as with Olympic athletes, an extra flash of red is significant. In fact, researchers from theUniversityofGlasgowsay that the birds’ brightly coloured beaks are an indicator of health. Jonathan Blount, a biologist, andhis colleagues think they have found proof that bright red or orange beaks attract females because they mean that the males are healthier. Nothing in nature is simple, however, because in species such as the blue footed booby, a completely different colour seems to give the male birds the same advantage with females.Meanwhile, what about those athletes who win in their events while wearing red? Do their clothes give them an unintentional advantage? Robert Barton accepts that “that is the implication” of their findings. Is it time for sports authorities to consider new regulations on sports clothing?12. According to their research. Hill and Barton conclude that ________.A. the colour of clothing has an effect on most sport eventsB. red should be the choice of colour for clothing in sportsC. red plays a role when competitors are equally capableD. athletes perform better when surrounded by bright red13. The underlined word “tip” in Paragraph 2 is closest in meaning to ________.A. achieveB. advanceC. keepD. change14. The example of the blue footed booby proves that ________.A. male birds use different body parts to draw attentionB. red is not the only colour to attract female birdsC. blue gives female birds the same advantageD. blue can indicate how healthy a bird is15. What can be inferred from the passage?A. The colour red gives male animals the most advantage.B. Male zebra finches prefer to have red plastic rings on their legs.C. Rules on sports clothing are going to be changed.D. Athletes wearing red may have an advantage over their opponents.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2019-2020学年江苏如东高级中学高三英语下学期期末试卷及参考答案
2019-2020学年江苏如东高级中学高三英语下学期期末试卷及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30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AFour Best Hikes in the WorldThere's nothing like getting out and getting some fresh air on a hike. No matter whether your idea of a hike is a leisure walk or climbing the highest mountain on Earth, we've got you covered. Below are four best hikes inthe world.Torres del Paine W CircuitLocation (位置): Patagonia. ChileDistance: 37 + milesTime: 5~6 daysBest time to go: October to JanuaryThe W Circuit is one of the most recommended hikes you'll find. Not only will you appreciate the diverse landscapes and striking granite pillars (花岗岩柱子), but you'll probably meet some new friends along the way.Grand Canyon Rim - to - Rim HikeLocation: Arizona, the United StatesDistance: 48 milesTime: 1~3 daysBest time to go: May to June, September to OctoberThere's no better way to experience one of the greatest wonders in the world. Located in one of the USA's most beautiful parks, the views are ly appealing. Just make sure you're prepared for the challenge.Trek to PetraLocation: JordanDistance: 47 milesTime: 5~ 6 daysBest time to go: October to AprilTake the road less traveled through the Kingdom of Jordan and experience one of the seven wonders of theworld. Hike through canyons, gorges and ridges, and see tombs and temples along the way all while avoiding crowds of tourists.Yosemite Grand TraverseLocation: California, the United StatesDistance: 60 milesTime: 6~7 daysBest time to go: July to SeptemberKnown for some of the best hiking in the world, Yosemite National Park is famous for its views and huge sequoia (红杉) trees. Praised byNational Geographic, the Yosemite Grand Traverse will take you through waterfalls and green mountaintops.1.Which of the following is the best time for the hike in Patagonia, Chile?A.AprilB.MayC.AugustD.December2.Where should you go for a less crowded hike?A.JordanB.Patagonia, ChileC.Arizona, the United StatesD.California, the United States3.What can