黑龙江省哈尔滨师范大学附属中学2016届高三英语上学期12月月考试题
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哈师大附中2020级高三11月份阶段性测试英语试题2022/11本试卷分为第I卷(选择题)和第II卷(非选择题)两部分。
满分150分。
考试时间为100分钟。
第I卷第一部分:阅读理解(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。
AKing’s College Summer SchoolKing’s College Summer School is an annual training program for high school students at all levels who want to improve their English. Courses are given by the teachers of King’s College and other colleges in New York. Trips to museums and culture centers are also organized. This year’s summer school will be from July 25 to August 15. MoreA.Only top students can take part in the program.B.Only the teachers of King’s College give courses.C.King’s College Summer School is run every other year.D.Visits to museums and culture centers are part of the program.2.If you are to live with your relatives in New York, how much will you have to pay the school?A.$200B.$400C.$500D.$9003.What information can you get from the text?A.The program will last two months.B.You can write to Thompson only in English.C.You can get in touch with the school by e-mail or by telephone.D. As a Chinese student, you can send your application on July 14, 2023.BIt was rush hour in the morning. Heather Sant, 36, was driving her white 2012 Mazda on the road with her nine-year-old daughter and ten-year-old son in the back. Suddenly, a red pickup truck cut them off. Sant turned the wheel hard to the right, sending the Mazda slide off the road and down an embankment (路堤) that ended in a drop-off after about 50 feet. If the car didn’t stop, it would take off and dive into the road some 20 feet below. Then a bit of luck: as the car raced toward the edge, its undercarriage got stuck, stopping it suddenly.The people inside, however, were far from safe. The car had come to rest on top of a retaining wall (护墙), balancing on the edge of disaster. One sudden move by anyone inside could throw the car off balance and send it over. Jacob Rodriguez watched everything in terror from the nearby truck company where he works. As an experienced sailor, just like every time he met shipwreck, he whispered a sailing saying “Ship, shipmates, self”, then immediately he and four other men ran to the car. They leaped onto the trunk to balance the weight as the terrified kids in the back seat watched.Meanwhile, Julio and his nephew Marco, were driving to their jobs at nearby Premier Automotive. Julio jumped out of the car to help while Marco went to the shop, grabbed a strong rope, and returned to the car. He tied the Mazda to an F-350 truck that had been driven over by one of the other rescuers. With the car secured, the group carefully opened the back doors and helped the children out. But their departure shifted the cars weight, causing it to tip forward. The men, still on the trunk, begged Sant to jump into the back seat. She did and then inched out of the back door. Finally, the men carefully got off the trunk. Everyone was safe.