南师附中2016~2017学年度上学期高一期末分班考试

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湖南师范大学附属中学2016-2017学年高一上学期期末考试英语试题

湖南师范大学附属中学2016-2017学年高一上学期期末考试英语试题

湖南师大附中2016-2017学年度高一第一学期期末考试英语试题-(这是边文,请据需要手工删加)(这是边文,请据需要手工删加)湖南师大附中2016-2017学年度高一第一学期期末考试英语时量:120分钟满分:150分得分____________第一部分听力(共两节,满分30分)做题时,先将答案标在试卷上。

录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节,满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

例:How much is the shirt?A. £19.15.B. £9.18.C. £9.15.答案是C。

1. When will the party be held?A. Next Wednesday.B. Next Thursday.C. Next Friday.2. What is the man going to do?A. Have an interview.B. Have dinner with his friend.C. Have a party.3. What does the woman want to borrow from the man?A. A computer.B. A bike.C. A car.4. Which subject is difficult for the boy?A. Math.B. English.C. Physics.5. Where are the speakers?A. In a hotel.B. In a shop.C. In a restaurant.第二节,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。

南师附中新高一分班测试模拟物理练习一(无答案)

南师附中新高一分班测试模拟物理练习一(无答案)

南师附中新高一分班测试模拟物理练习一(无答案)绝密★启用前南师附中分班考试模拟卷(二)物理本试卷分第Ⅰ卷(选择+填空)和第Ⅱ卷(解答题)两部分,共100分,考试用时100分钟。

第I卷(选择+填空共50分)一、选择题(本大题共12小题,每小题3分,共36分)。

1.(3分)一氢气球下系一小重物,重物只在重力和绳的拉力作用下做匀速直线运动,不计空气阻力和风力的影响,重物匀速运动的方向如图中箭头所示的虚线方向,图中气球和重物G在运动中所处的位置正确的是()A .B .C .D .2.(3分)雨滴从高空由静止开始下落,下落过程中空气对雨滴的阻力随雨滴的速度增大而增大,下列图象中可能正确反映雨滴下落运动情况的是()A .B .C .D .3.(3分)如图所示,演员正在进行杂技表演.由图可估算出他将一只鸡蛋抛出的过程中对鸡蛋所做的功最接近于()A.0.3J B.3J C.30J D.0J 4.(3分)有一架飞机沿水平方向向左做匀速直线运动,每隔1秒钟从飞机上轻轻释放一小球,若不考虑空气阻力和风力的影响,当三个小球落至地面前,站在地面上的人看到这三个小球在空中的排列情况应是下图中的哪一个()A .B .C .D .5.(3分)如图所示,甲、乙两位小朋友站在水平地面上做手拉手比力气游戏,结果甲把乙拉了过来,对这个过程中作用于双方的力的关系,下列说法正确的是()A.甲拉乙的力一定比乙拉甲的力大B.甲对乙的拉力与乙对甲的拉力是一对平衡力C.甲拉乙的力和乙拉甲的力一定大小相等D.只有在甲把乙拉动的过程中,甲的力才比乙的力大,在可能出现的短暂相持过程中,两个力才一样大6.(3分)如图所示,电源电压不变,滑动变阻器R1最大阻值为20Ω,灯丝电阻R L=8Ω,R2<R L,当滑动变阻器滑片P置于b端时,小灯泡的实际功率为2W,则滑片P在a端时,小灯泡的实际功率可能是()A.10W B.15W C.25W D.30W7.(3分)如图所示,一个木块A放在长木板B上,弹簧秤一端接A,另一端固定在墙壁上,长木板B放在水平地面上,在恒力F作用下,长木板B以速度υ匀速运动,水平弹簧秤的示数为T,下列关于摩擦力的说法正确的是()A.木块受到的摩擦力大小等于FB.长木板受到的摩擦力大小等于TC.若长木板以2υ的速度运动时,长木板受到的摩擦力大小等于2FD.若用2F的力作用在长木板上,木块受到的摩擦力大小仍等于T8.(3分)如图所示,木块m放在木板AB上,在木板的A端用一个竖直向上的力F使木板绕B 端逆时针缓慢转动(B端不滑动).在此过程中,m与AB保持相对静止,则()A.木块m对木板AB的压力增大B.木块m受到的静摩擦力逐渐减小C.竖直向上的拉力F保持不变D.拉力F与其力臂的积逐渐增大9.(3分)同一木块甲,先后两次分别在物体乙和丙的作用下,都恰能停留在水面下,如图所示,则下面说法错误的是()A.两种情况下,甲受的浮力一定相等B.乙的质量一定比丙的质量大C.乙的质量一定比甲的质量小D.乙的密度一定比甲的密度大10.(3分)某实物投影机有10个相同的强光灯L1~L10(24V/200W)和10个相同的指示灯X1~X10(220V/2W),将其连接在220V交流电源上,电路见图,若工作一段时间后,L2灯丝烧断,则()A.X1功率减小,L1功率增大B.X1功率增大,L1功率增大C.X2功率增大,其它指示灯的功率减小D.X2功率减小,其它指示灯的功率增大11.(3分)在凸透镜前有一个物体,经凸透镜折射后在光屏上能成一个缩小的像.若物体以速度υ沿垂直于主光轴的方向缩短(物长不为零),则()A.物体在缩短的过程中,像也以υ的速度缩短B.物体在缩短的过程中,像以小于υ的速度缩短C.物体在缩短的过程中,像以大于υ的速度长高D.物体在缩短的过程中,像以小于υ的速度长高12.(3分)如图所示电路,T (图形符号:)为稳压管,这个稳压管相当于一个特殊的电阻,它的特点是:当通过它的电流强度I1=0.1安到0.5安之间的任意值时,它两端的电压值就恒为10伏,为了保证在将变阻器R的有效阻值由20欧连续变化到50欧过程中,变阻器R两端的电压恒为10伏,则电路总的输入电流I的值应()A.控制在0.1安到0.5安之间 B.控制在0.6安到0.7安之间C.控制在0.3安到1.0安之间 D.控制在0.6安到1.0安之间南师附中新高一分班测试模拟物理练习一(无答案)二、填空题(本大题共5小题,每小题2分,共14分)。

南京市2016-2017学年度第一学期期末检测卷(高一数学)参考答案终稿

南京市2016-2017学年度第一学期期末检测卷(高一数学)参考答案终稿

南京市2016-2017学年度第一学期期末检测卷高一数学参考答案及评分标准 2017.01说明:1.本解答给出的解法供参考.如果考生的解法与本解答不同,可根据试题的主要考查内容比照评分标准制订相应的评分细则.2.对计算题,当考生的解答在某一步出现错误时,如果后续部分的解答未改变该题的内容和难度,可视影响的程度决定给分,但不得超过该部分正确解答应得分数的一半;如果后续部分的解答有较严重的错误,就不再给分.3.解答右端所注分数,表示考生正确做到这一步应得的累加分数.4.只给整数分数,填空题不给中间分数.一、填空题:本大题共14小题,每小题5分,共70分.1.{0,1,2} 2.(-∞,1) 3.2π3 4.-513 5.126.9 7.-148.5 9.c <a <b 10.1 11.3 12.4 13.(0,13)∪(3,+∞) 14.(0,14) 二、解答题:本大题共6小题,共90分.15. 解:(1)因为sin α+cos αsin α-2cos α=2,化简得sin α=5cos α. ……………………………2分 当cos α=0时不符合题意,所以cos α≠0,所以tan α=5. ………………………………………………6分(2)cos(π2-α)·cos(-π+α)=-sin αcos α ……………………………8分 =-sin α·cos αsin 2α+cos 2α=-tan αtan 2α+1…………………………………………12分 =- 526. ……………………………………………14分 16.解:(1)因为a =(-2,1),b =(3,-4),所以a +b =(1,-3),2a -b =(-7,6), ……………………4分所以(a +b )·(2a -b )=1×(-7)+(-3)×6=-25. ……………………6分(2)由(1)可知a +b =(1,-3),且a =(-2,1),所以|a |=5,|a +b |=10,a ·(a +b )=-5. ……………………9分设向量a 与a +b 的夹角为θ,则cos θ=a ·(a +b )|a |·|a +b |=-22. ……………………11分 因为θ∈[0,π],所以θ=3π4,即向量a 与a +b 的夹角为3π4. ……………………14分 17.解:(1)依题意,y =x (a -2x )(2a -2x ),x ∈(0,1]. ………………………………4分(2)y =V (x )x=(a -2x )(2a -2x ) …………………………………6分 =4x 2-6ax +2a 2.因为对称轴x =34a ,且a >2 ,所以x =34a >32>1, …………………………8分 所以当x =1,y min =4-6a +2a 2. ………………………12分答:当x =1时,y 最小,最小值为4-6a +2a 2. …………………………14分18. 解:(1)由T =2πω,得2πω=π,所以ω=2. 因为点P (π6,2)是该函数图象的一个最高点,且A >0,所以A =2.…………2分 此时f (x )=2sin(2x +φ).又将点P (π6,2)的坐标代入f (x )=2sin(2x +φ), 得2sin(π3+φ)=2,即sin(π3+φ)=1, 所以π3+φ=2k π+π2,k ∈Z ,即φ=2k π+π6,k ∈Z . ………………………4分 又因为|φ|<π2,所以φ=π6. 综上,f (x )=2sin(2x +π6). ………………………6分 (2) 因为x ∈[-π2,0],所以2x +π6∈[-5π6,π6], ………………………8分 所以sin(2x +π6)∈[-1,12],即2sin(2x +π6)∈[-2,1], 所以函数y =f (x )的值域为[-2,1]. ………………………10分(3)y =g (x )=2sin[2(x -θ)+π6]=2sin(2x -2θ+π6). ………………………12分 因为0≤x ≤π4,所以π6-2θ≤2x -2θ+π6≤2π3-2θ, 所以⎩⎨⎧π6-2θ≥2k π-π2,2π3-2θ≤2k π+π2,k ∈Z , 解得-k π+π12≤θ≤-k π+π3,k ∈Z . ………………………14分 因为0<θ<π2,所以k =0,所以π12≤θ≤π3. ………………………16分 19.解:(1)因为AB →=CB →-CA →, ………………………2分所以AB →2=(CB →-CA →)2=CB →2-2CB →·CA →+CA →2=22-2×2×1×12+12=3, 所以|AB →|=3. ………………………4分(2)解法1:①当λ=12时,AE →=12CB →-CA →,CD →=12(CB →+CA →). ……………………6分 所以AE →·CD →=(12CB →-CA →)·12(CB →+CA →)=12×(12CB →2-12CB →·CA →-CA →2) =12×(12×22-12×2×1×12-12)=14. …………………8分 ②假设存在非零实数λ,使得AE →⊥CD →.因为BE →=λBC →,所以AE →=CE →-CA →=(1-λ)CB →-CA →. …………………10分因为AD →=λAB →,所以CD →=CA →+AD →=CA →+λAB →=CA →+λ(CB →-CA →)=λCB →+(1-λ)CA →. ……………………12分所以AE →·CD →=[(1-λ)CB →-CA →]·[λCB →+(1-λ)CA →]=λ(1-λ)CB →2+(λ2-3λ+1)CB →·CA →-(1-λ)CA →2=λ(1-λ)×22+(λ2-3λ+1)×2×1×12-(1-λ)×12 =-3λ2+2λ=0. ………………………14分解得λ=23或λ=0. 因为点在三角形的边上,所以λ∈[0,1],故存在非零实数λ=23,使得AE →⊥CD →. ………………………16分 解法2:由(1)得CA =1,CB =2,AB =3,满足CB 2=AB 2+CA 2, 所以∠CAB =90︒.如图,以A 原点,AB 边所在直线为x 轴,AC 边所在的直线为y 轴,建立平面直角坐标系,则A (0,0),B (3,0),C (0,1). ……………6分 ①当λ=12时,AE →=(32,12),CD →=(32,-1), 则AE →·CD →=14. ………………………10分 ②假设存在非零实数λ,使得AE →⊥CD →.因为AE →=(3(1-λ), λ),CD →=(3λ,-1),所以AE →·CD →=-3λ2+2λ=0, ………………………14分解得λ=0或λ=23. 因为点在三角形的边上,所以λ∈[0,1],所以存在非零实数λ=23,使得AE →⊥CD →. ………………………16分 20.解:(1)F (x )=f (x )-g (x )=x -a -a |x |.①当a =12时,由F (x )=0,得x -12-12|x |=0. 当x ≥0时,x -12-12x =0,解得x =1,满足条件. 当x <0时,x -12+12x =0,解得x =13,不满足条件. 综上,函数y =F (x )的零点是1. ………………………2分②F (x )=0,则x -a -a |x |=0,即a (1+|x |)=x .因为1+|x |≠0,所以a =x 1+|x |. ………………………4分 设φ(x )=x 1+|x |, 当x >0时,φ(x )=x 1+x =1-11+x,所以φ(x )∈(0,1). ………………………6分 因为φ(-x )=-φ(x ),所以φ(x )是奇函数,所以当x <0时,φ(x )∈(-1,0).又因为φ(0)=0,所以当x ∈R ,φ(x )∈(-1,1),所以a ∈(-1,1). ………………………8分(2)设函数h (x )的最大值和最小值分别是M ,N .因为对任意x 1,x 2∈[-2,2],| h (x 1)-h (x 2)|≤6成立,所以M -N ≤6. ………………………10分解法1:因为h (x )=f (x )+g (x )=x -a +a |x |,x ∈[-2,2],所以h (x )=x -a +a |x |=⎩⎨⎧(a +1)x -a ,x ≥0,(1-a )x -a ,x <0.①当a >1时,因为a +1>0,所以h (x )在(0,+∞)单调增;因为1-a <0,所以h (x )在(-∞,0)单调减.因为h (2)=a +2,h (-2)=a -2,所以h (2)>h (-2),所以M =h (x )max =h (2)=a +2,N =h (x )min =h (0)=-a ,所以a +2-(-a )≤6,解得a ≤2.又因为a >1,所以1<a ≤2. ………………………12分②当a =1时,h (x )=⎩⎨⎧2x -1,x ≥0,-1, x <0,所以M =h (x )max =h (2)=3,N =h (x )min =-1,所以3-(-1)≤6恒成立,所以 a =1符合题意.③当-1<a <1时,因为a +1>0,所以h (x )在(0,+∞)单调增;因为1-a >0,所以h (x )在(-∞,0)单调增.所以M =h (x )max =h (2)=a +2,N =h (x )min =h (-2)=a -2,所以(a +2)-(a -2)=4≤6恒成立,所以-1<a <1符合题意.④当a =-1时,h (x )=⎩⎨⎧1, x ≥0,2x +1,x <0,所以M =h (x )max =1,N =h (x )min =h (-2)=-3,所以1-(-3) =4≤6恒成立,所以a =-1符合题意. ……………………14分⑤当a <-1时,因为a +1<0,所以h (x )在(0,+∞)单调减;因为1-a >0,所以h (x )在(-∞,0)单调增.所以M =h (x )max =h (0)=-a ,因为h (2)=a +2,h (-2)=a -2,所以h (2)>h (-2) ,所以N =h (x )min =h (-2)=a -2,所以-a -(a -2)≤6,解得a ≥-2.又因为a <-1,所以-2≤a <-1.综上,a 的取值范围为[-2,2]. ……………………16分解法2:因为h (x )=f (x )+g (x )=x -a +a |x |,x ∈[-2,2],所以h (x )=x -a +a |x |=⎩⎨⎧(a +1)x -a ,x ≥0,(1-a )x -a ,x <0.可知函数的图象是由两条折线段构成.所以函数的M 和N 分别为h (-2)=-2+a ,h (0)=-a ,h (2)=2+a 三个值当中的两个. 显然2+a >-2+a .当a ≤-1时,2+a ≤-a ;当a >-1时,2+a >-a .当a ≤1时,-2+a ≤-a ;当a >1时,-2+a >-a .所以,①当a >1时,M =2+a ,N =-a ,M -N =2+2a ,因为M -N ≤6,所以a ≤2.又因为a >1,所以1<a ≤2. …………………12分②当-1<a ≤1时,M =2+a ,N =-2+a ,M -N =4.因为M -N ≤6恒成立,所以-1<a ≤1满足条件. …………………14分③当a ≤-1时,M =-a ,N =-2+a ,M -N =2-2a .因为M -N ≤6,所以a ≥-2.又因为a ≤-1,所以-2≤a ≤-1.综上,a 的取值范围为[-2,2]. ………………………16分解法3:因为h (x )=f (x )+g (x )=x -a +a |x |,x ∈[-2,2],所以h (x )=x -a +a |x |=⎩⎨⎧(a +1)x -a ,x ≥0,(1-a )x -a ,x <0.①当0≤x≤2,h(x)=(1+a)x-a.若a>-1,则1+a>0,所以h(x)=(1+a)x-a是增函数.所以h(x)max=h(2)=2+a,h(x)min=h(0)=-a.若a<-1,则1+a<0,所以h(x)=(1+a)x-a是减函数.所以h(x)max=h(0)=-a,h(x)min=h(2)=2+a.若a=-1,h(x)=1,所以h(x)max=h(x)min=1.②当-2≤x<0,h(x)=(1-a)x-a.若a<1,则1-a>0,所以h(x)=(1-a)x-a是增函数.所以h(x)<h(0)=-a,h(x)min=h(-2)=-2+a.若a>1,则1-a<0,所以h(x)=(1-a)x-a是减函数.所以h(x)max=h(-2)=-2+a,h(x)>h(0)=-a.若a=1,h(x)=-1,所以h(x)max=h(x)min=-1.………………12分显然2+a>-2+a.因为当a≤-1时,2+a≤-a;当a>-1时,2+a>-a;当a≤1时,-2+a≤-a;当a>1时,-2+a>-a.………………………14分所以,(Ⅰ)当a>1时,M=2+a,N=-a,M-N=2+2a.因为M-N≤6,所以a≤2.又因为a>1,所以1<a≤2.(Ⅱ)当-1<a≤1时,M=2+a,N=-2+a,M-N=4.因为M-N≤6恒成立,所以-1<a≤1满足条件.(Ⅲ)当a≤-1时,M=-a,N=-2+a,M-N=2-2a.因为M-N≤6,所以a≥-2.又因为a≤-1,所以-2≤a≤-1.综上,a的取值范围为[-2,2].………………………16分。