you do along the Yosemite Grand Traverse?A.Plant sequoia treesB.Appreciate waterfallsC.Visit local templesD.Climb granite pillarsBChimps use loud calls and gestures to make their feelings known but until now, the exact meaning for individual movements has remained a mystery. Now researchers believe they have translated the key gestures used in the chimp community and identified their intentions for the first time.From 4,351 gestures, experts were able to identify 66 that are used for 19 specific message meanings, including showing a foot to tell a child they can climb on their back. The researchers were able to narrow down these 66 gestures to 36 that are used intentionally to achieve 15 purposes. The translations were made by Dr Catherine Hobaiter and her colleagues at St Andrews University in Scotland.Dr Hobaiter used behavior sampling and filmed all recorded cases of gestural communication. Other gestures include stomping their feet to ask another chimp to stop what they are doing, and slapping objects together to ask another to follow them. Of the 19 meanings,17 encouraged interactions to start, or to develop, such as “move closer,” and “change play”. Some of the gestures were found to have more than one meaning. and only 10 of the66 gestures were used for only a single meaning.Researchers collected a total of 471 video clips from two social groups of chimps at a shelter near Kinshasa, Democratic Republic of Congo. As well as identifying what the gesture means, they also discovered the technique needed to increase the chances of success.“Human children use gestures to communicate before they produce their first words, and their earliest gestures typically appear around 10 months of age,” explained the researchers. “In great apes, there is good evidence that language-trained individuals are capable of acquiring and understanding signals, but this is far less clear in their natural communication. ”4. Chimps slap the objects to____________.A. tell others to stop what they are doingB. ask others chimps to join themC. gather other chimps to move closerD. encourage interactions to start5. What did researchers find after studying 471 video clips?A. Chimps trained in language are good at understanding signals.B. Two social groups of chimps live at a shelter near Kinshasa.C. Language-trained individuals do well in natural communication.D. Chimps’earliest gestures appear around 10 months of age.6. How is the last paragraph developed?A. By analyzing causes.B. By examining differences.C By making comparisons. D. By following time order.7. What can be a suitable title for the text?A A New Research on Chimps B. Human Children and ChimpsC. Getting the Chimps Trained for LanguageD. Translating the Sign Language of ChimpsCThe China International Search and Rescue Team(CISAR) was formed in 2001 and is now made up of several hundred rescue workers and about 20 police dogs. The team brings help and hope to those whose lives are changed by astorm, flood, earthquake, or any other natural disasters.After long and careful training, the team went on its first international rescue tasks in 2003. That year, the Chinese team helped save lives after earthquakes inAlgeriaandIran. It was the first time that a Chinese team had worked outsideChinawhose members won high praise for bravery and skill.Since then, the CISAR has completed many tasks. The list of people to whom help has been given is long. Theteam treated more than 3,000 people who were wounded in the 2006 earthquake inIndonesia, helped 2,500 wounded people after the earthquake that hitHaitiin 2010, and spent several months giving aid to over 25,000 people suffering from the 2010 floods inPakistan. On April 26, 2015, a group of 62 people from CISAR went toNepalafter the 8.1 magnitude earthquake that happened there.Rescue workers are trained to find people, treat wounds, and hand out food, water, and