“Another foot,” Rodriguez told USA Today, “and this would be a different story.”4. What most probably caused the traffic accident?A. The Mazda was overloaded.B. Sant was really speeding.C. The wheel failed to work.D. The pickup cut in abruptly.5. Why did the men ask Sant to move to the back seat?A. To keep the balance of the car.B. To fasten the rope to the truck.C. To open the back door of the car.D. To help the men get off the trunk.6. What does Rodriguez want to stress in the last paragraph?A. The timely help offered by rescuers.B. The excellent cooperation of all the people.C. The serious danger the rescuers faced.D. The dangerous position where the car stopped.7. Which of the following statements is consistent with the theme of the passage?A. Persistence will pay off.B. Doing is better than saying.C. One in trouble, all to help.D. Many hands make light work.CThroughout the history of the arts, the nature of creativity has remained constant to artists. No matter what objects they select, artists are to bring forth new forces and forms that cause change — to find poetry where no one has ever seen or experienced it before.Landscape is another unchanging element of art. It can be found from ancient times through the 17th-century Dutch painters to the 19th-century romanticists and impressionists. In the 1970s Alfred Leslie, one of the new American realists, continued this practice. Leslie sought out the same place where Thomas Cole, a romanticist, had produced paintings of the same scene a century and a half before. Unlike Cole who insists on a feeling of loneliness and the idea of finding peace in nature, Leslie paints what he actually sees. In his paintings, there is no particular change in emotion, and he includes ordinary things like the highway in the background. He also takes advantage of the latest developments of color photography to help both the eye and the memory when he improves his painting back in his workroom.Besides, all art begs the age-old question: What is real? Each generation of artists has shown their understanding of reality in one form or another. The impressionists saw reality in brief emotional effects, the realists in everyday subjects and in forest scenes, and the Cro-Magnon cave people in their habitat site drawings of the animals in the ancient forests. To sum up, understanding reality is a necessary struggle for artists of all periods.Over thousands of years the function of the arts has remained relatively constant. Past or present, Eastern or Western, the arts are a basic part of our immediate experience. Many and different are the faces of art, and together they express the basic need and hope of human beings.8. What does the underlined word “poetry” most probably mean?A. An unusual quality.B. A