考点：圆的一般方程和标准方程的互化.
5．A
【解析】 试题分析：正三角形的高为a CM 23=,在直观图中的长度为a M C 4
3/=,故C '''?A B 的高a a h 8645sin 430/==,故其面积216
68621a a a S =?=,故应选A. 考点：平面图形直观图的画法规则及运用.
6．C
【解析】
试题分析：将1EA 平移到1GB ,连1FB ,则1EGB
∠就是异面直线所成的角,因为2,511==GB FB ,而311122=++=+=CF CG FG ,故2
1221GB FG FB +=,即0190=∠EGB ,故应选C.
考点：异面直线所成角的概念及求法.
7．D
【解析】
试题分析：设点P(2,5)关于直线0x y +=的对称点为),(n m Q ,由题设125=--m n 且02
522=+++n m ,解之得2,5-=-=n m ,故应选D. 考点：点对称问题的求解思路和方法.
8．C
【解析】
试题分析：取11D B 的中点为H ,连H C 1,因为⊥1BB 平面?H C D C B A 11111,平面1111D C B A ,故11BB H C ⊥,又111B D H C ⊥,故⊥H C 1平面11BB D D ,则BH C 1∠就是直线1BC 与平面11BB D D 所成角,因a BC a H C 2,2211==,故a BH 2
6=,故BH C 1∠
应选C.
D 1A B
C
考点：线面角的定义及求法.
【易错点晴】本题以正方体这一简单几何体为背景,考查的是直线与平面所成角的余弦值的求法问题及直线与平面的位置关系等知识的综合运用的综合问题.求解时充分借助题设条件和线面角的定义,运用线面的垂直关系找出直线1BC 在平面11BB D D 的射影,进而确定BH C 1∠就是直线1BC 与平面11BB D D 所成角,然后在直角BH C 1?中求出a BC a H C 2
,2211==,故a BH 2
6=,故BH C 1∠. 9．A
【解析】
试题分析：由平面几何的知识可知当圆心⊥OP 过点P 的直线时,这被分成的两部分面积最
大,因为1=OP k ,故所求直线的斜率1-=k ,故其方程为)1(1--=-x y ,即20x y +-=,应选A.
考点：圆的标准方程和直线的点斜式方程.
【易错点晴】本题以直线与圆表示的区域为前提和背景,考查的是圆的标准方程及直线与圆的位置关系等知识的综合运用的综合问题.求解时充分借助题设条件中的有效信息,利用圆心⊥OP 过点P 的直线的所截圆所得的弦长最短.过圆心的直线截圆所得弦长最长这些结论可知1=OP k ,故所求直线的斜率1-=k ,故其方程为)1(1--=-x y .
10．B
【解析】
试题分析：因为点P 到面CD B A 11的距离是个定值,所以点P 到平面QEF 的距离是定值;又由于QFE ?面积是定值,因此三棱锥QEF P -的体积也是定值.故应选B.
考点：空间直线与平面的位置关系及几何体的体积面积的综合运用.
【易错点晴】化归与转化的数学思想是高考所要考查的四大数学思想之一.本题以正方体这一简单几何体为背景,考查的是距离角度体积面积的定值问题的判定方法问题.求解时,首先要搞清楚QFE ?面积是定值,其次是点P 到面CD B A 11的距离是个定值;这样就容易判定三棱锥QEF P -的体积也是定值,从而选填答案B.
11
．56 【解析】
试题分析:如图,从三视图所提供的信息可以看出该几何体是一个正方体截取一个三棱锥角所剩余的几何体,其体积6
51112131111=
????-??=V ,表面积239243112131132+=?+???+??=S ,
故应填56.
1
A C
考点：三视图的识读和理解．
12．80；13496+
【解析】
试题分析: 从三视图所提供的信息可以看出该几何体是以正方体和四棱锥的下上组合体,其体积
803443
1444=???+??=V ,表面积
349634215421134445+=??+??++??=S ,故应填80；13496+.