other supplies. They have to be able to do work that is difficult under conditions which can be very dangerous. After a disaster, there is usually no electricity or water, and there may be diseases and other dangers. Rescue workers get to save lives, but they must also bury the dead. That means they have to be strong in both body and mind.Rescue workers must have big hearts, too. It takes a lot of love and courage to risk one’s own life to save someone else’s. The members of the CISAR have plenty of both and are always ready to go wherever help is needed.8. What is the function of the numbers in Paragraph 3?A. To advertise for the CISAR.B. To add some basic information.C. To praise Recue Workers’ contributions.D. To stress the dangers Rescue Workers face.9. What is the author’s attitude towards Rescue Workers?A. Hopeful.B. Respectful.C. Curious.D. Supportive.10. What are the last two paragraphs mainly about?A. The duty rescue workers must perform.B. The qualities rescue workers must own.C. The difficulties rescue workers must go through.D. The willingness rescue workers should require.11. What may be the best title of the passage?A. China to the RescueB. How to train CISARC. Welcome to CISARD. Rescue on requestDIt’s a little before8 a.m. when Mathias Schergen pushes open the side door at Chicago’s Jenner Elementary Academy for the Arts. He walks down the hall toward the office to sign in. It’s the same routine he’s had asJenner’s art teacher for nearly a quarter century. “It’s going to be a good day,” a colleague calls out. “It’s a good day.” They hug. It seems like a typical Friday. Except it’s not. After 23 years at Jenner Elementary, Schergen is retiring. Even on his last day, there are still art projects to finish.Schergen leaves behind a richlegacyat this school. He’s won grants (拨款) for art projects. He turned an empty classroom into a museum. He’s pushed his students to make art about their lives. And he was awarded a Golden Apple — the most honorable teaching award in Chicago. But it wasn’t always easy. For years, Schergen taught in one of the city’s toughest neighborhoods. “When I first got my room, I noticed there were bullet holes in the window. That made me nervous,” he says. So he stuffed Beanie Babies in the holes to make it “look kind of funny”. “I didn’t even tell my wife for a whole year,” he says. “I didn’t want her to know.”With one hour to go, Schergen piles the chairs and sweeps the floor. He cleans out the sink for the last time. Fifth-grader Deontae Barnes, one of his best helpers, has watched him say goodbye all day. He wanders in the doorway. “Ah, come here, son,” Schergen says, signaling him over. He bends down for a hug. “Thank you for making these last days special and being a help to me.”When Deontae leaves, a reporter asks Schergen: When your kids ask why you’re retiring, what do you tell them? “I just tell them that grown people have dreams too,” he says. “I have other things in my life I have to do. It’s time. It’s just time.”12. Why is it a special Friday for Schergen?A. He was retiring on that day.B. He won an honor for his school.C. He was interviewed by a reporter.D. He received a Golden Apple award.13. What does the underlined word “legacy” in Paragraph 2 refer to?A. Art projects.B. Great achievements.C. Respect from students.D. Change in teaching.14. What made Schergen nervous when he first got to the school?A. Safety concerns in the school.B. The poorly-equipped classroom.C. Being misunderstood by his family.D. Students’ poor academic performance.15. What is the best title for the text?A.A Typical Day for an Art TeacherB. Time for Art ProjectsC. A Teacher’s Final Day at SchoolD. The Last Art Class第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
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如东高级中学2019---2020学年第二学期高一年级期末热身练
高一数学2020-07-11
一、单项选择题：本题共10小题，每小题5分，共50分．
1.下列结论中错误
．．的是
A.