natural scene.C. A collection of poems.D. An object for artistic creation.9. What is the author’s opinion of artistic reality?A. It will not be found in future works of art.B. It does not have a long-lasting standard.C. It is expressed in a fixed artistic form.D. It is lacking in modern works of art.10. What does the author suggest about the arts in the last paragraph?A. They express people’s curiosity about the past.B. They make people interested in everyday experience.C. They are considered important for variety in form.D. They are regarded as a mirror of the human situation.11. Which of the following is the main topic of the passage?A. History of the arts.B. New developments in the arts.C. Basic questions of the arts.D. Use of modern technology in the arts.DFor most of our history, humans have been short, a study has found. Until around 150 years ago, few people grew taller than 170 centimeters.Christiane Scheffler at the University of Potsdam and Michael Hermanussen in Altenhof have spent several years studying the height of people from a wide range of populations. In their latest paper, they combined an existing data of more than 6000 prehistoric human skeletons with multiple studies of more recent historical populations from Europe and the US. They also included their own data on 1666 present-day school children from Indonesia.In the prehistoric populations, the maximum height for men was 165 to 170 centimeters, while women topped out at 160cm. Today, men in England have an average height of around 175 cm, while for women it is about 162 cm.But there is significant variation between modern countries. The Indonesian school children in the study were shorter than similarly aged children from the US, despite being well-nourished. Scheffler and Hermanussen argue that height can be a signal of dominance, so in societies where it’s possible to move up through the social classes, evolution favors individuals who reach a greater height.Subramanian at Harvard University isn’t convinced by the pair’s interpretation. His team previously showed that the best predictor of a child’s height is the height of their parents. This suggests that the influence of other factors, such as social mobility, is limited.After assessing nearly 163,000 children living in 55 low and middle-income countries, Subramanian’s team found that 42.9 percent had poor nutrition but no sign of stunting (阻碍发育) or other physical indicators of this fact. This implies there is a lot of hidden malnutrition (营养不良) that doesn’t reveal itself through stunting. A person’s nutritional condition should be assessed by looking at their diet not their height, says Subramanian.12. How did Scheffler and Hermanussen conduct the research?A. Studying the skeletons of prehistoric human.B. Connecting existing data with recent research.C. Assessing children living in various income