高一分班测试(英语)含答案

高一分班测试(英语)含答案

绝密★启用前南京师范大学附属中学高一分班测试模拟练习一(英语)本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,共100分,考试用时60分钟。

第I卷(选择题共80分)一、单项选择(共20题,每小题1分,满分20分)1. —Who is your English teacher?—Mr. Brown, __________ European working in __________ Shandong University.A. an; theB. a; theC. an; /D. a; /2. I had hoped to visit his family last night, __________ the traffic was quite heavy.A. andB. orC. butD. so3. Being hard-working is a kind of quality and that's _________ it takes to do anything well.A. howB. whichC. thatD. what4. — Let's hurry. Professor Beach is coming.— Oh, I was afraid that we ___________.A. already miss himB. will miss himC. have already missed himD. had already missed him5. My mother always says to me: “__________ time flies! You have grown up.”A. howB. whatC. asD. when6. The old lady was the only person that died in the fire. __________else was saved by the firemen.A. EveryoneB. AnyoneC. SomeoneD. No one7. —What do you think of the concert?—I really enjoy it. I didn't expect it was ___________ wonderful.A. mostB. moreC. asD. very8. As a teacher, Mr. Black __________ all himself ___________ his students.A. devote; to teachB. devotes; to teachingC. devoted; to teachD. devote; to teaching 9. No one in the factory but Tom and I __________ that the doorkeeper lives here.A. knowsB. knowC. have knownD. am to know10. He paid the boy 10 dollars for washing ten windows, most of __________ had not been cleaned for years.A. themB. thoseC. thatD. which11. He served in the factory in __________ when he was in___________.A. 1940’s; his twentiesB. the 1940’s; the twentiesC. the 1940s’; the twentiesD. the 1940’s; his twenties12. The mother hurried into Jack’s room and found him __________ at the desk.A. seatedB. seatingC. was seatingD. to be seating13. Scientists say it may be five or six years ___________ it is possible to test the medicine on people.A. sinceB. afterC. beforeD. when14. I’m so sorry, but it is ___________ my ability to work out the problem.A. overB. aboveC. outD. beyond15. —I’d like to take a week’s holiday.—___________. We are too busy.A. Don’t worryB. Don’t mention itC. Forget itD. Good luck16. — Shall I tell Sally about it?— No, you ___________. I’ve told her already.A. shouldn’tB. mustn’tC. needn’tD. ca n’t17. —Do you know ___________ the MP3 player last week?—Sorry, I have no idea.A. how much did she pay forB. how much she paid forC. how much will she pay forD. how much she will pay for18. There was a rainstorm yesterday. The flood ___________ the old bridge over the small river.A. washed awayB. went awayC. kept awayD. put away19. I have no one ___________ me, for I am a new comer here.A. helpedB. helpingC. to helpD. to have helped20. —Have you read the news that a ninety-year-old man will attend the national entrance exam (高考) once again.—He is a great man, I think. You know, ___________.A. all roads lead to RomeB. it’s never too late to learnC. one cannot be in two placesD. a light heart lives long二、完形填空(共30题,每小题1分,满分30分)AMost of what I need to know about how to live and how to be, I learned in Kindergarten (幼儿园). These are the things I learned: 21 everything. Play fair. Don’t hit people. Clean up your own mess. Put things back 22 you found them. Don’t take things that aren’t yours. 23 you’re sorry when you hurt somebody. Wash your hands before you eat. Flush. Warm cookies and cold milk are 24 for you. Learn some and think some and draw and paint and sing and dance and play and work every day 25 .Take a nap(午睡) every afternoon. When you go out into the world, 26 traffic, hold hands and stick together. The root goes 27 and the plant goes up and 28 really knows how or why, but we are all like that.And then 29 the book about Dick and Jane and the first 30 you learned, the biggest word of all:Look. Everything you 31 is in there somewhere.32 what a better world it would be if we all—the whole world—had cookies and milk about 3 o’clock every afternoon and then 33 down with our blankets for a nap. 34 if we had a basic policy in our countries to always put things back where you found them and cleaned up our own messes. And it is still true. No matter how 35 you are, when you go out into the world, it is better to hold hands and stick together.21. A. Master B. Make C. Take D. Share22. A. which B. that C. where D. when23. A. Talk B. Say C. Speak D. Tell24. A. fit B. convenient C. good D. comfortable25. A. some B. any C. all D. little26. A. care for B. make for C. watch for D. ask for27. A. away B. down C. off D. out28. A. somebody B. everybody C. nobody D. anybody29. A. read B. write C. collect D. remember30. A. word B. rule C. order D. song31. A. can know B. may know C. dare know D. need to know32. A. Draw B. Feel C. Imagine D. Design33. A. fell B. lay C. put D. set34. A. Or B. But C. Otherwise D. So35. A. wise B. old C. energetic D. goodBAs we all know, English is nowadays more and more important and popular around the world. But 36 English is a language with so many confusing rules may be a question to many people. So let’s look back upon the 37 of it, which may be divided into three part s——Old English, Middle English and Modern English.As for Old English, three Germanic groups from the 38 mainland——the Angles, the Saxons and the Jutes——invaded (侵略) Britain. They 39 had their own language and Angles-Saxon developed from 40 these languages with Celtic which 41 before the middle of the 5th century. At the end of the 9th century, the Vikings began to invade Britain, and brought 42 them their languages, which also mixed with Angle-Saxon and 43 what we now call Old English. This language different from the English we speak nowadays 44 made up of an Angle-Saxon base plus words from the language of Denmark and Norway.Middle English is the name 45 to the English used from around the 12th and 16th century. Many factors 46 the development of this new type of English. French had an 47 on the English language, which created even more pairs of similar words like “reply” and “answer”.Modern English began during the Renaissance(文艺复兴)in the 16th century. 48 this, Modern English includes many Latin and Greek words. Pronunciation also experienced huge changes during this period The question of 49 English will continue changing in the future is easy to answer. Even today, if a person from the south of England goes to the north, he or she may find it difficult 50 what people there are saying because English is still changing.36. A. how B. when C. where D. why37. A. history B. time C. plan D. century38. A. European B. Asian C. African D. Australian39. A. either B. each C. every D. none40. A. mixture B. mixed C. mixing D. mix41. A. was talked B. was said C. was told D. was spoken42. A. in B. with C. about D. on43. A. caused B. created C. organized D. realized44. A. is B. has been C. is being D. was45. A. giving B. gave C. given D. give46. A. led to B. devoted to C. got to D. gave to47. A. affect B. relation C. effect D. effort48. A. Because of B. Because C. As D. Since49. A. what B. whether C. that D. when50. A. understanding B. understand C. to understand D. understoodIII. 阅读理解(共15题,每小题2分,满分30分)AMost people may not think that there are any problems with watermelons, but some Japanese sellers do. There is often wasted space when they store watermelons, because round watermelons do not sit nicely on shelves. So some clever farmers have solved the watermelon problem. They make their watermelons grow in square glass boxes so the watermelons become square. The only problem now is the price. The square watermelons cost about 80 dollars each.Many people know the saying “Roses are red, violets are blue, sugar is sweet and so are you”. Well, today the poem needs to change to “Roses are red, roses are blue”! Australian scientists have just recently discovered that when they put a chemical from a blue flower into a rose, they can create a nearly 100% blue rose. Some of science’s inventions are beautiful, not just useful!Since the mid-1990s, many Chinese farmers have had amazing rice harvests. For many years Chinese scientists have been working on developing new kinds of stronger rice and now they have produced a new kind, called super rice. Super rice not only can protect itself against insects and diseases, but it also produces more than any other kind of rice. Today farmers can grow almost one and a half times more rice than they could grow in the past.Doctors might be able to use vegetables to fight cancer now. US scientists have discovered a way to put a chemical into tomato plants while they are growing. When cancer patients eat the tomatoes, the tomatoes will fight the cancer in their bodies. These cancer-fighting vegetables are not yet for sale, because scientists have not finished testing them. If study finds no problems with the tomatoes, then patients can begin to use this easy and delicious new treatment.51. The square watermelons look like ____________.A. C. D.52. Which of the following is true?A. Super rice can keep itself from insects.B. Blue roses are more beautiful than red ones.C. The tomatoes can fight many kinds of diseases.D. The square watermelons cost less than ordinary ones.53. The main idea of this passage is ____________. A. how scientists discover new plantsB. how science makes plants produce moreC. how science has changed plants and people’s livesD. how technology makes our world more beautifulBFor most of us, Hawaii begins to show her charm little by little. A friend describes a sunset of Waikiki. We hear the twang (弹拨) of a guitar. We see golden beaches and golden people; we see sun, sand, sea, and surf…or somewhere between the blue skies and the palm trees. We’re completely attracted.The Hawaiian Islands are one of the most beautiful places on the earth. The weather is friendly. The temperature changes from 60-90 degrees all year long. It’s a little warmer in summer, and a little cooler in winter, but every day is a beach day for somebody.The environment is friendly. The beauty of Hawaii is almost unparalleled. Amazing mountains were created millions of years ago by volcanic activity that pushed these islands three miles away from the ocean floor. Wave action created coral reefs (珊瑚礁) and then broke them to create miles of white sand beach. The position (地理位置) at the center of the Pacific Ocean made sure that almost every plant and animal that would find a home here would come as an invited visitor.There are no strangers in Paradise. Perhaps the most beautiful part of Hawaii is the sincere warmth of our people. We call it the spirit of Aloha. It has allowed people from all over the world to find common ground, and a new home, in this most gentle place.54. What is the best title for the passage?A. The beauty of Hawaii.B. Friendly people in Hawaii.C. Hawaii —our paradise.D. Everyday is a beach day.55. The underlined word “unparalleled” probably means “________’’.A. unmatchedB. unnoticedC. uncomfortableD. unimportant56. What is implied (暗示) but not described is that________.A. the people in Hawaii are very friendlyB. the writer’s friend has been to HawaiiC. the weather is very pleasant hereD. Hawaii is at the center of Pacific57. The underlined sentence means________.A. the strangers aren’t allowed to come hereB. people in Hawaii know each other very wellC. Hawaii makes people feel at homeD. the strangers find it hard to live in HawaiiCAn American exploration company, the Nauticos Corporation, has found the wreck(残骸) of an ancient (古代的)Greek ship in a very deep area of the Mediterranean Sea. It may be the deepest ancient shipwreck ever found. The discovery questions a long-held idea that ancient sailors didn’t have skills needed to guide ships in open seas.The ancient ship lay more than 3000 meters below the surface of the Mediterranean. The wreck was more than 489 kilometers from the Island of Corsica. Scientists believe the ship is about 2300 years old.The company was looking for an Israeli submarine(潜水艇) that had disappeared more than thirty years ago. Nauticos official Thomas Dettweiler said two company ships were searching for the Israeli submarine. He supervised(监督)the operation from a ship with sonar equipment. A sonar system uses sound waves to find and identify objects under the water. A second ship carried a vehicle for deep-sea searches.During the search, one ship’s sonar began making strange sounds. Mr. Dettweiler thought this meant they had found the lost submarine. He ordered the other ship to send down the deep-sea vehicle to search.But a video camera on the vehicle did not show a submarine. Instead, it showed many large clay containers lying on the bottom of the sea. These containers are called amphoras(古罗马和希腊的两耳细颈酒罐). Mr. Dettweiler was excited. He says he knew that the 2000 amphoras they had found meant they had discovered an important shipwreck. In ancient times such containers held wine, olive oil and other goods.Nauticos and the scientists hope to do more research in the area soon. Mr. Dettweiler believes four other ancient ships may be nearby. He says finding shipwrecks from different time periods could be especially important. It could be the first evidence of continued open-sea trading in the ancient world.58. Which of the following is NOT true?A. It is still believed that ancient sailors didn’t know how to guide ships in open sea.B. The ancient ship was discovered by an American exploration company.C. The ancient ship was found by chance.D. Four other ancient ships may be nearby.59. A sonar system __________.A. helps people to find any objectB. can only be used to look for shipsC. is a technology related to sea wavesD. can only be used under the water60. How did they know they had found an ancient shipwreck?A. The sonar system began to make sounds.B. The video camera showed the picture of a wrecked ship.C. They found amphoras.D. They found wine, olive oil and other products under the sea.61. Mr. Dettweiler hopes to find more ancient ships because _____.A. the ancient ships are usefulB. these ancient ships show the importance of open-sea tradingC. they are important in the Greek historyD. they may prove there was open-sea trading in ancient times.DLooking back on my childhood, I am sure that naturalists are born and not made. Although we were all brought up in the same way, my brothers and sisters soon left their pressed flowers and insects. Unlike them, I had no ear for music and languages. I was not an early reader and I could not do mental arithmetic (心算).Before World War I we spent our summer holidays in Hungary. I don’t have a clear memory of the house we lived in, of my room and my toys. Nor do I recall clearly the large family of grandparents, aunts, uncles and cousins who gathered next door. But I do have a clear memory of the dogs, the farm animals, the local birds, and above all, the insects.I am a naturalist, not a scientist. I have a strong love of the natural world and my love had led me into different observations (观察). I love discussing my favorite topics and enjoy burning the midnight oil reading about other people’s observations and discoveries. Then something happens that brings these observations together in my mind. Suddenly you think you see the answer to the riddle, because it all seems to fit together. This has resulted in my publishing 300 papers and books, which some might honor with the title of scientific research.But curiosity, keen eyes, a good memory and enjoyment of the animal and plant world do not make a scientist: one of the excellent and essential qualities required is self-discipline (自律), a quality I don’t have. A scientist, up to a certain point, can be made. A naturalist is born. If you can combine the two, you get the best of both worlds.62. The first paragraph tells us the writer ______.A. was interested in flowers and insects in his childhoodB. lost his hearing when he was a childC. didn’t like his brothers and sistersD. was born into a naturalist’s family63. The writer can’t remember his relatives clearly because _______.A. he didn’t live very long with themB. the family was too largeC. he was too young when he lived with themD. he was fully busy observing nature64. It can be implied from the passage that the writer was _______.A. a scientist as well as a naturalistB. a naturalist but not a scientistC. no more than a born naturalistD. first of all a scientist65. The writer says that he is a naturalist rather than a scientist probably because he thinks he _______.A. has a great deal of trouble doing mental arithmeticB. doesn’t have some of the qualities required of a scientistC. just reads about other people’s observations and discoveriesD. thinks of how to solve problems in a most natural way第II卷非选择题(共20分)IV. 单词拼写(共10题,每小题0.5分,满分5分)根据所给汉语或首字母写出相应的单词,注意形式变化,每空限填一词66. Just as the old saying goes “don’t judge a book by its cover”, we shouldn’t judge a person by his/ her a__________.67. The President of the United States p____________ the heroes in World War II with medals.68. The ___________ (惊讶)look on his face showed that he knew nothing about the birthday party for him.69. Panda and other endangered animals need p__________ by all the people in China.70. More Hope Schools have been built with the money d_________ by people across the country.71. Though there were four tickets ___________ (可获得的), only two were free.72. It is said that fighting AIDS not only needs the g___________ efforts but also all the Chinese people’s help.73. As we all know, his s__________ on how to improve the living quality of students have been accepted by the headmaster.74. Three hours had passed but the speaker still didn’t come. The students became i__________.75. The victims were sent to hospital i__________ after the accident happened.V. 根据汉语完成句子(共10空,每空0.5分,满分5分)注意形式变化,每空限填一词76. 经过激烈的讨论,每个小组都不怎么费力地提出了问题的解决办法。