1
B C
考点：三视图的识读和理解．
【易错点晴】本题考查的是三视图与原几何体的形状的转化问题.解答时先依据题设中提供的三视图,将其还原为立体几何中的简单几何体,再依据几何体的形状求其表面积和体积.在本题求解过程中,从三视图中可以推测这是一个该几何体是以正方体和四棱锥的下上组合体,其体积803443
1444=???+??=V ,表面积
3496342
15421134445+=
??+??++??=S . 13．(1,-，0x =
【解析】
试题分析:因9)3()1(22=+++y x ,故圆心)3,1(--C ;又因为3=CO k ,所以33-=l k ,直线的方程为x y 3
3-=,即03=+y x ,故应填)3,1(--C ,03=+y x . 考点：直线的方程及圆的标准方程．
14．3
3 【解析】
试题分析:如图,由题意平面11BC A 与12条侧棱所成的角都相等,且都等于α.因⊥D B 1平面11BC A ,且31=D B ,故331=O B ,所以33sin sin 1==∠αBO B ,故应填3
3.
1A C
考点：线面角的定义及求解．
15．]60,20[00
【解析】
试题分析:由题设,AC 不动,CB 与AC 的夹角不变,都是040;若AB 不动,则AC 与AB 的夹角不变,都是010,这样异面直线1CB 与1AC 所成角的最大值是000602040=+,最小值为000202040=-,故范围是]60,20[00,故应填]60,20[00.
考点：异面直线所成角的定义及求解．
16．83- 【解析】 试题分析:如图,建立如图所示的空间直角坐标系xyz O -,则
111222(0,0,0),(,,),(,,)333333D E a a a F a a a , (,,)B a a a ，设),,(z y x P ,由题设])32()32()32[(4)31()31()31(222222a z a y a x a z a y a x -+-+-=-+-+-,即09
159149149142222=+---
++a za ya xa z y x ,也即2222274)97()97()97(a a z a y a x =-+-+-,由此可知点Q P ,都是在球心为)337,337,337(C ,半径为2的球面上,又4=PQ ,故点Q P ,是球的直径的两个端点;所以+=心+=,所以()()MP MQ MC CP MC CQ ?=++
224MC CP CQ MC =+?=- ,而M 在正方体的表面上,故当点M 在正方体的顶点1B 上时,
34||min =MC ,此时 MP MQ ? 的值最小为38434-=-,应填83
-.
考点：空间向量的数量积公式及有关知识的综合运用．
【易错点晴】本题借助几何体的几何特征和题设条件, 巧妙地构建空间直角坐标系xyz O -,借助空间向量的有关知识将问题合理转化为点Q P ,都是在球心为)337,337,337(C ,半径为2的球面上,进而确定点Q P ,是球的直径的两个端点;所以+=心+=,
所以4))((22-=?+=
++=?MC CQ CP MC CQ MC CP MC MQ MP ,最终将问题转化为求
MC 的最小值的问题,进而使得问题获解.
17．（I ） 220x y --=；（II
【解析】
试题分析：（I ）借助题设条件运用直线的点斜式方程求解；（II ）借助题设运用圆心距与半径弦长之间的公式求解.
试题解析：
（Ⅰ）已知圆()2
2:19C x y -+=的圆心为()1,0C ， 因直线过点,P C ，所以直线l 的斜率为2，
直线l 的方程为()21y x =-,即220x y --=
（Ⅱ）当直线l 的倾斜角为45
时，斜率为1，直线l 的方程为22y x -=-,
即0x y -=
圆心C 到直线l
，
又圆的半径为3，弦AB
考点：直线与圆的方程及弦心距与半径的关系等有关知识的综合运用．
18．（1）2a =；（2）3
24. 【解析】
试题分析：（1）借助题设条件运用两直线互相垂直建立方程求解；（2）借助题设运用两直线平行建立方程求出a ,再用两平行直线之间的距离公式求解.