areas.D. Analyzing the results of other scientists’ research.13. What is Subramanian’s attitude towards the explanation of Scheffler and Hermanussen?A. Worried.B. Cautious.C. Supportive.D. Doubtful.14. What can we infer from Subramanian’s study?A. Poor nutrition delays physical development.B. A balanced diet contributes to growing taller.C. A human’s height has little to do with nutrition.D. High social classes can reach a greater height.15. What is the text mainly about?A. The significance that lies in nutrition.B. The factors that influence human’s height.C. The importance that humans attach to height.D. The reasons why prehistoric humans were short.第二节(共5小题;每小题2分,满分10分)根据短文内容,从短文后的选项中选出填入空白处的最佳选项。
文科数学参考答案13. 存在x ∈R ,x 2+x +1≥0 14. 2 15. 16. (2)(3)三.解答题 17. 解:(Ⅰ)设公差为,由已知得, …………2分则()()232331142d d d ⨯⎛⎫+=++ ⎪⎝⎭,整理得, ∵,∴, …………4分 ∴ …………6分(Ⅱ)由(1)得()2122n n n S n n -=+⋅= …………7分 ∴()()21111141212122121n b n n n n n ⎛⎫===- ⎪-+--+⎝⎭…………9分 ∴1111111213352121n T n n ⎛⎫=-+-++- ⎪-+⎝⎭ 11122121n n n ⎛⎫=-= ⎪++⎝⎭ …………12分 18.解:(Ⅰ)…………3分由茎叶图知:第一组学生身高更集中. …………4分 (Ⅱ)第一组男生身高的众数为168,中位数为171.5 …………6分 (Ⅲ)第一组身高位于的男生共5人,记为A,B,C,D,E ,其中位于共有两人,记为D,E第二组身高位于共3人,记为F,G,H, 其中位于有1人记为H从两组中各选出一个(A,F),(A,G),(A,H), (B,F),(B,G),(B,H) ,(C,F),(C,G),(C,H), (D,F),(D,G),(D,H), (E,F),(E,G),(E,H)共15种.这两人均位于共有(D,H ),(E,H)2种.设”这2人身高均位于”为事件A…………12分 17 18 15 16 8 9 6 9 2 3 6 2 5 5 第一组 第二组 9 8 8 6 5 2 1 0 2 119.(Ⅰ)证明:∵PH ⊥平面ABCD ,AB 平面ABCD ,∴PH ⊥AB …………2分∵AB ⊥AD ,AD ∩PH=H ,AD ,PH 平面PAD∴AB ⊥平面PAD …………4分又AB 平面PAB ,∴平面PAB ⊥平面PAD . …………5分(Ⅱ)解:取CD 中点E ,PC 中点F , 连HF ,HE ,EF则EF PD∵PH ⊥平面ABCD ,AD 平面ABCD ,∴PH ⊥AD∵PA=PD ,∴H 是AD 中点, ∴EH AC∴∠FEH 是异面直线AC 与PD 所成角或其补角. …………7分Rt △ABC 中,AB=BC=1,∴AC=,∴EH= …………8分Rt △APD 中,PA=PD ,AD=2,∴PD=,∴EF= …………9分∵BC ∥AH 且BC=AH=1,∴四边形ABCH 是平行四边形,∴CH=AB=1Rt △PHC 中,PH=1,∴PC=,∴HF= …………10分△EFH 中,EF=EH=HF ,∴∠FEH=60o ∴异面直线AC 与PD 所成角的大小为60o . …………12分20.解:(Ⅰ),且的面积为,,2135,22PF PF ∴=∴==3524,222a a =+=∴= …………3分椭圆方程为 …………4分(Ⅱ)设,不妨设 线段的垂直平分线方程为()y x x y x y +--=--00001122, …………5分 线段的垂直平分线方程为 …………6分()y x x y x y x +-⎧-=--⎪⎨⎪=⎩000011220 ()()x x y x y y y y +--∴=--+=+200000001112222 …………9分点到轴的距离 …………10分在上为减函数 ,当时, …………12分21.解:(Ⅰ)110,()(ln )0x f e a b e e>=+=, …………1分 , …………2分①当时,令得,,的增区间为令得,,的减区间为 …………4分②当时,令得,,的增区间为令得得,,的减区间为 …………6分 (Ⅱ)()()1ln xf x g x x a=-=当时,221211()()ln ln ln x g x g x x x x -=-= …………8分 只需221211222211212(1)2()ln 1()x x x x x x x x x x x -->=++ ,令, ,…………10分 2222(1)(21)()0(1)t t t F t t t -+-'=>+在上单调递增, ,原命题成立 …………12分22.(Ⅰ)证明:是圆的切线 且……2分又 ……4分即是等腰三角形 ……5分(Ⅱ)证明: 由(1)知 ……6分又CEDBEF BFD ∠=∠=∠且 ……8分 又 ……10分23. 解:(Ⅰ)曲线022:22=--+y x y x C 的极坐标方程为0sin 2cos 22=--θρθρρ,即 ……2分设点的极坐标为,则,θθρsin 2cos 22+= ,, ……4分与不重合,与不重合()0sin cos >+=∴ρθθρ ……5分 (Ⅱ)由(Ⅰ)可设点的极坐标为,点的极坐标为,则2313sin 3cos 1+=+==ππρOA , 2136sin 6cos 2+=+==ππρOB , ………7分 2123231222316cos 2222122212=⋅⎪⎪⎭⎫ ⎝⎛+⋅-⋅⎪⎪⎭⎫ ⎝⎛+=-+=πρρρρAB ………9分 的周长为2231++=++AB OB OA ………10分 24.解:(Ⅰ)由已知得 ,则即 ………5分(Ⅱ)恒成立,即恒成立-+-≥---=-……7分x x a x x a a33()3当且仅当时取等号故的取值范围为或………10分。