江苏省南京师范大学附属中学2016-学年高一上学期期中考试数学试题word版含答案.doc

江苏省南京师范大学附属中学2016-学年高一上学期期中考试数学试题word版含答案.doc

江苏省南京师范大学附属中学2016-2017学年高一上学期期中考试数学试题W o r d版含答案.d o cwork Information Technology Company.2020YEAR南京师大附中2016-2017学年度第1学期高一年级期中考试数学试卷一、填空题1.设集合A = {1,2,3} , B = {2,3,5},则 A U B = .2.函数11)(-=x x f 的定义域是 .3.若函数x a x f )1()(-=在(-∞,+∞)上单调递增,则实数a 的取值范围是 .4.若幂函数)(x f y =的图象过点(4,2),则)16(f 的值是. 5.若3log ,4,3log 5.0232===c b a ,则将a,b,c 按从小到大的的顺序排列是 .6.己知)(x f y =是定义在R 上的偶函数,若x≥0时,1)(-=x x f ,则x < 0时,)(x f = .7.若函数32)(+=x x f ,函数))27((,)(31g f x x g =的值是 .8.已知函数⎩⎨⎧-≤=1>,1,)(x x x e x f x ,若2)(=x f ,则x 的值是. 9.已知函数b a x f x +=)( (a > 0,a≠1)的图象如图所示,则a b 的值是 .10.若集合A = [-2,2],B = (a, +∞),A ∩B = A ,则实数a 的取值范围是 .11.函数12141)(+⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=X X x f 在[-3,2]的最大值是 .12.若二次函数)(x f 满足)()0(<1),2()2(a f f f x f x f ≤-=+)(且,则实数a 的取值范围是. 13.已知函数X X x f --=22)(,若对任意的x ∈[1,3],不等式0>)4()(2x f tx x f -++恒成立,则实数t 的取值范围是 . 14.已知函数)(||12)(R x x x x f ∈+-=,区间M = [a,b](a < b),集合N = {y|y = f (x),x ∈M}.若M = N ,则b-a 的值是 .二、解答题15.(本题满分8分)己知全集 U = R ,集合 A = {x|3 ≤ x < 7}, B = {x | 2 < log 2 x < 4}.(1)求A U B ;(2)求(CuA )∩B.16.(本题满分8分)计算:(1);)827()3()32(3202--+- (2)3log 34222log 3log -⨯17.(本题满分10分)某旅游景区的景点A 处和B 处之间有两种到达方式,一种是沿直线步行,另一种是沿索道乘坐缆车, 现有一名游客从A 处出发,以50m/min 的速度匀速步行,30min 后到达B 处,在B 处停留20min 后, 再乘坐缆车回到A 处.假设缆车匀速直线运动的速度为150m/mm . ⑴求该游客离景点A 的距离y(m)关于出发后的时间x(mm)的函数解析式,并指出该函数的定义域;⑵做出(1)中函数的图象,并求该游客离景点A 的距离不小于1000m 的总时长.18.(本题满分10分)己知 a > 0 且 a ≠1,若函数)1(log )(-=x x f a ,)5(log )(x x g a -=.(1)求函数)()()(x g x f x h -=的定义域;(2)讨论不等式)()(x g x f ≥成立时x 的取值范围.19.已知 a ∈ R ,函数 121)(+-=x a x f . (1)用函数单调性定义证明:)(x f 在(-∞,+∞)上单调递增;(2)若)(x f 为奇函数,求:①a 的值;②)(x f 的值域.20.(本题满分12分)对于两个定义域相同的函数)(x f 、)(x g ,若存在实数m,n ,使)()()(x ng x mf x h +=,则称函数)(x f 是由“基函数)(x f ,)(x g ”生成的.(1)若x x x f 3)(2+=和43)(+=x x g 生成一个偶函数)(x h ,求)2(h 的值;(2)若132)(2-+=x x x h 是由ax x x f +=2)(和b x x g +=)(生成,其中a,b ∈R 且a b≠0,求ba 的取值范围; (3)利用“基函数)14(log )(4+=x x f ,)1)(-=x x g ”生成一个函数)(x h ,使得)(x h 满足:①是偶函数,②有最小值1,求)(x h 的解析式.。

江苏省南京师范大学附属中学2015-2016学年高一上学期期中考试物理试题

江苏省南京师范大学附属中学2015-2016学年高一上学期期中考试物理试题

高一年级物理试卷一、单项选择题(每小题3分,共15分.每小题只有一个选项符合题意) 1.下列说法中正确的是( )A .研究奥运会冠军刘翔的跨栏技术时可将刘翔看作质点B .在某次铅球比赛中,某运动员以18.62米的成绩获得金牌,这里记录的成绩是比赛中铅球经过的路程C .瞬时速度可理解为时间趋于零时的平均速度D .“北京时间10点整”指的是时间,一节课40min 指的是时刻 2.关于摩擦力,下列说法正确的是( ) A .静摩擦力的方向总是跟物体运动趋势的方向相反 B .两物体间的弹力和摩擦力的方向一定互相垂直C .运动着的物体不可能受静摩擦力作用,只能受滑动摩擦力作用D .相互接触的物体间压力增大,摩擦力一定增大3.如图所示,在水平面上有一个质量为m 的小物块,在某时刻给它一个初速度,使其沿水平面做匀减速直线运动,其依次经过A 、B 、C 三点,最终停在O 点.A 、B 、C 三点到O 点的距离分别为1L 、2L 、3L ,小物块由A 、B 、C 三点运动到O 点所用的时间分别为1t 、2t 、3t .则下列结论正确的是( )A .312123==L L L t t t B .312222123==L L L t t t C .312123<<L L L t t tD .312222123<<L L L t t t 4.一物体作匀加速直线运动,通过一段位移∆x 所用的时间为1t ,紧接着通过下一段位移∆x所用时间为2t ,则物体运动的加速度为( ) A .()()1212122∆-+x t t t t t t B .()()121212∆-+x t t t t t t C .()()121212∆+-x t t t t t t D .()()121212∆+-x t t t t t t5.甲、乙两车在同一水平道路上,一前一后相距6m =s ,乙车在前,甲车在后,某时刻两车同时开始运动,两车运动的过程如图所示,则下列表述正确的是( )A .当4s =t 时两车相遇B .当4s =t 时两车间的距离最大C .两车有两次相遇D .两车有三次相遇二、多项选择题(4小题,共16分;每道题有两个或者两个以上的选项正确;每小题4分,选不全得2分,不选或者选错得0分) 6.关于重力和弹力,下列说法正确的是( )A .物体各部分都受重力作用,但可以认为物体各部分所受重力集中于一点B .任何几何形状规则的物体的重心必与其几何中心重合C .用一根竹竿拨动水中的木头,木头受到竹竿的弹力,是由于木头发生形变产生的D .挂在电线下的电灯受到向上的拉力,是因为电线发生微小形变产生的7.如图所示,汽车以10m/s 的速度匀速驶向路口,当行驶至距路口停车线20m 处时,还有3s绿灯就要熄灭.而该汽车在绿灯熄灭时刚好停在停车线处,则汽车运动的速度-时间图像可能是图中的( )DCBA8.如图所示,质量为m 的木块在水平拉力F 作用下向右滑行.木板处于静止状态,已知木块与木板间的动摩擦因数为1μ,木板与地面间的动摩擦因数为2μ,木板质量为M .则( )A .木块受到木板的摩擦力的大小为1mg μ,方向水平向左B .木板受到地面的摩擦力的大小为1mg μ,方向水平向左C .木板受到地面的摩擦力的大小为()2+m M g μD .无论怎样改变F 的大小,地板都不可能运动9.从地面竖直上抛一物体A ,同时在离地面某一高度处有另一物体B 自由落下,两物体在空中同时到达同一高度时速率都为v ,空气阻力均忽略不计,则下列说法中正确的是( )A .物体A 上抛的初速度和物体B 落地时速度的大小相等 B .物体A 、B 相遇时速度方向可能相同C .物体A 能上升的最大高度和B 开始下落的高度相同D .两物体在空中同时达到同一高度处是B 物体开始下落时高度的一半 三、实验题(10、11两小题,共18分)10.科学探究活动通常包括以下几个环节:A .提出问题 B .猜想与假设 C .制定计划与设计实验 D .进行实验与收集证据 E .分析与论证 F .评估 G .交流与合作等.一位同学在学习了滑动摩擦力之后,认为滑动摩擦力的大小可能与两物体接触面积的大小有关,于是他通过实验探究这个问题.⑴这位同学认为,滑动摩擦力的大小与两物体接触面积的大小成正比;这属于上述科学探究的( )环节(填环节前面的字母).⑵为完成本实验,需要自己选用器材其中他选用的木块应是下列选项中的( ) A .各面粗糙程度相同的正方体 B .各面粗糙程度不同的正方体C .各面粗糙程度相同,长、宽、高不同的长方体D .各面粗糙程度不同,长、宽、高不同的长方体⑶在本实验中,该同学设计了两种实验方案,下述两种方案中,你认为更合理、更易于操作的是()(填“A”或“B”)A.木板水平固定,通过弹簧秤水平拉动木块,如图甲所示B.木块与弹簧秤相连,弹簧秤水平固定,通过细绳水平拉动木板,如图乙所示11.在探究小车速度随时间变化规律实验中:⑴操作步骤中释放小车和接通电源的先后顺序应该是()A.先释放小车后接通电源B.先接通电源后释放小车⑵某同学获得一条点迹清楚的纸带,如图所示,已知打点计时器每隔0.02s打一个点,该同学选择了0、1、2、3、4、5、6七个计数点,每相邻两个计数点间还有四个点图中没有画出.测量数据如图所示,单位是mm.(a)根据(填公式)可判断小车做运动.(b)计算小车通过计数点3的瞬时速度为;小车通过计数点5的瞬时速度为;(保留3位有效数字)(c)小车的加速度为(保留2位有效数字)四、计算题(5小题,共51分;请写出必要的公式和计算步骤,否则不能得分)12.(9分)汽车以20m/s的速度在平直公路上匀速行驶,刹车后经4s速度变为4m/s,求:⑴刹车过程中的加速度;⑵刹车后2s时的速度;⑶刹车后6s通过的位移;13.(9分)某一做直线运动的物体,其速度时间图象如图所示,求:⑴哪一时刻物体距出发点最远?最远距离是多少? ⑵4s 内物体运动的路程?⑶4s 内物体运动的平均速度大小?14.(10分)比萨斜塔是世界建筑史上的一大奇迹.如图所示,已知斜塔第一层离地面的高度1 6.8m =h ,为了测量塔的总高度,在塔顶无初速度释放一个小球,小球通过斜塔第一层的时间10.2s =t ,重力加速度g 取210m/s ,不计空气阻力.求:⑴小球经过斜塔第一层1h 的过程中,平均速度多大? ⑵小球落地时速度多大?⑶斜塔离地面的总高度为多少?15.(11分)如图所示,在水平作用力F 的作用下,木板B 在水平地面上向左匀速运动,其水平表面上的木块A 静止不动,与之相连的固定在竖直墙上的弹簧秤的示数为3.2N .已知木块A 重8N ,木板B 重12N .⑴若此时弹簧秤弹簧的伸长量为1.6cm ,则该弹簧秤弹簧的劲度系数为多少? ⑵木块A 与木板B 之间的动摩擦因数为多少?⑶若已知木板与水平地面间的动摩擦因数为0.3,则水平力F 为多大?16.(11分)一辆摩擦车正在以120m/s =v 的速度匀速行驶,突然,司机看见正前方65m=x (距车队队尾)处有一列汽车车队以8m/s 的速度正在匀速行驶,相邻汽车间距为20m ,摩擦车司机立即采取制动措施.若从司机看见车队开始计时(0=t ),摩擦车的速度-时间图象如图所示(汽车和摩擦车大小不计).图看不清楚⑴求摩擦车制动时的加速度大小.⑵摩擦车最多能追上几辆汽车?⑶摩擦车从赶上车队到离开车队,共经历多长时间?。