试题解析：
（1）当0b =时，1l ：10ax +=，由12l l ⊥知(2)0a -=， 2分 解得2a =；
（2）当3b =时，1l ：310ax y ++=，当12//l l 时，有3(2)0,310,
a a a --=??-≠?解得3a =，
此时，1l 的方程为：3310x y ++=， 2l 的方程为：30x y ++=即3390x y ++=，
则它们之间的距离为3
d ==. 考点：直线的方程及直线与直线的位置关系等有关知识的综合运用．
19．（1）上面几何体是圆锥，下面几何体是长方体，且圆锥底面圆和长方体上底两边相切；
（2）221)π+，6π+.
【解析】
试题分析：（1）借助题设三视图所提供的图形信息和数据信息判定求解；（2）借助题设运用面积公式和体积公式探求.
试题解析：
（1）从三视图中可以看出，该几何体是组合体，而且上面几何体是圆锥，
下面几何体是长方体，且圆锥底面圆和长方体上底两边相切．
（2）圆锥母线长10132=+ 表面积()22110+-=
-+=π锥底长方体圆锥侧s s s s
体积为π+=6V ， 故所求几何体的表面积是()22110+-π2()m ,体积是π+63()m
考点：三视图的理解和识读和几何体的面积与体积公式等有关知识的综合运用． 20．（1）5m ；（3）5
16)58()54(22=-+-y x .
试题分析：（1）借助题设条件运用圆的标准方程的条件求解；（2）借助题设运用OM ON ⊥建立方程求解；（3）依据题设运用中点坐标公式求圆心和半径.
试题解析：
（1）由0422
2=+--+m y x y x 得:2,4,D E F m =-=-= 2242040D E F m +-=-> 5（2）由题意???=+--+=-+0
420
4222m y x y x y x 把y x 24-=代入04222=+--+m y x y x
得081652
=++-m y y 5
1621=+y y ，5821m y y += ∵OM ON ⊥得出：02121=+y y x x
∴016)(852121=++-y y y y ∴5
8=m （3）设圆心为),(b a
5
82,5421121=+==+=y y b x x a 半径5
54=r 圆的方程5
16)58
()54
(22=-+-y x 考点：圆的标准方程和一般方程之间的互化等有关知识的综合运用．
【易错点晴】圆的标准方程和一般方程是圆的方程的两种形式,是中学数学中的重要内容和工具,也高考和各级各类考试的重要内容和考点.本题以圆的一般形式为背景考查的是圆的一般式与标准式的互化问题.求解时先借助圆的一般式的约定,建立不等式2242040D E F m +-=->,求得58=
m ；第三问则运用圆的标准方程求圆的圆心和半径从而使得问题获解.
21．（1）证明见解析；（2）证明见解析；（3）1λ.
试题分析：（1）借助题设条件运用线面平行的判定定理推证；（2）借助题设运用面面垂直的判定定理推证；（3）依据题设运用二面角的定义求解探求.
试题解析：
（1）令PD 中点为F ，连接EF ，AF
点,E F 分别是PD PC 、的中点， ∴EF //12
CD ,EF ∴//AB . ∴四边形FABE 为平行四边形.
//BE AF ∴,AF ?平面PAD , BE ?平面PAD
PAD BE 面//∴
（2）在梯形ABCD 中,过点B 作BH CD ⊥于H ,
在BCH ?中，1BH CH ==,045BCH ∴∠=.
又在DAB ?中，1AD AB ==,045ADB ∴∠=,
045BDC ∴∠=,090DBC ∴∠=
∴BD BC ⊥.