黑龙江省哈尔滨师范大学附属中学 2016届高三上学期期中考试数学(理)试题考试时间:120分钟 满分:150分第Ⅰ卷(选择题 共60分)一、 选择题(本大题共12小题,每小题5分,共60分. 在每小题给出的四个选项中,只有一项是符合题目要求的) 1.已知集合,,那么集合 A . B . C . D . 2.已知不共线的向量,,,则 A . B . C . D .3.等差数列中,35710133()2()24a a a a a ++++=,则这个数列的前13项和为 A .13 B .26 C .52 D .156 4.右图是一个几何体的三视图,根据图中数据,可得该几何体的体积是 A . B . C . D . 5.将函数的图象向右平移个单位长度,所得图象 关于点对称,则的最小值是 A . B .1 C . D . 2 6.设,则sin()cos()sin()cos()αππααππα-+-=+--A .B .1C .3D . -1 7.设是由正数组成的等比数列,为其前项和,已知,则 A . B . C . D . 8.定义在上的奇函数满足且,则A . -2B .0C .2D .4 9.已知函数命题:(0,),()02p x f x π∀∈<,则A .是真命题,00:(0,),()02p x f x π⌝∃∈≥ B .是真命题,:(0,),()02p x f x π⌝∀∈>C .是假命题,:(0,),()02p x f x π⌝∀∈≥ D .是假命题,00:(0,),()02p x f x π⌝∃∈≥ 10.已知函数(12)3,1()ln ,1a x a x f x x x -+<⎧=⎨≥⎩的值域为,则实数的取值范围是A .B .C .D .11.在中,角的对边分别是,若cos (2)cos c a B a b A -=-,则的形状是A .等腰三角形B .直角三角形C .等腰直角三角形D .等腰或直角三角形12.已知函数,若,且对任意恒成立,则的最大值为P A B C D EA .3B .4C .5D .6第Ⅱ卷(非选择题 共90分)二、填空题(本大题共4小题,每小题5分,共20分) 13.等差数列中,12342,4a a a a +=+=,则 .14.设为锐角,若则 . 15.已知向量,,在轴上存在一点使有最小值,则点的坐标是 .16.在平面直角坐标系中,使角的顶点与原点重合,角的始边与轴的非负半轴重合.已知点是角终边上一点,,定义.对于下列说法:①函数的值域是; ②函数的图象关于原点对称;③函数的图象关于直线对称; ④函数是周期函数,其最小正周期为; ⑤函数的单调递减区间是32,2,.44k k k Z ππππ⎡⎤-+∈⎢⎥⎣⎦其中正确的是 .(填上所有正确命题的序号)三、解答题(本大题共6小题,共70分. 解答应写出文字说明、证明过程或演算步骤) 17.(本题满分12分)已知数列的前项和为,且1110,910n n a a S +==+. (Ⅰ)求证:是等差数列; (Ⅱ)设,求数列的前项和.18.(本题满分12分)已知向量函数.(Ⅰ)求函数的图象的对称中心和单调递增区间;(Ⅱ)在中,角的对边分别是,且()3,1,f C c ab ===且,求的值.19.(本题满分12分)四棱锥P -ABCD 中,直角梯形ABCD 中,AD ⊥CD ,AB ∥CD ,∠APD =60°,P A =CD =2PD =2AB =2,且平面PDA ⊥平面ABCD ,E 为PC 的中点.(Ⅰ)求证:PD ⊥平面ABCD ;(Ⅱ)求直线PD 与平面BDE 所成角的大小.20.(本题满分12分)如图,在长方体ABCD -A 1B 1C 1D 1中,AB =AA 1=1,E 为BC 中点. (Ⅰ)求证:C 1D ⊥D 1E ;A 1B 1C 1D 1ABC DE(Ⅱ)在棱AA 1上是否存在一点M ,使得BM ∥平面AD 1E ? 若存在,求的值;若不存在,说明理由;(Ⅲ)若二面角B 1-AE -D 1的大小为90°,求AD 的长.21.(本题满分12分) 设函数()()1ln 2++=x a x x f ,其中.(Ⅰ)当时,求曲线在原点处的切线方程;(Ⅱ)试讨论函数极值点的个数;(Ⅲ)求证:对任意的,不等式恒成立.请从下面所给的22 , 23 ,24三题中任选一题做答,如果多做,则按所做的第一题计分. 22.(本题满分10分)选修4—1:几何证明选讲已知AB 是半圆O 的直径,AB =4,点C 是半圆O 上一点,过C 作半圆O 的切线CD ,过点A 作AD ⊥CD 于D ,交半圆于E ,DE =1.(Ⅰ)求证:AC 平分∠BAD ; (Ⅱ)求BC 的长.23.(本题满分10分)选修4—4:坐标系与参数方程在平面直角坐标系中,以坐标原点为极点,轴的正半轴为极轴建立极坐标系,半圆C 的极坐标方程为2sin ,0,.2πρθθ⎡⎤=∈⎢⎥⎣⎦(Ⅰ)求C 的参数方程;(Ⅱ)设点D 在C 上,C 在D 处的切线与直线垂直,根据(Ⅰ)中的参数方程,确定点D 的坐标. 24.(本题满分10分)选修4—5:不等式选讲 (Ⅰ)已知不等式的解集是,求实数的值; (Ⅱ)已知实数满足,求的最大值.参考答案1-6:BABADC 7-12:BAACDB13、 6 14、 15、(3,0) 16、 ①③④ 17.(1)当时,由,得 ,相减得:当时,11210100109a S a ==+=,∴,n n n a a a lg 1)10lg(lg 1+==∴+, ,又是首项为1,公差为1的等差数列. 6‘(2)()⎪⎭⎫ ⎝⎛+-=+=111212n n n n b n ,则11111212231n T n n ⎛⎫=-+-++- ⎪+⎝⎭L = 12‘18、解:(1)2()2cos 2cos212==+f x x x x x 2‘ 令,,对称中心为4‘ 令222,262πππππ-≤+≤+∈k x k k Z ,,36ππππ-≤≤+∈k x k k Z增区间:,36ππππ⎡⎤-+∈⎢⎥⎣⎦k k k Z 6‘(2)()2sin 2136π⎛⎫=++= ⎪⎝⎭f C C ,, ,132,666πππ∴<+<C , 8‘ ()2222222cos 2=+-=+-=+-c a b ab C a b a b ab ,,,且,12‘19、解:(1)2,1,60,==∠=oQ PA PD PAD2222cos 3∴=+-⋅∠=AD PA PD PA PD PAD ,,,又平面,平面平面,平面平面,平面6‘ (2),以分别为轴,轴,轴,建立空间直角坐标系EDC B AD 1C 1B 1A 1MNz yx MA 1B 1C 1D1ABC D E1(0,0,0),(0,0,1),(0,1,),2D P EB 1(0,1,),2∴==uuu r uu ur DE DB ,设平面的一个法向量为,则1020⎧+=⎪+=y z y ,令, cos ,∴〈〉==uu u r r DP n ,设直线与平面所成的角为,,直线与平面所成的角为 12‘20.