江苏省南京师范大学附属中学2016-2017学年高一分班测试模拟练习二英语试题Word版缺答案

江苏省南京师范大学附属中学2016-2017学年高一分班测试模拟练习二英语试题Word版缺答案

绝密★启用前南京师范大学附属中学高一分班测试模拟练习二(英语)命题人:王文君审核人:吴钰鸣魏晨本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,共120分,考试用时60分钟。

第I卷英语知识运用(共三大题,满分96 分)一.单项填空(共15小题,每小题2分,共30分)从每小题所给的A、B、C、D四个选项中选出一个可以填入空白处的最佳选项。

1.The“Chinese Dream”is _______ dream to improve peo ple’s happiness and _______ dream of harmony, peace and development.A. the; aB. a; aC. a; theD. the; the2.—What do you think of the 3-D printer?—It’s great although it still needs further _______.A. inventionB. educationC. informationD.development3.Let’s go _______, OK?A. interesting somewhereB. interesting anywhereC. somewhere interestingD. anywhere interesting4.—Look! Someone _______ the floor. It's clean.—Well, it wasn't me. I went shopping just now.A. is sweepingB. was sweepingC. had sweptD. has swept5.Learning to write is learning to think. You don’t know things clearly _______ you can write them down.A. unlessB. ifC. sinceD. whether6.—What do you think of our performance on the New Year’s Evening party?—Perfect! I don’t think I can see a _______ one.A. worseB. goodC. betterD. bad。

南师附中高一分班测试英语模拟卷与答案

南师附中高一分班测试英语模拟卷与答案

南京师范大学附属中学高一分班测试模拟试卷(英语)一、语法单选(每小题1分,共15分)1. The Great Wall is the last place ________ we will visit.A. whereB. whichC. thatD. in which2. What big fish! Where did you get ________? It is the first time I ________ such fish.A it…s ee B. it…have seenC. them…seeD. them… have seen3. In the dark street, he had no one to ________ for help.A. turn upB. turn aroundC. turn toD. turn down4. The dark clouds are gathering as if it ________ rain.A. willB. wouldC. were going toD. is going to5. You can’t imagine the trouble she ________ her missing child.A. had foundB. had findingC. had to findD. has found6. Luckily we had a map, without ________ we would have got lost.A. whichB. itC. thatD. what7. The cow ________ by the fire rushed out into the street.A. frighteningB. frightenedC. was frightenedD. be frightened8. It is surprising for them to complete ________ work in ________.A. so little…such a short timeB. so much… so short timeC. such little…so short a timeD. so much…such a short time9. In the accident, only 3 of the miners escaped ________ by the gases.A. killedB. killingC. to be killedD. being kill10. You can make yourself ________ in English quite well if you keep on speaking the language.A. understoodB. understandC. to understandD. understanding11. --- Tom is ill in hospital, and tomorrow ________ Sunday.--- I’m sorry to hear that. I ________ go and see him. We are good friends.A. will be, am going toB. is, willC. is going to be, shallD. is, am to12. The students who are sitting at the back of the classroom, please keep quiet, ________?A. do youB. don’t theyC. aren’t youD. will you13. Having sold most of his furniture, Alan hardly had ______ left in the house.A. everythingB. nothingC. anythingD. something14. The fire was finally controlled, but not ______ great damage had been caused.A. afterB. beforeC. sinceD. while15.Whenever and wherever we are, keep in mind that we should listen more to others. It’swrong to express ourselves without thinking. Just as we all know, __________.A. a good beginning makes a good endingB. a still tongue makes a wise headC. a great hope makes a great manD. a good husband makes a good wife二、完形填空(每小题1分,共20分)One of my fondest memories as a child is going by the river and sitting on the bank. There I would 16 the peace and quiet, watch the water rush 17 and listen to the singing of birds and the rustling of 18 in the trees. I would also watch the bamboo trees 19 under pressure from the wind and watch them 20 gracefully to their original position after the wind had 21 .When I think about the bamboo tree's ability to bounce back to its original position, the word "resilience"(恢复能力)comes to mind. When used in 22 a person this word means the ability to readily 23 from shock, depression or any other situation that stretches the limits of a person's 24 .Have you ever felt like you are at your 25 point? Thankfully, you have survived the experience to live to talk about it.During the 26 you probably felt a mix of emotions that threatened your health. You felt emotionally drained(耗尽的), 27 exhausted and you most likely stood 28 physical symptoms.Life is a 29 of good times and bad times, happy moments and unhappy moments. The next time you are experiencing one of those bad times or unhappy 30 that take you close to your breaking point, bend, 31 don't break. Try your best not to let the situation get the best of you.A 32 of hope will take you through the unpleasant ordeal (考验). With 33 for a better tomorrow or a better situation, things may not be as bad as they seem to be. The unpleasant ordeal may be easier to 34 if the final result is worth having.If life gets 35 and you are at your breaking point, show resilience. Like the bamboo tree,bend, but don't break!16. A. see B. hear C. enjoy D. touch17. A. downstream B. happily C. uphill D. closely18. A. fruits B. branches C. roots D. leaves19. A. move B. bend C. fall D. decline20. A. go B. turn C. return D. suffer21. A. died down B. died off C. died away D. died out22. A. honor of B. reward to C. reference to D. favor of23. A. recover B. suffer C. come D. escape24. A. thoughts B. mind C. body D. emotions25. A. starting B. breaking C. standing D. tiring26. A. practice B. experiment C. victory D. experience27. A. possibly B. terribly C. mentally D. probably28. A. unpleasant B. unreasonable C. exciting D. good29. A. result B. change C. wonder D. mixture30. A. events B. moments C. adventures D. changes31. A. but B. however C. though D. and32. A. little B. number C. measure D. few33. A. idea B. hope C. imagination D. search34. A. deal with B. look into C. depend on D. get stuck35. A. acceptable B. wrong C. tough D.easy三、阅读理解(每小题2分,共30分)ALast week my youngest son and I visited my father at his new home in Tucson, Arizona. He moved there a few years ago, and I was eager to see his new place and meet his friends.My earliest memories of my father are a tall, handsome, successful man devoted to his work and his family, but uncomfortable with his children. As a child I loved him. He seemed unhappy with me if I didn’t got straight A’s and u nhappy with my boyfriends if their fathers were not as “successful” as he was. Whenever I went out with him on weekends, I used to try hard to think out things to say, feeling on guard.On the first day of my visit, we went out with one of my father’s fr iends for lunch at anoutdoor cafe. We walked along that afternoon, did some shopping, ate on the street table, and laughed over my son’s funny facial expressions. Gone was my father’s critical (挑剔的) air and strict rules. Who was this person I knew as my father, who seemed so friendly and interesting to be around? What had held him back before?The next day dad pulled out his childhood pictures and told me quite a few stories about his own childhood. Although our times together became easier over the years, I never felt closer to him at that moment. After so many years, I’m at last seeing another side of my father. And in so doing, I’m delighted with my new friend. My dad, in his new home in Arizona, is back to me from where he was.36. Why did the author feel bitter(痛苦) about her father as a young adult?A. He was silent most of the time.B. He was too proud of himself.C. He didn’t love his children.D. He expected too much of her.37. When the author went out with her father on weekends, she would feel ________.A. nervousB. sorryC. tiredD. safe38. What does the author think of her father after her visit to Tucson?A. More critical.B. More humorous.C. Easy-going and friendly.D. Strict and hard-working.39. The underlined words in the last paragraph refer to ________.A. the author’s sonB. the author’s fatherC. the friend of the author’s fatherD. the cafe owner40. We can infer from the passage that ________.A. the writer's father used to be strict with her when he was a childB. the writer's father worked hard but cared less for his familyC. it was possibly the first time that the writer had visited her father's new homeD. as a child, the writer loved her fatherBEveryone has got two personalities—the one that is shown to the world and the other that is secret and real. You don’t show your secret personality when you’re awake because you can control your behavior, but when you’re asleep, your sle eping position shows the real you. In a normal night, of course, people frequently change their position. The important position is the one that you go to sleep in.If you go to sleep on your back, you’re a very open person. You normally trust people and y ou are easily influenced by fashion or new ideas. You don’t like to upset people, so you never express your real feelings. You’re quite shy and you aren’t very confident.If you sleep on your stomach, you are a rather secretive(不坦率的)person. You worry a lot and you’re always easily upset. You’re very stubborn(顽固的), but you aren’t very ambitious. You’re usually live for today not for tomorrow. This means that you enjoy having a good time.If you sleep curled up (卷曲),you are probably a very nervous person. You have a low opinion of yourself and so you’re often defensive. You’re shy and you don’t normally like meeting people. You prefer to be on your own. You’re easily hurt.If you sleep on your side, you have usually got a well-balanced personality. You know your strengths and weaknesses. You’re usually careful. You have a confident personality. You sometimes feel anxious, but you don’t often get depressed. You always say what y ou think even if it annoys people.41. According to the writer, you naturally show your secret and real personality .A. only in a normal nightB. only when you go to sleepC. only when you refuse to show yourself to the wordD. only when you change sleeping position42. Which is NOT mentioned in the second paragraph about a person’s personality?A. He or she is always open with others.B. He or she always likes new ideas earlier than others.C. He or she is always easily upset.D. He or she tends to believe in others.43. Point out which sentence is used to show the personality of a person who is used to sleeping on his or her stomach?A. He or she is careful not to offend others.B. He or she doesn’t want to stick to his or her opinion.C. He or she can’t be successful in any business.D. He or she likes to bring others happiness.44. Which of the following may be the reason for you not to make friends with a person who sleeps curled up?.A. He or she would rather be alone than communicate with you.B. He or she is rarely ready to help you.C. He or she prefers staying at home to going out.D. He or she wouldn’t like to get help from you.45. It appears that the writer tends to think highly of the person who sleeps on one side because .A. he or she always shows sympathy for peopleB. he or she is confident, but not stubbornC. he or she has more strengths than weaknessesD. he or she often considers annoying peopleCFirst of all, students need to realize that conflict is unavoidable. A report on violence among middle school and high school students indicates that most violent incidents between students begin with a relatively minor insult (侮辱). For example, a fight could start over the fact that one student eats a peanut butter sandwich each lunchtime. Laughter over the sandwich can lead to insults, which in turn can lead to violence. The problem isn't in the sandwich, but in the way students deal with the conflict.Once students recognize that conflict is unavoidable, they can practice the golden rule of conflict resolution (解决) stay calm. Once the student feels calmer, he or she should choose words that will calm the other person down as well. Rude words, name-calling, and accusation only add fuel to the emotional fir On the other hand, soft words spoken at a normal sound level can put out the fire before it explodes out of control.After both sides have calmed down, they can use another key strategy for conflict resolution; listening. Listening allows the two sides to understand each other. One person should describe his or her side, and the other person should listen without interrupting. Afterward, the listener can ask non-threatening questions to clarify the speaker's position. Then the two people should change roles.Finally, students need to consider what they are hearing. This doesn't mean trying to figure out what's wrong with the other person. It means understanding what the real issue is and what both sides are trying to accomplish. For example, a shouting match over a peanut butter sandwich might happen because one person thinks the other person is unwilling to try new things. Students need to ask themselves questions such as these: How did this start? What do I really want? Whatam I afraid of? As the issue becomes clearer, the conflict often simply becomes smaller. Even if it doesn't, careful thought helps both sides figure out a mutual solution.There will always be conflict in schools, but that doesn't mean there needs to be violence. After students in Atlanta started a conflict resolution program, according to Educators for Social Responsibility, "64 percent of the teachers reported less physical violence in the classroom; 75 percent of the teachers reported an increase in student cooperation; and 92 percent of the students felt better about themselves". Learning to resolve conflicts can help students deal with friends,. teachers. parents, bosses, and coworkers. In that way, conflict resolution is a basic life skill that should be taught in schools across the country.46. This article is mainly about__________.A. the lives of school childrenB. the cause of arguments in schoolsC. how to analyze youth violenceD. how to deal with school conflicts47. From Paragraph 2 we can learn that________ .A. violence is more likely to occur at lunchtimeB. a small conflict can lead to violenceC. students tend to lose their temper easilyD. the eating habit of a student is often the cause of a fight48. Why do students need to ask themselves the questions stated in Paragraph 5?A. To find out who to blame.B. To get ready to buy new things.C. To make clear what the real issue is.D. To figure out how to stop the shouting match.49. After the conflict resolution program was started in Atlanta, it was found that______.A. there was a decrease in classroom violenceB. there was less student cooperation in the classroomC. more teachers fell better about themselves in schoolsD. the teacher-student relationship greatly improved50. The writer’s purpose for writing this article is to_______.A. complain about problems in school educationB. teach students different strategies for school lifeC. advocate teaching conflict management in schoolsD. inform teachers of the latest studies on school violence第II卷非选择题(共35 分)IV. 单词拼写(共10小题;每小题1分,满分10分)根据所给首字母或汉语,写出相应的单词。