面PCD ⊥面ABCD ,面PCD ?面ABCD CD =,PD CD ⊥,PD ?面PCD , PD ∴⊥面ABCD ,
PD BC ∴⊥,
BD PD D ?=,BD ?平面PBD ,PD ?平面PBD
∴BC ⊥平面PBD ,
BC ?平面PBC ,
∴平面PBC ⊥平面PBD
（3）作QR CD ⊥于R ，作RS BD ⊥于S ，连结QS
由于QR ∥PD ，∴AB QR CD ⊥平面
∴∠QSR 就是二面角Q BD C --的平面角
∵面PBD ⊥面ABCD ，且二面角Q BD P --为?45
∴∠QSR=?45 ∴SR=QR
设SR=QR=x ，则RC= 2x ， DR=2 2x -，=2x
∵QR ∥PD ∴=1PQ DR PC DC
∴1λ
考点：空间直线与平面的平行于垂直位置关系的判定定理等有关知识的综合运用．
【易错点晴】空间直线与平面的位置关系的判定和性质一直是立体几何中的常见题型.本题以一个四棱锥为背景.考查的是空间中直线与平面的平行和垂直的判定和性质的运用问题.
求解第一问时充分运用直线与平面平行的判定定理,探寻面内的直线与面外的直线平行；第二问中的面面垂直问题则运用转化与化归的思想将其化为直线与平面的垂直问题来推证；第
λ.
三问则依据二面角的定义建立方程从而求出参数1
2021年高二10月第一次月考英语试题
2021年高二10月第一次月考英语试题 一、单项选择（1*10=10分） 1. Our class ___________ twenty four boys and thirty girls. A. consists for B. consists of C. consists with D. consists in 2. Autumn has e. The garden is covered with ________ leaves. A. falling B. fallen C. being fallen D. being falling 3. Please remained ____________ before our guests leave. A. sit B. sat C. seated D. seating 4. They are discussing the project _____________ next week. A. to be carried out B. being carried out C. carrying out D. will be carried out 5. He explained it again to made himself ______________. A. understand B. to understand C. understanding D. understood 6. We have to buy a new air conditioner, for the old one has broken ________. A. off B. up C. down D. out 7. Kate keeps a diary in English to have her written English ____________.
2021-2021年高二10月月考英语试题(有答案)
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高二英语10月月考试题24
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6.What is the man? A.A hotel clerk. B.A shop assistant. C.A bank clerk. 7.How much will the woman pay for the bankbook? A.50 dollars. B.15 dollars. C.5 dollars. 听下面一段对话，回答第10至第12三个小题。 8.What do we know about the woman? A.She is learning German after work. B.She knows several languages. C.She is very good at German. 9.What does the man usually do in his spa re time? A.He studies. B.He works. C.He watches movies. 10.Why will the man call Henry? A.To ask him to teach him French. B.To borrow some books. C.To ask for advice. 听下面一段对话，回答四个小题。 11.Where was the man when he got the phone call? A.In a bus.
湖北省宜昌市第二中学2021-2022高二数学10月月考试题
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8.中国古代数学著作算法统宗中有这样一个问题：“三百七十八里关，初行健步不为难， 次日脚痛减一半，六朝才得到其关，要见次日行数里，请公仔细算相还”其意思为：“有一个人走378里路，第一天健步行走，从第二天起脚痛每天走的路程为前一天的一半，走了6天后到达目的地”，请问从第几天开始，走的路程少于30里( ) A. 3 B. 4 C. 5 D. 6 9.“”是“直线与直线相互垂直”的 ( ) A. 充分必要条件 B. 充分而不必要条件 C. 必要而不充分条件 D. 既不充分也不必要条件 10.已知等差数列满足，则n的值为( ) A. 8 B. 9 C. 10 D. 11 11.已知等比数列中的各项都是正数，且成等差数列，则 A. B. C. D. 12.意大利数学家列昂那多斐波那契以兔子繁殖为例，引入“兔子数列”：1，1，2，3，5， 8，13，21，34，55，，即若此数列被2整除后的余数构成一个新数列，则数列的前2021项的和为 A. 672 B. 673 C. 1346 D. 2021 二、填空题（本大题共4小题，共20.0分） 13.等差数列的前n项和分别为，且，则______ ． 14.已知三个数，1，成等差数列；又三个数，1，成等比数列，则值为______．
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