方法一: 证明:(1)连D 1C ,长方体中,EC ⊥平面DCC 1D 1,∴EC ⊥DC 1∵AB=AA 1,∴正方形DCC 1D 1中,D 1C ⊥DC 1又EC∩D 1C=C ,∴DC 1⊥平面ECD 1 ∵D 1E 面ECD 1,∴C 1D ⊥D 1E 4‘解:(2)存在点M 为AA 1中点,使得BM ∥平面AD 1E .证明:取A 1D 1中点N ,连BM ,MN ,NB∵E 为BC 中点,∴ND 1 BE∴四边形BED 1N 是平行四边形,∴BN ∥D 1E 又BN 平面AD 1E ,D 1E 平面AD 1E∴BN ∥平面AD 1E∵MN AD 1,MN 平面AD 1E ,AD 1平面AD 1E ∴MN ∥平面AD 1E ∵BN∩MN=N ,∴平面BMN ∥平面AD 1E∵BM 平面BMN ,∴BM ∥平面AD 1E 此时, 8‘ 方法二:证明:(1)以D 为原点,如图建立空间直角坐标系D-xyz ,设AD=a , 则D(0,0,0),A(a ,0,0),B(a ,1,0),B 1(a ,1,1),C 1(0,1,1),D 1(0,0,1),E(,1,0),∴11(0,1,1),(,1,1)2aC D D E =--=-uuu r uuu r ,∴,∴C 1D ⊥D 1E 4‘解:(2)设,则,∴,1(,1,0),(,0,1)2aAE AD a =-=-uu u r uuu r ,设平面AD 1E 的法向量 ,则1020a AE x y AD ax z ⎧⋅=-+=⎪⎨⎪⋅=-+=⎩uu u r uuu r,∴平面AD 1E 的一个法向量∵BM ∥平面AD 1E ,∴ ,即,∴即在存在AA 1上点M ,使得BM ∥平面AD 1E ,此时.8‘解:(3)设平面B 1AE 的法向量 ,1(,1,0),(0,1,1)2aAE AB =-=uu u r uuu r则1020a AE x y AB y z ⎧''⋅=-+=⎪⎨⎪⋅=+=⎩uu u r uuu r,∴平面B 1AE 的一个法向量∵二面角B 1-AE-D 1的大小为90°,∴ ⊥ ,∴22420a a ⋅=+-=∵a >0,∴a =2,即AD=2. 12‘21.解:(1)当时,,则,曲线在原点处的切线方程为 2‘(2)()1,122122'->+++=++=x x a x x x a x x f ,令()1,222->++=x a x x x g 当时,,所以0,则0,所以在上为增函数,所以无极值点; 当时,,所以0,则0,所以在上为增函数, 所以无极值点; 当时,,令0,则, 当时,,,此时有2个极值点; 当时,,,此时有1个极值点; 综上:当时,无极值点; 当时,有2个极值点;当时,有1个极值点; 8‘ (3)对于函数,令函数()332()ln(1)h x x f x x x x =-=-++则()32213(1)3211x x h x x x x x +-'=-+=++,,所以函数在上单调递增, 又时,恒有 即恒成立. 取,则有()()321111111ln +-+>⎪⎭⎫ ⎝⎛++n n n 恒成立, 即不等式恒成立. 12‘22.解:(1)连接OC, 因为OA=OC,所以∠OAC=∠OCA因为CD 为半圆O 的切线,所以OC ⊥CD,因为AD ⊥CD,所以OC ∥AD,所以∠OCA=∠CAD,∠OAC=∠CAD, 所以AC 平分∠BAD………………5分 (2)连接CE,有(1)知∠OAC=∠CAD,所以BC=CE.因A,B,C,D 四点共圆,故∠ABC=∠CED,因为AB 是半圆O 的直径, 所以∠ACB 是直角, Rt △CDE 相似于Rt △ACB,DE:CE=CB:AB,BC=2.………………10分23. 解 (I)半圆C 的普通方程为; []2220,0,1,x y y x +-=∈ ………………2分半圆C 的参数方程为cos ,,1sin .22x y αππαα=⎧⎛⎫⎡⎤∈-⎨⎪⎢⎥=+⎣⎦⎝⎭⎩为参数 ………………5分(II)设点D 对应的参数为,则点D 的坐标为且 由(1)可知半圆C 的圆心是C(0,1),因半圆C 在D 处的切线与直线垂直,故直线DC 的斜率与直线的斜率相等,(1sin )1tan cos ααα+-==即,,,226πππαα⎡⎤∈-∴=⎢⎥⎣⎦………………8分所以点D 的坐标为3,22⎛⎫⎪ ⎪⎝⎭………………10分24.解 (I)28,80,8+≤++≥≥-x t t t t 得所以 ,828,44,t x t t t x --≤+≤+--≤≤由的解集是得(II)由柯西不等式得()()222221491234923y z y z x xx y z ⎛⎫⎛⎫++++≥++=++ ⎪ ⎪⎝⎭⎝⎭ ()228,77x y z x y z ≥++≤++ 当且仅当320123zy x ==>即22224949y z y z x x ==++=>0且, 亦即x y z ===时()。
哈尔滨市第六中学2016届高三上学期12月月考数学试卷(文)第Ⅰ卷(选择题 共60分)一、选择题:本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一个是符合题目要求的.1.若复数z 满足)1(21i z i +-=⋅,则z 的共轭复数的虚部是( ) A .i 21- B .i 21 C .21- D .212.设{}62|≤≤=x x A ,{}32|+≤≤=a x a x B ,若A B ⊆,则实数a 的取值范围是( )A .[]3,1 B .),3[+∞ C .),1[+∞ D .()3,1 3.下列四种说法中,正确的个数有 ( )① 命题",x R ∀∈均有"0232≥--x x 的否定是:0",x R ∃∈使得200320"x x --≤;② “命题Q P ∨为真”是“命题Q P ∧为真”的必要不充分条件;③ R m ∈∃,使mmmxx f 22)(+=是幂函数,且在),0(+∞上是单调递增;④ 不过原点(0,0)的直线方程都可以表示成1=+bya x ; A .3个 B .2个 C .1个 D .0个4.如图是底面积为3,体积为3的正三棱锥的主视图(等腰三角形) 和左视图,此正三棱锥的左视图的面积为 ( ) A .332B .3C .3D .325.设z x y =+,其中实数x ,y 满足2000x y x y y k +≥⎧⎪-≤⎨⎪≤≤⎩,若z 的最大值为6,则z 的最小值为( )A .3-B .2-C .1-D .0 6.