2017—2018学年度江苏省南京师大附中高一上学期期末考试化学试卷

2017—2018学年度江苏省南京师大附中高一上学期期末考试化学试卷

2017—2018学年度江苏省南京师大附中高一上学期期末考试化学试卷可能用到的相对原子质量:H-1 C-12 N-14 O-16 S-32 Cl-35.5 Na-23 Al-27 Ca-40 Fe-56第I卷(选择题共44分)一、选择题(本题包括10小题,每小题2分,共20分。

每小题只有一个选项符合题意。

)1、硅被誉为无机非金属材料的主角。

下列物品用到硅单质的是A、陶瓷餐具B、石英钟表C、计算机芯片D、光导纤维2、下列气体中,不会造成空气污染的是A、N2B、NOC、NO2D、SO23、下列物质中,不属于电解质的是A、NaOHB、蔗糖C、H2SO4D、NaCl4、下列气体不能用排空气法收集的是A、NHB、Cl2C、NO2D、NO5、从碘水中提取碘可以采用的方法是A、萃取B、沉淀C、蒸馏D、过滤6、下列事故或药品的处理正确的是A、少量浓硫酸沾在皮肤上,立即用氢氧化钠溶液冲洗B、大量氯气泄漏时,用肥皂水浸湿毛巾捂住嘴和鼻,并迅速离开现场C、不慎洒出的酒精在桌上燃烧起来,应立即用大量水扑灭D、将含硫酸的废液倒入水槽,用水冲入下水道7、下列有关物质用途的叙述不正确的是A、用氧化亚铜作制造红色玻璃的颜料B、氢氧化钠可作色素的吸附剂C、用铝制餐具长时间存放碱性食物D、明矾可用作净水剂8、下列物品的鉴别方法不正确的是A、用焰色反应鉴别NaCl、KCl和Na2SO4B、用氢氧化钠溶液鉴别MgCl2溶液和AlCl3溶液C、利用丁达尔效应鉴别Fe(OH)3胶体和FeCl3溶液D、用品红试液鉴别SO2和CO2两种气体9、向下列各物品的水溶液中滴加稀H2SO4或MgCl2溶液时,均有白色沉淀生成的是A、BaCl2B、Ba(OH)2C、Na2CO3D、KOH10、制备硅单质时,主要化学反应如下:(1)SiO2+2C 高温Si +2CO↑(2) Si +2Cl2加热SiCl4(3)SiCl4+2H2高温Si +4HCl下列对上述三个反应的叙述中,不正确的是A、(1)(3)为置换反应B、(1)(2)(3)均为氧化还原反应C、(2)为化合反应D、三个反应的反应物中硅元素均被还原二、选择题(本题包括8小题,每小题3分,共24分。

福建省师大附中2016-2017学年高一物理上学期期末考试

福建省师大附中2016-2017学年高一物理上学期期末考试

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2017年江苏南京师范大学附属中学新高一分班测试模拟试卷与答案(数学)

2017年江苏南京师范大学附属中学新高一分班测试模拟试卷与答案(数学)

南京师范大学附属中学高一分班测试模拟试卷(数学)一、填空题(每题5分,共70分)1.如图,在四边形ABCD 中,E 、F 分别是AB 、AD 的中点。

若EF =2,BC =5,CD =3, 则tan C 的值为 .2.如图,P A 、PB 是⊙O 切线,A 、B 为切点,AC 是直径,∠P = 40°,则∠BAC 的值为 .3.在两个袋内,分别装着写有1、2、3、4四个数字的4张卡片,今从每个袋中各任取一张卡片,则所取两卡片上数字之积为偶数的概率是 .4.如图,矩形纸片ABCD 中,已知AD =8,折叠纸片使AB 边与对角线AC 重合,点B 落在点F 处,折痕为AE ,且EF =3,则AB 的长为 .5.若直角坐标系内两点P 、Q 满足条件①P 、Q 都在函数y 的图象上②P 、Q 关于原点对称,则称点对(P ,Q )是函数y 的一个“友好点对”(点对(P ,Q )与(Q ,P )看作同一个“友好点对”)。

已知函数⎪⎩⎪⎨⎧>≤++=02101422x xx x x y ,,,则函数y 的“友好点对”有 个.6.已知a 、b 是一元二次方程2210x x --=的两个实数根,则代数式()()2a b a b ab -+-+ 的值等于7.有一个六个面分别标上数字1、2、3、4、5、6的正方体,甲、乙、丙三位同学从不同B CE1x m n +=的解x 满足1+<<k x k ,k 为整数,则k =8.如图,直角梯形纸片ABCD 中,AD //BC ,∠A =90º,∠C =30º.折叠纸片使BC 经过点D ,点C 落在点E 处,BF 是折痕,且BF =CF =8,则AB 的长为9.记函数y 在x 处的值为()f x (如函数2y x =也可记为2()f x x =,当1x =时的函数 值可记为(1)1f =)。

已知||)(x x x f =,若c b a >>且0=++c b a ,0≠b ,则 )()()(c f b f a f ++的所有可能值为10.有一塔形几何体由若干个正方体构成,构成方式如图所示,上层正方体下底面的四个顶点是下层正方体上底面各边的中点。

【物理】南师附中分班考模拟题1卷

【物理】南师附中分班考模拟题1卷

南京师范大学附属中学高一分班测试模拟试卷(物理)一、选择题.1.(3分)如图所示,用高压放电的电火花产生一个冲击波,再用椭球形凹面镜使声波反射后集中到胆结石上,就能使胆结石粉碎.这一现象主要说明()A.声音是一种波B.只有用凹面镜才能反射声波C.声波具有能量D.只有把声波集中起来,它才具有能量2.(3分)古人在夕阳西下的时候吟出“柳絮飞来片片红”的诗句.洁白的柳絮这时看上去却是红色的,这是因为柳絮()A.发出红光B.发出红外线C.反射夕阳的红光 D.折射夕阳的红光3.(3分)同种材料制成的几个圆柱体,它们的质量、粗细、高度各不相同,把它们竖直放在水平桌面上,则()A.较高的圆柱体对桌面的压强大B.较粗的圆柱体对桌面的压强大C.质量较大的圆柱体对桌面的压强大D.它们对桌面的压强一样大4.(3分)下列过程中,有一个力的作用效果与其他三个不同类,它是()A.把橡皮泥捏成不同造型B.进站的火车受阻力缓缓停下C.苹果受重力竖直下落D.用力把铅球推出5.(3分)随着人们生活水平的逐步提高,家用电器的不断增多,在家庭电路中,下列说法正确的是()A.灯与控制它的开关是并联的,与插座是串联的B.使用测电笔时,不能用手接触到笔尾的金属体C.电路中电流过大的原因之一是使用的电器总功率过大D.增加大功率用电器时,只需换上足够粗的保险丝即可6.(3分)我国的登月工程分三个阶段进行,其中发射首颗月球探测卫星“嫦娥一号”是这一工程的第一个阶段,关于“嫦娥一号”卫星的说法正确的是()A.卫星探测到的信息是靠超声波传回地面的B.卫星从地面发射升空的过程中,运动状态不断改变C.卫星从地面发射升空的过程中,主要是机械能转化为内能D.卫星从地面发射升空的过程中,重力势能增大,动能增大,机械能增大7.(3分)回顾如图所示“什么情况下磁可以生电”的探究过程,你认为以下说法正确的是()A.让导线在磁场中静止,蹄形磁铁的磁性越强,电流表指针偏转角度越大B.用匝数较多的线圈代替单根导线,且使线圈在磁场中静止,这时电流表指针偏转角度大些C.蹄形磁铁固定不动,使导线沿水平方向运动时,电流表指针会发生偏转D.蹄形磁铁固定不动,使导线沿竖直方向运动时,电流表指针会发生偏转8.(3分)关于温度、热量、内能,以下说法正确的是()A.0℃的冰没有内能B.水沸腾时继续吸热,温度保持不变C.物体的温度越低,所含的热量越多D.物体的内能与温度有关,只要温度不变,物体的内能就一定不变9.(3分)小明将标有“6V 6W”的灯L1与标有“6V 3W”的灯L2串联(假设灯丝电阻不变),如图所示.下列说法错误的是()A.该电路能探究串联电路的电压特点B.该电路能探究电流与电压的关系C.该电路能探究电流做功的多少与电压的关系D.该电路可以测出灯泡L1的额定功率二、填空与作图题(10-13每题3分,14题3分和15题2分,共17分)10.(3分)星期天小明和妈妈来到市中心的广场游玩,看到音乐喷泉的上空出现了一弯绚丽的彩虹,如图.妈妈解释说:彩虹是由于太阳光经过喷泉形成的小水珠时发生了,从而产生了色散现象形成的.小明用带有红色滤色镜的照相机给妈妈照了一张像留念,则相片上的彩虹是一条色的光带.喷泉形成的水珠象珠玉落在玉盘上,溅起层层水花,光在这样的水面上发生了反射,所有池边霓虹灯不能在水面成像,只留下亭亭的影子.11.(3分)在综合实践活动中,小明制作了如图所示的简易棉花糖机.取少量蔗糖置于侧壁扎有几排小孔的易拉罐内,闭合开关,电动机带动易拉罐转动,并用酒精灯对易拉罐底部进行加热,这是采用方法改变其内能的.加热过程中,蔗糖颗粒变成液态糖浆,糖浆从易拉罐侧壁的小孔中被甩出后,由于糖浆具有继续前进,遇冷变成丝状糖絮,用筷子收集到棉花糖.上述制取棉花糖的整个过程中,蔗糖发生的物态变化有.12.(3分)如图是小红和小明站在冰面上静止时的情景.小明发现冰刀下的冰由固态变成了液态,这是(填物态变化名称)现象,这一过程需要(吸热/放热).小明在后面推了小红一下,使小红向前滑去,同时小明向后运动,表明物体间力的作用是的.13.(3分)如图是研究通电导体产生热量的多少与电阻的关系的装置,A、B两个相同的瓶内分别装有质量和初温都(相同/不相同)的煤油,将铜丝和镍铬合金丝串联接入电路中,这是为了控制相同,两电阻丝产生热量的多少是通过反映出来的.14.(3分)某同学研究曲面玻璃对光的折射规律时作了如下的试验.如图为一半圆柱形透明玻璃的横截面,水平面AOB镀银(反射光),O表示半圆截面的圆心,一束光在横截面内从M点入射,从N点射出,请帮他完成光路图(虚线为该点的法线).15.(2分)如图甲将伞撑开时,伞的每条骨架都可近似看成一个杠杆,乙图是其中一条骨架的示意图,伞面对骨架的压力为阻力F2,支点为O,请你在乙图中画出支撑杆AB对骨架的动力F1的大致方向及阻力臂l2.三、试验与探究题(5分+5分+5分+3分=18分)16.(5分)在进行光学实验的探究过程中:(1)小明同学用如图1所示的装置研究光的反射规律,其中有一个可折转的光屏,这个光屏在实验中除了能呈现光外,另一个重要作用:.(2)小明利用透明玻璃板、两段完全相同的蜡烛等器材探究平面镜成像的特点.选取两段完全相同的蜡烛的目的是.(3)小明同学在做“探究凸透镜成像规律”实验时,蜡烛、凸透镜、光屏的位置如图2所示,点燃蜡烛后,光屏上得到了(倒立/正立)、(放大/缩小)的实像,(照相机/投影仪),就是利用这一原理制成的.若将蜡烛向左移,要使光屏上出现清晰的像,光屏将向(左/右)移.17.(5分)在测量小塑料块的密度的实验中,小明和小影只找到一架天平(附砝码),没有量筒,他思考了一段时间后终于想出了办法.(1)如图甲所示,小明在使用天平过程中的错误是:.(2)小明纠正错误后调平天平,按照以下步骤继续实验.①将小烧杯装半杯水,放在左盘中,称出它的质量为100克;②将小塑料块放入小烧杯的水中,浮在水面上,天平平衡时右盘中所加砝码和游码的位置如图乙所示,则小塑料块的质量为g.③用细针缓慢地将小塑料块按入小烧杯的水中,不动(小塑料块未触底),小影加减砝码,调平天平,砝码和游码表示的质量和为120克.则小塑料块受到的浮力为,体积为,密度是kg/m3.18.(5分)利用电源、滑动变阻器和定值电阻R0(阻值已知)以及两个电流表A1、A2来测量未知电阻Rx的阻值.同学们设计了如图A、B、C三个电路.(1)经过讨论,同学们认为测量未知电阻R x阻值的最佳方案为图电路.(2)小宇在某次实验中,测得电流表A1的示数为I1,A2的示数为I2,则待测电阻R x=.(3)在某次测量中,小宇发现无论怎样移动滑动变阻器的滑片P,两个电流表A1、A2的示数始终相等,即I1=I2,由此判断出R x处的故障是.(4)实验中滑动变阻器的作用是和.19.(3分)湖北嘉得利汽车公司组织了一次新车型节油大赛,比赛分两组:第一组使用A型车(排量1.2升),第二组使用B型车(排量2.0升),A、B型车外观和体积相同,属于同一产品系列.(1)你认为下列评价方式中不能作为比赛评价标准A.消耗相同体积的汽油,比谁行驶的路程长B.行驶相同的路程,比谁消耗的汽油体积少C.行驶相同的路程,比谁用的时间短D.用“”的比值大小作比较(2)下表是三位车手的比赛成绩,第一组比赛中最节油的选手是.(3)从本次比赛数据看,型车比较节油.四、综合应用.(6分+7分=13分)20.(6分)鸡蛋的营养丰富、味道鲜美,深受人们喜爱.小明家新买了一个煮蛋器,其外观如图所示.该煮蛋器一次可煮6只鸡蛋,铭牌如图所示.某次,妈妈将6只质量均为50g的鸡蛋装入煮蛋器,通电3min后,煮蛋器中水开始沸腾了,妈妈断开电源,再过2min,妈妈从煮蛋器中取出鸡蛋,小明发现,鸡蛋已经完全煮熟了.则:(1)本次煮鸡蛋的过程中,共用电多少焦?(2)若鸡蛋的初温为20℃,煮蛋器中水的沸点为100℃,则本次煮鸡蛋的过程中鸡蛋吸收的热量的多少?该煮蛋器煮鸡蛋的效率为多大?[鸡蛋的比热容取2.1×103J/(kg•℃)](3)妈妈为什么在鸡蛋煮了3min后再断电2min,然后再从煮蛋器中取出鸡蛋?说说你的看法.(说一点即可)21.(7分)如图1是我国首台自主设计、自主集成的“蛟龙”号载人潜水器,在北京时间2011年11月26日6时12分首次下潜至5038.5米,顺利完成本次5000米级海试任务,这标志着中国已成为世界上掌握大深度载人深潜技术的国家.(1)“蛟龙号”潜水器的体积约为21m3,当它在水下缓慢地匀速下潜时的总质量多大?(不计算水的阻力)(2)完成任务后,用如图2的滑轮组(滑轮重不计)来提起潜水器.若出水后拉力随时间的变化关系如图17甲所示,潜水器的速度υ和上升的高度h随时间t变化的关系图象分别如图3(丙)和(丁)所示.求:①在1~2s内,拉力F的功率P.②在0~1s内,拉力F做的功W.。