B. 若是第二象限角，则为第一象限或第三象限角
C. 若角的终边过点，则
D. 若扇形的周长为6，半径为2，则其圆心角的大小为1弧度
2．经过两直线与的交点，且在两坐标轴上的截距相等的直线方程是
A. B.
C. D. 或
3.如果平面直角坐标系内的两点，关于直线l对称，那么直线l的方程为
A. B. C. D.
4.甲、乙、丙、丁四名同学在某次军训射击测试中，各射击10次．四人测试成绩对应的条形图如下：
以下关于这四名同学射击成绩的数字特征判断不正确的是
A. 平均数相同
B. 中位数相同
C. 众数不完全相同
D. 方差最大的是丁
5. 过点引直线，使，到它的距离相等，则这条直线的方程是
A. B.
C. 或
D. 或
6.在中，角A、B、C的对边分别为a，b，c，的面积为S，若，则角C的值为
A. B. C. D.
7.如图，的二面角的棱上有A，B两点，直线AC，BD分别在这个二面角的两个半平面内，且都垂直于已知，，，则CD的长为
A. B. 7 C. D. 9
（第7题图）（第9题图）
8.已知向量，，，若，则与的夹角为
A. B. C. D.
9. 如上图，四边形ABCD中，，，将四边形ABCD沿对角线BD折成四面体，使平面平面BCD，则下列结论正确的是
A. B.
C. 与平面所成的角为
D. 四面体的体积为
10. 中国传统文化中很多内容体现了数学的“对称美”如图所示的太极图是由黑白两个鱼形纹组成的圆形图案，充分体现了相互转化、对称统一的形式美、和谐美给出定义：能够将圆O的周长和面积同时平分的函数称为这个圆的“优美函数”．
给出下列命题：正弦函数可以是无数个圆的“优美函数”；
函数可以是无数个圆的“优美函数”；
函数可以是某个圆的“优美函数”；
函数是“优美函数”的充要条件为函数的图象是中心对称图形
其中正确命题的序号是
A. B. C. D.
二．多项选择题：本题共2小题，每小题5分，共10分.在每小
题给出的选项中，有多项符合题目要求.全部选对的得5分，部
分选对的得3分，有选错的得0分.
11.下列关于空间几何体的说法正确的是
A. 棱柱的两个底面是全等的多边形，且对应边互相平行
B. 棱柱的侧棱都相等，侧面都是平行四边形
C. 各个面都是三角形的几何体是三棱锥
D. 圆柱的任意两条母线互相平行
12.如图，，，，是以OD为直径的圆上一段圆弧，是以BC为直径的圆上一段圆弧，是以OA为直径的圆上一段圆弧，三段弧构成曲线则下述正确的是
A. 曲线W与x轴围成的面积等于
B. 曲线W上有4个整点横纵坐标均为整数的点
C. 所在圆的方程为：
D. 与的公切线方程为：
三．填空题：本大题共6小题，每小题5分，共30分．请将答案填入答题纸相应答题上．
13.已知水平放置的按“斜二测画法”得到如图所示的直观图，其中，
，则原的面积为_________．
14.如上图所示，有A，B，C，D四个海岛，已知B在A的正北方向15海里处，C在A的东偏北方向，又在D的东偏北方向，且B，C相距21海里，D在A的正东方向，则C，D两岛间的距离是_________．
15. 若a，1，，则函数有零点的概率为________．
16. 已知三棱锥的四个顶点均在同一个球面上，底面ABC满足，，若该三棱锥体积的最大值为3，则其外接球的体积为_________．
四、解答题：本大题共6题，共70分．解答应写出文字说明，证明过程或演算步骤．
17.【本题满分10分】已知向量，，且函数．
若，求的值；
在中，且，求面积的最大值．
18.【本题满分12分】如图所示，在三棱柱中，侧面为菱形，，
，侧面为正方形，平面平面点M为的中点，点N为AB的中点．
证明：平面；
求三棱锥的体积．
19. 【本题满分12分】为了分析某高三学生的学习状态，对其下一阶段的学习提供指导性建议现对他前7次考试的数学成绩x、物理成绩y进行分析下面是该学生7次考试的成绩．
数学成绩分888311792108100112
物理成绩分949110896104101106
他的数学成绩与物理成绩哪个更稳定请给出你的证明
已知该学生的物理成绩y与数学成绩x是线性相关的，若该学生的数学成绩达到116分，请你估计他的物理成绩大约是多少
20. 【本题满分12分】如图，在四棱锥中，ABCD平面ABCD，四边形ABCD是菱形，ABCD，，且交于点O，E是PB上任意一点．
求证：；
已知二面角的余弦值为，若E为PB的中点，
求EC与平面PAB所成角的正弦值．
21. 【本题满分12分】如图，在平面直角坐标系xOy中，已知圆
，直线．
若直线过点，且与圆O相交所得弦的长度为，
求直线的方程；
若点B为直线l上的动点，直线BM与圆O相切于点M．
记的外接圆为动圆C，证明：动圆C过异于原点O的定点，并求出定点坐标；
设半径为4的圆D与圆O外离，过点B圆D作的切线，切点为
若对任意的点B，都有成立，求圆D的方程．
22. 【本题满分12分】
若在定义域内存在实数，使得成立，则称函数有“飘移点”．函数是否有“飘移点”？请说明理由；
证明函数在上有“飘移点”；
若函数在上有“飘移点”，求实数a的取值范围．。