将函数sin(6)4y x π=+的图象上各点的横坐标伸长到原来的3倍(纵坐标不变),再向右平移8π个单位,所得函数图像的一个对称中心是 ( )A .,016π⎛⎫⎪⎝⎭B .,09π⎛⎫ ⎪⎝⎭C .,04π⎛⎫ ⎪⎝⎭D .,02π⎛⎫⎪⎝⎭7.若数列1a ,21a a ,32aa ,…,1n n a a -是首项为1,公比为2-的等比数列,则4a 等于 ( )A .8-B .22-C .22D .8 8.数列{}n a 满足11=a ,对任意的*N n ∈都有n a a a n n ++=+11,则122016111a a a +++=( ) A .20152016 B .40322017 C .40342017 D .201620179.定义在R 上的奇函数()f x 满足()()1fx f x +=-,当10,2x ⎛⎤∈ ⎥⎝⎦时,()()2l o g 1f x x =+,则()f x 在区间31,2⎛⎫⎪⎝⎭内是 ( ) A .减函数且()0f x < B .减函数且()0f x > C .增函数且()0f x > D .增函数且()0f x <10.若函数()()()()2010x a x f x x ax x ⎧-≤⎪=⎨++>⎪⎩的最小值为()0f ,则实数a 的取值范围是 ( )A .[]1,2-B .[]1,0-C .[]1,2D .[]0,2 11.在ABC ∆中,,,a b c 分别为角,,A B C 的对边,若2cos 22B a cc+=,则ABC ∆的形状为( ) A .正三角形 B .直角三角形C .等腰三角形D .等腰三角形或直角三角形12.给出以下命题,其中正确的命题的个数是 ( )① 存在两个不等实数,αβ,使得等式sin()sin sin αβαβ+=+成立; ② 若数列{}n a 是等差数列,且(*)m n s t a a a a m n s t N +=+∈、、、,则m n s t +=+;③ 若n S 是等比数列{}n a 的前n 项和,则61261812S ,,S S S S --成等比数列;④ 若n S 是等比数列{}n a 的前n 项和,且(n n S Aq B A B =+∈*其中、是非零常数,n N ), 则0A B +=;⑤ 已知ABC ∆的三个内角,,A B C 所对的边分别为,,a b c ,若222a b c +>, 则ABC ∆一定是锐角三角形;A .1个B .2个C .3个D .4个第Ⅱ卷(非选择题 共90分)二、填空题:本大题共4小题,每小题5分,共20分.将答案填在机读卡上相应的位置. 13.对某同学的6次物理测试成绩(满分100分)进行统计,作出的茎叶图如图所示,给出关于该同学物理成绩的以下说法:①中位数为84;②众数为85; ③平均数为85; ④极差为12;其中,正确说法的序号是____________;14.某程序框图如图所示,该程序运行后输出的S 的值是__________; 15.ABC ∆的外接圆圆心为O ,半径为2,0OA AB AC ++=,则CB 在CA 方向上的投影为____________; 16.已知正三角形C AB 的三个顶点都在半径为2的 球面上,球心O 到平面C AB 的距离为1, 点E 是线段AB 的中点,过点E 作球O 的截面, 则截面面积的最小值是_________;三、解答题:本大题共6小题,共70分.解答时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12分)已知向量,2sin ),cos ,(cos ),sin ,(sin C n m A B n B A m =⋅==且A 、B 、C 分别为△ABC 的三边a 、b 、c 所对的角. (1)求角C 的大小;(2)若18)(,sin ,sin ,sin =-⋅AC AB CA B C A 且成等差数列,求c 边的长. 18.(本小题满分12分)甲乙两个学校高三年级分别有1200人,1000人,为了了解两个学校全体高三年级学生在该地区六校联考的数学成绩情况,采用分层抽样方法从两个学校一共抽取了110名学生的数学成绩,并作出了频数分布统计表如下: 甲校:乙校:(1)计算x ,y 的值.(2)若规定考试成绩在内为优秀,请分别估计两个学校数学成绩的优秀率;(3)由以上统计数据填写下面2×2列联表,并判断是否有90%的把握认为两个学校的数学成绩有差异.参考公式:22()()()()()n ad bc K a b c d a c b d -=++++临界值表P (K ≥k 0) 0.10 0.05 0.010 k 02.7063.8416.635分组 [70,80)[80,90)[90,100)[100,110)频数 3 4 8 15 分组 [110,120)[120,130)[130,140)频数 15x32分组 [70,80)[80,90)[90,100)[100,110)频数 1 2 8 9 分组 [110,120)[120,130)[130,140)频数1010y3甲校 乙校 总计 优秀 非优秀 总计19.(本小题满分12分)如图,已知棱柱1111D C B A ABCD -的底面是菱形,且⊥1AA 面ABCD , F AA AD DAB ,1,601==︒=∠为棱1AA 的中点,M 为线段1BD 的中点. (1)求证:平面⊥FB D 1平面11B BDD ; (2)求三棱锥BDF D -1的体积.20.(本小题满分12分)已知椭圆12222=+by a x (0>>b a )的离心率为22,且短轴长为2.(1)求椭圆的方程;(2)若与两坐标轴都不垂直的直线l 与椭圆交于B A ,两点,O 为坐标原点, 且32=⋅OB OA ,32=∆AOB S ,求直线l 的方程.21.(本小题满分12分)已知函数R x a x e x f x ∈+-=,)(2的图像在点0=x 处的切线为bx y =. (1)求函数)(x f 的解析式;(2)当R x ∈时,求证:x x x f +-≥2)(;(3)若kx x f >)(对任意的),0(+∞∈x 恒成立,求实数k 的取值范围;请考生在第22、23、24题中任选一题作答,如果多做,则按所做的第一题计分.作答时,用2B 铅笔在答题卡上把所选题目对应的题号涂黑. 22.(本小题满分10分)选修4一1:几何证明选讲如图,在ABC ∆中, 90=∠B ,以AB 为直径的圆O 交AC 于D , 过点D 作圆O 的切线交BC 于E ,AE 交圆O 于点F .(1) 证明:E 是BC 的中点;(2)证明:AF AE AC AD ⋅=⋅.23.(本小题满分10分)选修4一4:坐标系与参数方程已知极坐标系的极点在平面直角坐标系的原点O 处,极轴与x 轴的正半轴重合,且长度单位相同;曲线C 的方程是)4sin(22πθρ-=,直线l 的参数方程为⎩⎨⎧+=+=ααsin 2cos 1t y t x (t 为参数,πα<≤0), 设)2,1(P ,直线l 与曲线C 交于B A ,两点. (1)当0=α时,求||AB 的长度; (2)求22||||PB PA +的取值范围.24.