湖南师大附中2016-2017学年高一上学期期末数学试卷 含解析

湖南师大附中2016-2017学年高一上学期期末数学试卷 含解析

2016-2017学年湖南师大附中高一(上)期末数学试卷一、选择题:本大题共11小题,每小题5分,共55分,在每小题给出的四个选项中,只有一项是符合题目要求的.1.已知两点A(a,3),B(1,﹣2),若直线AB的倾斜角为135°,则a的值为( )A.6 B.﹣6 C.4 D.﹣42.对于给定的直线l和平面a,在平面a内总存在直线m与直线l( )A.平行B.相交C.垂直D.异面3.已知直线l1:2x+3my﹣m+2=0和l2:mx+6y﹣4=0,若l1∥l2,则l1与l2之间的距离为()A.B.C.D.4.已知三棱锥P﹣ABC的三条侧棱PA、PB、PC两两互相垂直,且PA=2,PB=,PC=3,则这个三棱锥的外接球的表面积为()A.16πB.32πC.36πD.64π5.圆与圆的位置关系是( )A.相交B.外离C.内含D.内切6.已知α,β是两个不同的平面,m.n是两条不同的直线,则下列命题中正确的是()A.若m∥n,m⊂β,则n∥βB.若m∥α,α∩β=n,则m∥nC.若m⊥α,m⊥β,则α∥β D.若m⊥β,α⊥β,则m∥α7.在空间直角坐标系O﹣xyz中,一个四面体的顶点坐标为分别为(0,0,2),(2,2,0),(0,2,0),(2,2,2).画该四面体三视图中的正视图时,以xOz平面为投影面,则得到正视图可以为( ) A.B.C.D.8.若点P(3,1)为圆(x﹣2)2+y2=16的弦AB的中点,则直线AB 的方程为()A.x﹣3y=0 B.2x﹣y﹣5=0 C.x+y﹣4=0 D.x﹣2y﹣1=09.已知四棱锥P﹣ABCD的底面为菱形,∠BAD=60°,侧面PAD 为正三角形,且平面PAD⊥平面ABCD,则下列说法中错误的是()A.异面直线PA与BC的夹角为60°B.若M为AD的中点,则AD⊥平面PMBC.二面角P﹣BC﹣A的大小为45°D.BD⊥平面PAC10.已知直线l过点P(2,4),且与圆O:x2+y2=4相切,则直线l的方程为()A.x=2或3x﹣4y+10=0 B.x=2或x+2y﹣10=0C.y=4或3x﹣4y+10=0 D.y=4或x+2y﹣10=011.如图所示,在直角梯形BCEF中,∠CBF=∠BCE=90°,A、D分别是BF、CE上的点,AD∥BC,且AB=DE=2BC=2AF(如图1).将四边形ADEF沿AD折起,连结BE、BF、CE(如图2).在折起的过程中,下列说法中错误的是()A.AC∥平面BEFB.B、C、E、F四点不可能共面C.若EF⊥CF,则平面ADEF⊥平面ABCDD.平面BCE与平面BEF可能垂直二、填空题:本大题共3个小题,每小题5分,共15分.12.若直线x﹣y+1=0与圆(x﹣a)2+y2=2有公共点,则实数a取值范围是.13.已知一个圆柱的底面直径和母线长都等于球的直径,记圆柱的体积为V1,球的体积为V2,则= .14.已知三棱锥P﹣ABC的体积为10,其三视图如图所示,则这个三棱锥最长的一条侧棱长等于.三、解答题:本大题共3个小题,共30分.解答应写出文字说明,证明过程或演算步骤.15.已知△ABC的三个顶点的坐标分别为A(3,0),B(4,6),C(0,8).(1)求BC边上的高所在直线l的方程;(2)求△ABC的面积.16.已知圆C经过A(﹣2,1),B(5,0)两点,且圆心C在直线y=2x 上.(1)求圆C的标准方程;(2)设动直线l:(m+2)x+(2m+1)y﹣7m﹣8=0与圆C相交于P,Q两点,求|PQ|的最小值.17.如图,在三棱柱ABC﹣A1B1C1中,A1A⊥平面ABC,AB⊥AC,AB=AC=AA1,D为BC的中点.(1)证明:A1B⊥平面AB1C;(2)求直线A1D与平面AB1C所成的角的大小.一、本大题共2个小题,每小题5分,共12分.18.已知集合,则N∩∁R M= .19.已知函数f(x)在定义域R上单调递减,且函数y=f(x﹣1)的图象关于点A(1,0)对称.若实数t满足f(t2﹣2t)+f(﹣3)>0,则的取值范围是()A.(,+∞)B.(﹣∞,)C.(0,)D.(,1)∪(1,+∞)二、本大题共3个大题,共38分.20.如图,四棱锥S﹣ABCD的底面是正方形,每条侧棱的长都是底面边长的倍,P为侧棱SD上的点.(Ⅰ)求证:AC⊥SD;(Ⅱ)若SD⊥平面PAC,则侧棱SC上是否存在一点E,使得BE ∥平面PAC.若存在,求SE:EC的值;若不存在,试说明理由.21.设函数f(x)=mx2﹣mx﹣1,g(x)=.(1)若对任意x∈[1,3],不等式f(x)<5﹣m恒成立,求实数m的取值范围;(2)当m=﹣时,确定函数g(x)在区间(3,+∞)上的单调性.22.已知圆C:(x﹣a)2+(y﹣a﹣2)2=9,其中a为实常数.(1)若直线l:x+y﹣4=0被圆C截得的弦长为2,求a的值;(2)设点A(3,0),O为坐标原点,若圆C上存在点M,使|MA|=2|MO|,求a的取值范围.2016-2017学年湖南师大附中高一(上)期末数学试卷参考答案与试题解析一、选择题:本大题共11小题,每小题5分,共55分,在每小题给出的四个选项中,只有一项是符合题目要求的.1.已知两点A(a,3),B(1,﹣2),若直线AB的倾斜角为135°,则a的值为()A.6 B.﹣6 C.4 D.﹣4【考点】直线的倾斜角.【分析】利用斜率计算公式即可得出.【解答】解:∵过点A(a,3),B(1,﹣2)的直线的倾斜角为135°,∴tan135°==﹣1,解得a=﹣4.故选:D.2.对于给定的直线l和平面a,在平面a内总存在直线m与直线l( )A.平行B.相交C.垂直D.异面【考点】空间中直线与直线之间的位置关系.【分析】本题可采用分类讨论,对答案进行排除,分别讨论直线l和平面α平行,直线l和平面α相交,直线l⊂平面α,三种情况,排除错误答案后,即可得到结论.【解答】解:若直线l和平面α平行,则平面α内的直线与l平行或异面,不可能相交,可排除答案A;若直线l和平面α相交,则平面α内的直线与l相交或异面,不可能平行,可排除答案B;若直线l⊂平面α,则平面α内的直线与l相交或平行,不可能异面,可排除答案D;故选C.3.已知直线l1:2x+3my﹣m+2=0和l2:mx+6y﹣4=0,若l1∥l2,则l1与l2之间的距离为( )A.B.C.D.【考点】直线的一般式方程与直线的平行关系.【分析】由,解得m=±2,m=﹣2时舍去,可得m=2,再利用平行线之间的距离公式即可得出.【解答】解:由,解得m=±2,m=﹣2时舍去,∴m=2,因此两条直线方程分别化为:x+3y=0,x+3y﹣2=0.则l1与l2之间的距离==.故选:B.4.已知三棱锥P﹣ABC的三条侧棱PA、PB、PC两两互相垂直,且PA=2,PB=,PC=3,则这个三棱锥的外接球的表面积为()A.16πB.32πC.36πD.64π【考点】球的体积和表面积.【分析】三棱锥P﹣ABC的三条侧棱PA、PB、PC两两互相垂直,它的外接球就是它扩展为长方体的外接球,求出长方体的对角线的长,就是球的直径,然后求球的表面积.【解答】解:三棱锥P﹣ABC的三条侧棱PA、PB、PC两两互相垂直,它的外接球就是它扩展为长方体的外接球,求出长方体的对角线的长:=4所以球的直径是4,半径为2,球的表面积:4π×4=16π.故选A.5.圆与圆的位置关系是( )A.相交B.外离C.内含D.内切【考点】圆与圆的位置关系及其判定.【分析】将圆的一般方程转化为标准方程,根据两圆圆心之间的距离和半径之间的关系进行判断.【解答】解:圆的标准方程为(x﹣2)2+(y﹣3)2=1,圆心O1(2,3),半径r=1,圆的标准方程为(x﹣4)2+(y﹣3)2=9,圆心O2(4,3),半径R=3,两圆心之间的距离|O1O2|=4﹣2=2=R﹣r,∴两圆内切.故选:D.6.已知α,β是两个不同的平面,m.n是两条不同的直线,则下列命题中正确的是()A.若m∥n,m⊂β,则n∥βB.若m∥α,α∩β=n,则m∥n C.若m⊥α,m⊥β,则α∥β D.若m⊥β,α⊥β,则m∥α【考点】空间中直线与平面之间的位置关系.【分析】对于选项A,若m∥n,m⊂β则n∥β,可通过线面平行的判定定理进行判断对于选项B,可通过线面平行的性质定理进行判断;对于选项C,可通过面面平行的判定条件进行判断;对于选项D,可通过线面位置关系判断.【解答】解:A不正确,m∥n,m⊂β,由于n可能在β内,故推不出n∥β;B不正确,m∥α,α∩β=n,m不一定在β内,故不能推出m∥n;C正确,垂直于同一条直线的两个平面平行;D不正确,m⊥β,α⊥β,由于m⊂α的可能性存在,故m∥α不正确.故选:C.7.在空间直角坐标系O﹣xyz中,一个四面体的顶点坐标为分别为(0,0,2),(2,2,0),(0,2,0),(2,2,2).画该四面体三视图中的正视图时,以xOz平面为投影面,则得到正视图可以为()A.B.C.D.【考点】由三视图求面积、体积.【分析】由题意画出几何体的直观图,然后判断以zOx平面为投影面,则得到正视图即可.【解答】解:因为一个四面体的顶点在空间直角坐标系O﹣xyz中的坐标分别是(0,0,2),(2,2,0),(0,2,0),(2,2,2).几何体的直观图如图,所以以zOx平面为投影面,则得到正视图为:故选A.8.若点P(3,1)为圆(x﹣2)2+y2=16的弦AB的中点,则直线AB的方程为()A.x﹣3y=0 B.2x﹣y﹣5=0 C.x+y﹣4=0 D.x﹣2y﹣1=0【考点】待定系数法求直线方程.【分析】由垂径定理,得AB中点与圆心C的连线与AB互相垂直,由此算出AB的斜率k=﹣1,结合直线方程的点斜式列式,即可得到直线AB的方程.【解答】解:∵AB是圆(x﹣2)2+y2=16的弦,圆心为C(2,0),∴设AB的中点是P(3,1)满足AB⊥CP,因此,PQ的斜率k=﹣1,可得直线PQ的方程是y﹣1=﹣(x﹣3),化简得x+y﹣4=0,故选:C.9.已知四棱锥P﹣ABCD的底面为菱形,∠BAD=60°,侧面PAD 为正三角形,且平面PAD⊥平面ABCD,则下列说法中错误的是()A.异面直线PA与BC的夹角为60°B.若M为AD的中点,则AD⊥平面PMBC.二面角P﹣BC﹣A的大小为45°D.BD⊥平面PAC【考点】棱锥的结构特征.【分析】根据线面垂直,异面直线所成角的大小以及二面角的求解方法分别进行判断即可.【解答】解:对于A,∵AD∥BC,∴∠PAD为异面直线PA与BC 的夹角,为60°,正确;对于B,连PM,BM,则∵侧面PAD为正三角形,∴PM⊥AD,又底面ABCD是∠DAB=60°的菱形,∴三角形ABD是等边三角形,∴AD⊥BM,∴AD⊥平面PBM,故B正确;对于C,∵底面ABCD为菱形,∠DAB=60°平面PAD⊥平面ABCD,∴BM⊥BC,则∠PBM是二面角P﹣BC﹣A的平面角,设AB=1,则BM=,PM=,在直角三角形PBM中,tan∠PBM=1,即∠PBM=45°,故二面角P﹣BC﹣A的大小为45°,故C正确,故错误的是D,故选:D.10.已知直线l过点P(2,4),且与圆O:x2+y2=4相切,则直线l的方程为( )A.x=2或3x﹣4y+10=0 B.x=2或x+2y﹣10=0C.y=4或3x﹣4y+10=0 D.y=4或x+2y﹣10=0【考点】圆的切线方程.【分析】切线的斜率存在时设过点P的圆的切线斜率为k,写出点斜式方程再化为一般式.根据圆心到切线的距离等于圆的半径这一性质,由点到直线的距离公式列出含k的方程,由方程解得k,然后代回所设切线方程即可.切线斜率不存在时,直线方程验证即可.【解答】解:将点P(2,4)代入圆的方程得22+32=13>4,∴点P在圆外,当过点P的切线斜率存在时,设所求切线的斜率为k,由点斜式可得切线方程为y﹣4=k(x﹣2),即kx﹣y﹣2k+4=0,∴=2,解得k=.故所求切线方程为3x﹣4y+16=0.当过点P的切线斜率不存在时,方程为x=2,也满足条件.故所求圆的切线方程为3x﹣4y+16=0或x=2.故选A.11.如图所示,在直角梯形BCEF中,∠CBF=∠BCE=90°,A、D分别是BF、CE上的点,AD∥BC,且AB=DE=2BC=2AF(如图1).将四边形ADEF沿AD折起,连结BE、BF、CE(如图2).在折起的过程中,下列说法中错误的是( )A.AC∥平面BEFB.B、C、E、F四点不可能共面C.若EF⊥CF,则平面ADEF⊥平面ABCDD.平面BCE与平面BEF可能垂直【考点】平面与平面之间的位置关系.【分析】本题考查了折叠得到的空间线面关系的判断;用到了线面平行、面面垂直的判定定理和性质定理.【解答】解:在图2中取AC的中点为O,取BE的中点为M,连结MO,易证得四边形AOMF为平行四边形,即AC∥FM,∴AC∥平面BEF,故A正确;∵直线BF与CE为异面直线,∴B、C、E、F四点不可能共面,故B 正确;在梯形ADEF中,易得EF⊥FD,又EF⊥CF,∴EF⊥平面CDF,即有CD⊥EF,∴CD⊥平面ADEF,则平面ADEF⊥平面ABCD,故C正确;延长AF至G使得AF=FG,连结BG、EG,易得平面BCE⊥平面ABF,过F作FN⊥BG于N,则FN⊥平面BCE.若平面BCE⊥平面BEF,则过F作直线与平面BCE垂直,其垂足在BE上,矛盾,故D错误.故选:D二、填空题:本大题共3个小题,每小题5分,共15分.12.若直线x﹣y+1=0与圆(x﹣a)2+y2=2有公共点,则实数a取值范围是[﹣3,1] .【考点】直线与圆的位置关系.【分析】由题意可得,圆心到直线的距离小于或等于半径,即≤,解绝对值不等式求得实数a取值范围.【解答】解:由题意可得,圆心到直线的距离小于或等于半径,即≤,化简得|a+1|≤2,故有﹣2≤a+1≤2,求得﹣3≤a≤1,故答案为:[﹣3,1].13.已知一个圆柱的底面直径和母线长都等于球的直径,记圆柱的体积为V1,球的体积为V2,则= .【考点】棱柱、棱锥、棱台的体积.【分析】设出球的半径,然后求解圆柱的体积,球的体积,推出结果即可.【解答】解:设球的半径为r,由题意可得:球的体积为V2=;圆柱的底面直径和母线长都等于球的直径,记圆柱的体积为V1=πr2•2r,则==.故答案为:.14.已知三棱锥P﹣ABC的体积为10,其三视图如图所示,则这个三棱锥最长的一条侧棱长等于.【考点】棱柱、棱锥、棱台的体积.【分析】由已知中的三视图,画出几何体的直观图,数形结合求出各棱的长,可得答案【解答】解:由三棱锥的三视图可得几何体的直观图如下图所示: O是顶点V在底面上的射影,棱锥的底面面积S=×4×5=10,∵三棱锥P﹣ABC的体积为10,故棱锥的高VO=3,则VA=,VC=3,AC=5,BC=4,AB=,VB=,故最长的侧棱为,故答案为:三、解答题:本大题共3个小题,共30分.解答应写出文字说明,证明过程或演算步骤.15.已知△ABC的三个顶点的坐标分别为A(3,0),B(4,6),C(0,8).(1)求BC边上的高所在直线l的方程;(2)求△ABC的面积.【考点】待定系数法求直线方程.【分析】(1)求出BC的斜率,带入点斜式方程即可;(2)求出AC 的长,根据AC的方程,求出点B到直线AC的距离,从而求出三角形ABC的面积即可.【解答】解:(1)因为点B(4,6),C(0,8),则k BC==﹣,因为l⊥BC,则l的斜率为2.又直线l过点A,所以直线l的方程为y=2(x﹣3),即2x﹣y﹣6=0.(2)因为点A(3,0),C(0,8),则|AC|==,又直线AC的方程为+=1,即8x+3y﹣24=0,则点B到直线AC的距离d==,所以△ABC的面积S=|AC|×d=13.16.已知圆C经过A(﹣2,1),B(5,0)两点,且圆心C在直线y=2x上.(1)求圆C的标准方程;(2)设动直线l:(m+2)x+(2m+1)y﹣7m﹣8=0与圆C相交于P,Q两点,求|PQ|的最小值.【考点】圆的一般方程.【分析】(1)设圆C的方程为x2+y2+Dx+Ey+F=0,建立方程组,即可求圆C的标准方程;(2)直线l过定点M(3,2).由圆的几何性质可知,当l⊥CM时,弦长|PQ|最短.【解答】解:(1)设圆C的方程为x2+y2+Dx+Ey+F=0,则,解得D=﹣4,E=﹣8,F=﹣5.所以圆C的方程是x2+y2﹣4x﹣8y﹣5=0,即(x﹣2)2+(y﹣4)2=25.(2)直线l的方程化为(2x+y﹣8)+m(x+2y﹣7)=0.令,得x=3,y=2,所以直线l过定点M(3,2).由圆的几何性质可知,当l⊥CM时,弦长|PQ|最短.因为|CM|==则|PQ|min=2=417.如图,在三棱柱ABC﹣A1B1C1中,A1A⊥平面ABC,AB⊥AC,AB=AC=AA1,D为BC的中点.(1)证明:A1B⊥平面AB1C;(2)求直线A1D与平面AB1C所成的角的大小.【考点】直线与平面所成的角;直线与平面垂直的判定.【分析】(1)证明A1A⊥AC.AC⊥A1B.推出AB1⊥A1B.即可证明A1B⊥平面AB1C.(2)连结A1C,设AB1∩A1B=O,连CO,交A1D于G.说明G为△A1BC的重心.推出∠A1GO是A1D与平面AB1C所成的角.设AB=AC=AA1=1,在Rt△A1OG中,求解直线A1D与平面AB1C所成的角为60°.【解答】(1)证明:图1所示,因为A1A⊥平面ABC,则A1A⊥AC.又AC⊥AB,则AC⊥平面AA1B1B,所以AC⊥A1B.由已知,侧面AA1B1B是正方形,则AB1⊥A1B.因为AB1∩AC=A,所以A1B⊥平面AB1C.(2)解:图2所示,连结A1C,设AB1∩A1B=O,连CO,交A1D 于G.因为O为A1B的中点,D为BC的中点,则G为△A1BC的重心.因为A1O⊥平面AB1C,则∠A1GO是A1D与平面AB1C所成的角.设AB=AC=AA1=1,则A1B=BC=A1C=.得A1O=,A1G=A1D=sin 60°=,在Rt△A1OG中,sin∠A1GO=,则∠A1GO=60°.所以直线A1D与平面AB1C所成的角为60°..一、本大题共2个小题,每小题5分,共12分.18.已知集合,则N∩∁R M= [0,2].【考点】交、并、补集的混合运算.【分析】先分别求出集合M和N,由此能求出N∩∁R M.【解答】解:集合,∴M=(﹣∞,0)∪(2,+∞),N=[0,+∞),∴N∩C R M=[0,2].故答案为:[0,2].19.已知函数f(x)在定义域R上单调递减,且函数y=f(x﹣1)的图象关于点A(1,0)对称.若实数t满足f(t2﹣2t)+f(﹣3)>0,则的取值范围是()A.(,+∞) B.(﹣∞,) C.(0,)D.(,1)∪(1,+∞)【考点】抽象函数及其应用.【分析】判断函数的奇偶性,利用函数的单调性求出t的范围,化简,利用函数的单调性求解最值即可.【解答】解:因为y=f(x﹣1)的图象关于点A(1,0)对称,则y=f (x)的图象关于原点对称,即f(x)为奇函数.由f(t2﹣2t)+f(﹣3)>0,得f(t2﹣2t)>﹣f(﹣3)=f(3),因为f(x)在R上是减函数,则t2﹣2t<3,即t2﹣2t﹣3<0,得﹣1<t<3.因为y==1+;在区间(﹣1,3)上是减函数,则.故选:B.二、本大题共3个大题,共38分.20.如图,四棱锥S﹣ABCD的底面是正方形,每条侧棱的长都是底面边长的倍,P为侧棱SD上的点.(Ⅰ)求证:AC⊥SD;(Ⅱ)若SD⊥平面PAC,则侧棱SC上是否存在一点E,使得BE∥平面PAC.若存在,求SE:EC的值;若不存在,试说明理由.【考点】直线与平面垂直的性质;直线与平面平行的判定.【分析】(Ⅰ)先证明AC⊥面SBD,然后利用线面垂直的性质证明AC⊥SD;(Ⅱ)利用线面平行的性质定理确定E的位置,然后求出SE:EC的值.【解答】解:(Ⅰ)连BD,设AC交BD于O,由题意SO⊥AC,在正方形ABCD中,AC⊥BD,所以AC⊥面SBD,所以AC⊥SD.(Ⅱ)若SD⊥平面PAC,则SD⊥OP,设正方形ABCD的边长为a,则SD=,OD=,则OD2=P D•SD,可得PD==,故可在SP上取一点N,使PN=PD,过N作PC的平行线与SC的交点即为E,连BN.在△BDN中知BN∥PO,又由于NE∥PC,故平面BEN∥面PAC,得BE∥面PAC,由于SN:NP=2:1,故SE:EC=2:1.21.设函数f(x)=mx2﹣mx﹣1,g(x)=.(1)若对任意x∈[1,3],不等式f(x)<5﹣m恒成立,求实数m 的取值范围;(2)当m=﹣时,确定函数g(x)在区间(3,+∞)上的单调性.【考点】函数恒成立问题.【分析】(1)利用f(x)<5﹣m,推出m<,设h(x)=,则当x∈[1,3]时,m<h(x)恒成立.利用二次函数的单调性求解m的取值范围.(2)推出g(x)=﹣(+).设x1>x2>3,则g(x1)﹣g(x2)=(x1﹣x2)(﹣),利用函数的单调性的定义证明即可.【解答】解:(1)由f(x)<5﹣m,得mx2﹣mx﹣1<5﹣m,即m (x2﹣x+1)<6.因为x2﹣x+1=(x﹣)2+>0,则m<.设h(x)=,则当x∈[1,3]时,m<h(x)恒成立.因为y=x2﹣x+1在区间[1,3]上是增函数,则h(x)在区间[1,3]上是减函数,h(x)min=h(3)=,所以m的取值范围是(﹣∞,).(2)因为f(x)=mx(x﹣1)﹣1,则g(x)=mx﹣.当m=﹣时,g(x)=﹣(+).设x1>x2>3,则g(x1)﹣g(x2)=﹣﹣==(x1﹣x2)(﹣)因为x1﹣1>x2﹣1>2,则(x1﹣1)(x2﹣1)>4,得<,又x1﹣x2>0,则g(x1)﹣g(x2)<0,即g(x1)<g(x2),所以g(x)在区间(3,+∞)上是减函数.22.已知圆C:(x﹣a)2+(y﹣a﹣2)2=9,其中a为实常数.(1)若直线l:x+y﹣4=0被圆C截得的弦长为2,求a的值;(2)设点A(3,0),O为坐标原点,若圆C上存在点M,使|MA|=2|MO|,求a的取值范围.【考点】直线与圆的位置关系.【分析】(1)利用圆心到直线的距离公式,结合直线l:x+y﹣3=0被圆C截得的弦长为2,利用勾股定理,可求a的值;(2)求出M在圆心为D(﹣1,0),半径为2的圆上,根据点M在圆C上,可得圆C与圆D有公共点,从而可得不等式,解不等式,即可求a的取值范围.【解答】解:(1)由圆方程知,圆C的圆心为C(a,a+2),半径为3.设圆心C到直线l的距离为d,因为直线l被圆C截得的弦长为2,则d2+1=9,即d=2.所以即|a﹣1|=2,所以a=﹣1或a=3.(2)设点M(x,y),由|MA|=2|MO|,化简得x2+y2+2x﹣3=0.所以点M在圆D:(x+1)2+y2=4上.其圆心为D(﹣1,0),半径为2.因为点M在圆C上,则圆C与圆D有公共点,即1≤|CD|≤5.所以1≤≤5,解得﹣5≤a≤﹣2或﹣1≤a≤2.故a的取值范围是[﹣5,﹣2]∪[﹣1,2].2017年2月23日。

南京市2016—2017第一学期高一数学期末试卷

南京市2016—2017第一学期高一数学期末试卷

南京市2016—2017学年度第一学期期末检测卷高一数学注意事项:1 .本试卷共4页,包括填空题(第1题~第14题)、解答题(第15题~第20题)两部分.本试卷满分为160分,考试时间为120分钟.2 .答题前,请务必将自己的姓名、学校、班级、学号写在答题卡的密封线内.试题的答案写在答.题卡上对应题目的答案空格内•考试结束后,交回答题卡.一、填空题:本大题共14小题,每小题5分,共70分•请把答案填写在答题卡相应位置.上.1. 若集合A= { —1 , 0, 1, 2} , B= {x | x+ 1 >0},则A AB = ▲.2. 函数y= log 2(1 —x)的定义域为▲.n3. 函数f(x) = 3sin(3x + 4)的最小正周期为▲.4. 若角ct的终边经过点P(—5, 12),则cosa的值为▲.5. 若幕函数y= x a(妖R)的图象经过点(4, 2),贝U a的值为▲.26. 若扇形的弧长为6cm,圆心角为2弧度,则扇形的面积为▲cm .7. 设e2是不共线的向量.若向量e1—4e?与k e1+ e?共线,则实数k的值为▲.&定义在区间[0, 5 n上的函数y = 2sinx的图象与y= cosx的图象的交点个数为▲.9. 若a = log32, b= 20.3, c= Iog g2,则a, b, c 的大小关系用"v” 表示为▲.10. 若f(x) = 2x+ 2—x是偶函数,则实数a的值为▲.11. 如图,点E是正方形ABCD的边CD的中点.若K E - D B =—2,则只E - 1B E的值为▲.12. 已知函数f(x)对任意实数x€ R, f(x+ 2) = f(x)恒成立,且当x€ [ —1, 1)时,f(x)= 2x+a.若点P(2017, 8)是该函数图象上的一点,则实数a的值为▲.5 213. 设函数f(x)= ~2—3x2+ 2,则使得f(1) >f(log3X)成立的x的取值范围为▲xx—2m, x>m,14. 已知函数f(x)= —x,—m v x v m,其中m>0.若对任意实数x,都有f(x) v f(x+ 1)成滾 + 2m, x<—m,立,则实数m的取值范围为▲.、解答题:本大题共6小题,共90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.15. (本小题满分14分)曲 + C0S: = 2 .已知sin :■— 2cos:(1)求tan a(二)求cos$—:■)• cos(—二 + :•)的值.16. (本小题满分14分)已知向量a= (—2, 1), b= (3, —4).(1)求(a+ b) - (2a—b)的值;(2)求向量a与a + b的夹角.如图,在一张长为2a米,宽为a米(a> 2)的矩形铁皮的四个角上,各剪去一个边长是x米(O v x<1)的小正方形,折成一个无盖的长方体铁盒,设V(x)表示该铁盒的容积.(1) 试写出V(x)的解析式;(2) 记y=凶幻,当x为何值时,y最小?并求出最小值.x(第17题图)18. (本小题满分16分)n n 已知函数f(x)= Asin( 3X+枷A> 0, w> 0, |^|v $)的最小正周期为n且点P© 2)是该函数图象的一个最高点.(1)求函数f(x)的解析式;(2)若x引—2,, 0],求函数y = f(x)的值域;n(3)把函数y= f(x)的图像向右平移9(0 v 0v个单位,得到函数y = g(x)的图象.若函数ny=g(x)在[0, [I上是单调增函数,求9的取值范围.如图,在△ ABC 中,已知CA= 1, CB= 2, ACB= 60 .(1)求|為;(2)已知点D是边AB上一点,满足X b = ZAB,点E是边CB上一点,满足BE = ABC .①当A=1时,求只E • C D ;②是否存在非零实数人使得云E XC D ?若存在,求出的入值;若不存在,请说明理由.20. (本小题满分16分)已知函数f(x)= x—a, g(x) = a|x|, a€ R.(1)设F(x) = f (x) —g(x).1①若a= 2,求函数y= F(x)的零点;②若函数y= F(x)存在零点,求a的取值范围.(2)设h(x) = f(x) + g(x), x€ [ —2, 2].若对任意x〔, [ —2, 2], |h(x” —h(x2)|w6 恒成立,试求a的取值范围.。

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乙甲 x 1t 1 x/m O t/sB A 南师附中2016~2017学年度上学期高一期末分班考试物理试题一、选择题(本大题共12小题,每小题4分。

1~7小题只有一个选项正确,8~12小题有多个选项正确,全部选对得4分,漏选得2分,错选或不选得0分) 1.下列叙述正确的是( )A .每当上课铃声响起,南师附中的同学们就能听到亲切温馨的广播声“上课时间到了,请同学们迅速回到教室,准备上课。

”其实此处的“时间”准确地说应该是时刻B .1934年10月10日晚6点,中国工农红军第一方面军从江西瑞金出发开始了艰苦卓绝的二万五千里长征。

这里的“二万五千里”是指位移大小C .毛泽东主席有诗曰“坐地日行八万里,遥天巡看一千河”,此处八万里的路程是以地面为参考系的D .传统寓言故事“龟兔赛跑”中,虽然兔子睡了一觉导致乌龟先到终点,但是兔子的平均速度仍然大于乌龟的平均速度 2.下列说法正确的是( ) A .完全失重的物体重力为零B .速度变化率越大的物体加速度也一定越大C .形状规则的实心物体的重心一定在其几何中心上D .静摩擦力可以是动力,但滑动摩擦力一定是阻力3.如图所示,一轻绳吊着一重物悬挂在天花板上处于静止,则下列说法正确的是( ) A .重物对轻绳的拉力就是重物的重力B .轻绳对天花板的拉力是因为天花板发生了形变要恢复原状C .轻绳对重物的拉力与重物的重力是一对平衡力D .重物对轻绳的拉力与重物的重力是一对作用力与反作用力4.如图所示,A 、B 两均匀直杆上端分别用细线悬挂于天花板上,下端搁在水平地面上,处于静止状态,悬挂A 杆的绳倾斜,悬挂B 杆的绳恰好竖直,则下列说法中正确的有( ) A .地面对A 的作用力的方向竖直向上 B .A 有向左运动的趋势C .地面对B 的支持力可能为零D .地面对B 的摩擦力方向向右 5.下列各物理量的单位属于国际单位制中基本单位的是( ) A .克 B . 牛 C .伏 D .安 6.一辆小车放在水平桌面上,现在用两种方式使小车向右作加速运动。

第一种方式,对小车加上水平向右大小为10N 的拉力(见图甲);第二种方式,通过光滑定滑轮挂上10N 的重物(见图乙)。

则下列判断正确的是( ) A .两种方式中小车的加速度相同B .第一种方式小车的加速度小于第二种方式C .第二种方式中小车和重物的加速度相同D .第二种方式中细绳的张力小于10N7.如图所示为甲、乙两物体运动的x-t 图象,则下列说法正确的是( ) A .甲物体做变速直线运动,乙物体做匀变速直线运动 B .两物体的初速度都为零C .在0~t 1时间内两物体的平均速度大小相等D .甲、乙两物体相遇时速度相等8.当两个力F 1、F 2夹角α(0°< α <180°)保持不变,在同时增大两个分力时,关于它们的合力F 的大小变化叙述正确的是( )A .可能增大B .可能减小C .可能不变D .以上叙述都正确F =10NG =10N 甲 乙9.如图所示,倾角α=30°的斜面顶端有一光滑定滑轮,一根轻绳跨过定滑轮分别拴住物体A 和B ,与A 相连接的这段轻绳与斜面平行。

已知G A =14N ,G B =5N ,物体A 与斜面间的动摩擦因数732=,在竖直段轻绳上的某点C 施加一个方向始终水平的外力F ,使物体B 缓慢移动,而物体A 始终保持静止。

设最大静摩擦力等于滑动摩擦力,轻绳对物体A 的拉力大小为F T ,则此过程中( ) A .F 一直变大 B .F T 一直变大 C .F T 先减小后变大 D .若F 超过12N 时,物体A 将沿斜面向上运动10.在第9题中,物体A 与斜面间的静摩擦力为F f ,则( ) A .F f 一直变大 B .F f 先减小后变大 C .F f 方向始终不变D .F f 方向先沿斜面向上后沿斜面向下11.关于惯性,下列说法中正确的是( ) A .运动员在掷标枪时的助跑是为了利用惯性B .运动员在跳远时的助跑是为了增大起跳时的惯性C .战斗机在空战时,甩掉副油箱是为了减小惯性,提高飞行的灵活性D .踢出去的足球由于受到惯性作用仍然能够在草地上滚动12.如图所示,一水平轻弹簧一端固定,另一端与一物体相连(未拴接),弹簧原长l 0=20cm ,劲度系数k =100N/m ;物体的质量m =1kg ,与水平地面间的动摩擦因数μ=0.2。

现用外力F 使物体静止在A 点,且AO =10cm ,O 点是弹簧原长位置。

现撤去外力F ,物体运动到B 点静止,g =10m/s 2。

则下列说法正确的是( ) A .物体先做匀加速运动后做匀减速运动 B .撤去外力的瞬间物体的加速度大小为8m/s 2C .物体运动至距A 点8cm 时速度最大D .物体运动至O 点时速度最大二、填空题(本大题共3小题,第13题12分,第14、15题每题4分,共20分)13.有四个实验小组用如图甲所示的装置做“探究加速度a 与物体质量m 的关系”的实验。

(1)(多选)某同学想用钩码的总重力表示小车(包括车上的砝码)受到的合外力,为了减小这种做法带来的实验误差,你认为下列说法正确的是( )A .实验时要平衡摩擦力B .实验进行时应先释放小车再接通电源C .钩码的总重力要远小于小车(包括车上的砝码)的总重力D .钩码的总重力要远大于小车(包括车上的砝码)的总重力甲B A FCα 乙A B C D E 单位:cm1.213.045.458.48A O Ba m O a O a1 O m 1 O a 1 m A B CD m 1(2)某小组用接在50Hz 交流电源上的打点计时器测定小车的加速度时,得到一条如图乙所示的纸带,A 、B 、C 、D 、E 为5个相邻的计数点,每两个相邻计数点间有四个打印点没有画出,则小车的加速度大小为 m/s 2,打下D 点时小车的速度大小为 m/s 。

(结果均保留两位有效数字)(3)(多选)为了验证“加速度与物体质量成反比”的猜想,采用图象法处理数据。

图乙是四个小组描绘出的不同图象,其中能够准确验证上述猜想的图象有( )14.如图所示为杂技“顶竿”表演。

一个质量为m 的人(称为“底人”)站在地上,肩上扛一质量为0.1m 的竖直竹竿,当竿上一质量为0.5m 的人以加速度0.1g (g 为重力加速度)加速下滑时,底人对地面的压力大小为 。

15.如图所示,AO 、BO 、CO 、AD 是竖直面内四根固定的光滑细杆,与水平面的夹角依次是60°、45°、30°和90°,每根杆上都套着一个小滑环(可视为质点,图中未画出),四个滑环分别从各细杆顶端无初速释放(运动过程中各环互不影响),用t 1、t 2、t 3、t 4分别表示细杆AO 、BO 、CO 、AD 上的滑环到达底端所用的时间,则t 1、t 2、t 3、t 4的大小关系是 。

三、计算题(本题共4小题,共42分。

解答应写出必要的文字说明.方程式和重要演算步骤,只写出最后答案不能得分,有数值计算的题,答案中必须明确写出数值和单位)16.(10分)如图所示,甲、乙两个质量均为m 、半径均为r 的均匀光滑小球放置在底面直径为3r 的圆柱形容器内,容器放在水平地面上。

求甲球受到容器壁的压力大小和乙球对容器底面的压力大小。

BB AC DO30°45° 60°rr 3r 甲 乙1 v/(m ·s -1) O t/s234 2 4 6 A B αF 17.(10分)如图所示,A 、B 两物块在水平恒力F 的作用下,恰好能一起沿光滑水平面运动,A 与B 接触面光滑,且与水平面夹角α=53°,A 、B 的质量分别为m 和m 38。

求两物块从静止开始运动x =0.9m 的位移时的速度多大。

(g =10m/s 2)18.(10分)在水平地面上有质量为3kg 的物体,物体在水平拉力F 作用下由静止开始运动,3s 后拉力大小不变,方向改为与原来相反,该物体的速度图象如下图所示。

求水平拉力F 和物体与地面间的动摩擦因数μ各为多大。

19.(12分)如图所示,在倾角α=37°的斜面上有相距x =10m 的A 、B 两个物体,与斜面间的动摩擦因数都为μ=0.5。

B 物体从静止开始释放的同时A 物体正沿斜面向上运动,速度v 0=10m/s ,g =10m/s 2。

求:(1)A 物体上升到最高点时与B 物体之间的距离Δx ;(2)B 物体从开始释放起经过多少时间与A 物体相遇。

αBA v 0高一 班 姓名: 考号:南师附中2016~2017学年度上学期高一期末分班考试物理试题答题卷一、选择题(4分×12=48分) 题号 1 2 3 4 5 6 7 8 9 10 11 12 答案二、填空题(第13题12分,第14、15题每题4分,共20分)13. (1) (2) 、 (3) 14. 15.三、计算题(本题共4小题,共42分。

解答应写出必要的文字说明、方程式和重要演算步骤,只写出最后答案不能得分,有数值计算的题,答案中必须明确写出数值和单位)座位号16. 17.r r3r 甲乙 A B α F)(1 5m =1m ×210=2=101分t v x A )1( s 1=s 1010==∴101分a v t )(1 21=2222分t a x A )(1 21+=2222分t a t v x B B B )(1 2m /s =m /s 1×2==1分t a v B B 22-=A B x x x ∆) (1 s 2=s 24==∴2分B v x t ∆3s s 2s 121=+=+=∴t t t )(1 1m =m 1×2×21=21=2211分t a x B B )(1 4m =m )1-5-10(=--=∴11分B A x x x x ∆参考答案二、填空题(20分)13.(1) AC (2)0.60 、 0.27 (3)BC 14. 1.5mg 15.t 1=t 3>t 2>t 4 三、计算题(本题共4小题,每小题10分,共40分) 16.解:对甲球:mg F N1=tan α (2分)°30=⇒21=22-3=sin ααr r r (2分) mg F 33=∴1N (2分) 对甲、乙两球系统:F N2=2mg (2分)根据牛顿第三定律可知:乙球对容器底的压力大小等于2mg 。

(2分) 17.解:对于A 物体:F -F N sin α=ma (2分) F N cos α=mg (2分)对于A 、B 整体:F =(m +m 38)a (2分)解之得:a =0.5g =5m/s 2 (2分)s /m 3=s /m 9.0×5×2=2=∴ax v(2分)18.解:由v -t 图象可知,在0~3s 内,a 1=22112m/s =m/s 0-30-6=t v ∆∆ (2分) 根据牛顿第二定律:F -μmg =ma 1 (2分) 在3~4s 内,a 2=2222-6m/s =m/s 3-46-0=t v ∆∆ (2分) 根据牛顿第二定律:-F -μmg =ma 2 (2分) 解得:F =12N ,μ=0.2 (2分) 19.解:(1)A 向上运动时:m A g sin α+μm A g cos α=m A a 1 ∴ a 1 =g (sin α+μcos α)=10m/s 2(1分)m B g sin α-μm B g cosα=m B a B ∴ a B =g (sin α-μcos α)=2m/s 2 (1分)(2)A 物体向下运动时:m A g sin α-μm A g cos α=m A a 2 ∴ a 2 =g (sin α-μcos α)=2m/s 2 (1分)(1分)FF。

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