(本小题满分10分)选修4一5:不等式选讲已知函数|2|)(-=x x f ,m x x g ++-=|3|)(. (1)解关于x 的不等式01)(>-+a x f (R a ∈);(2)若函数)(x f 的图象恒在函数)(x g 图象的上方,求m 的取值范围.参考答案一、 选择题 :二、填空题:13. ①③ 14. 3018 15. 3 16.94π17.(本小题满分12分)解:(1))sin(cos sin cos sin B A A B B A n m +=⋅+⋅=⋅对于C B A C C B A ABC sin )sin(0,,=+∴<<-=+∆ππ,.sin C n m =⋅∴ 又C n m 2sin =⋅ ,.3,21cos ,sin 2sin π===∴C C C C(2)由B A C B C A sin sin sin 2,sin ,sin ,sin +=得成等差比数列,由正弦定理得.2b a c +=18,18)(=⋅∴=-⋅CB CA AC AB CA ,即.36,18cos ==ab C ab 由余弦弦定理ab b a C ab b a c 3)(cos 22222-+=-+=,36,3634222=⨯-=∴c c c ,.6=∴c18、(本小题满分12分) 解:(Ⅰ)甲校抽取110×12002200=60人, 乙校抽取110×10002200=50人,故x =10,y =7, ………4分 (Ⅱ)估计甲校优秀率为1525%60=, 乙校优秀率为2050=40%. ………8分(Ⅲ) k 2=2110(15302045)60503575⨯-⨯⨯⨯⨯≈2.83>2.706又因为 1-0.10=0.9,故有90%的把握认为两个学校的数学成绩有差异。
2016-2017学年黑龙江省哈尔滨师大附中高三(上)期中数学试卷(理科)一、选择题(本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的)1.(5分)复数的虚部()A.i B.﹣i C.1 D.﹣12.(5分)已知集合,则A∩B=()A.(1,+∞)B.[1,+∞)C.(﹣∞,0]∪(1,+∞) D.[0,1]3.(5分)已知函数f(x)是奇函数,且当x>0时,f(x)=x2+,则f(﹣1)=()A.﹣2 B.0 C.1 D.24.(5分)在区间[0,π]上随机取一个数x,使的概率为()A.B.C.D.5.(5分)若|+|=|﹣|=2||,则向量+与的夹角为()A.B.C. D.6.(5分)如果对于任意实数x,[x]表示不超过x的最大整数.例如[3.27]=3,[0.6]=0.那么“[x]=[y]”是“|x﹣y|<1”的()A.充分而不必要条件B.必要而不充分条件C.充分必要条件D.既不充分也不必要条件7.(5分)二项式(x2﹣)11的展开式中,系数最大的项为()A.第五项B.第六项C.第七项D.第六和第七项8.(5分)根据如图所示程序框图,若输入m=42,n=30,则输出m的值为()A.0 B.3 C.6 D.129.(5分)数列{a n}的前n项和为S n,若a1=1,a n+1=3S n(n≥1),则a6=()A.3×44B.3×44+1 C.44D.44+110.(5分)若α∈(,π)且3cos2α=4sin(﹣α),则sin2α的值为()A.B.﹣ C.﹣ D.11.(5分)身穿红、黄两种颜色衣服的各有两人,身穿蓝颜色衣服的有一人,现将这五人排成一行,要求穿相同颜色衣服的人不能相邻,则不同的排法共有()A.24种B.48种C.36种D.28种12.(5分)已知函数f(x)的导函数f′(x)=2+sinx,且f(0)=﹣1,数列{a n}是以为公差的等差数列,若f(a2)+f(a3)+f(a4)=3π,则=()A.2016 B.2015 C.2014 D.2013二、填空题(本大题共4小题,每小题5分,共20分.将答案填在答题卡相应的位置上)13.(5分)将高三(1)班参加体检的36名学生,编号为:1,2,3, (36)若采用系统抽样的方法抽取一个容量为4的样本,已知样本中含有编号为6号、24号、33号的学生,则样本中剩余一名学生的编号是.14.(5分)已知,则|a0|+|a1|+|a2|+…+|a9|=.15.(5分)袋子中装有大小相同的6个小球,2红4白,现从中有放回的随机摸球3次,每次摸出1个小球,则至少有2次摸出白球的概率为.16.(5分)已知x,y∈R,满足x2+2xy+4y2=6,则z=x2+4y2的取值范围为.三、解答题(本大题共5小题,共70分.解答应写出文字说明、证明过程或演算步骤)17.(12分)△ABC的内角A,B,C所对的边分别为a,b,c,向量=(a,c),=(1﹣2cosA,2cosC﹣1),(Ⅰ)若b=5,求a+c值;(Ⅱ)若,且角A是△ABC中最大内角,求角A的大小.18.(12分)中国乒乓球队备战里约奥运会热身赛暨选拨赛于2016年7月14日在山东威海开赛,种子选手A与非种子选手B1,B2,B3分别进行一场对抗赛,按以往多次比赛的统计,A获胜的概率分别为,且各场比赛互不影响.(Ⅰ)若A至少获胜两场的概率大于,则A入选征战里约奥运会的最终名单,否则不予入选,问A是否会入选最终的名单?(Ⅱ)求A获胜场数X的分布列和数学期望.19.(12分)已知各项为正数的数列{a n}的前n项和为S n,且满足(Ⅰ)求证:{a n}为等差数列,并求数列{a n}的通项公式;(Ⅱ)设,求证:.20.(12分)已知函数f(x)=x﹣2sinx.(Ⅰ)求函数f(x)在上的最值;(Ⅱ)若存在,使得不等式f(x)<ax成立,求实数a的取值范围.21.(12分)已知函数,其中a,b,c∈R.(Ⅰ)若a=b=1,求函数f(x)的单调区间;(Ⅱ)若a=0,且当x≥0时,f(x)≥1总成立,求实数b的取值范围;(Ⅲ)若a>0,b=0,若f(x)存在两个极值点x1,x2,求证;f(x1)+f(x2)<e.[选作题]22.(10分)已知函数f(x)=|x﹣a|﹣2.(Ⅰ)若a=1,求不等式f(x)+|2x﹣3|>0的解集;(Ⅱ)若关于x的不等式f(x)<|x﹣3|恒成立,求实数a的取值范围.[选作题]23.(Ⅰ)已知x2+y2=1,求2x+3y的取值范围;(Ⅱ)已知a2+b2+c2﹣2a﹣2b﹣2c=0,求证:.2016-2017学年黑龙江省哈尔滨师大附中高三(上)期中数学试卷(理科)参考答案与试题解析一、选择题(本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的)1.(5分)复数的虚部()A.i B.﹣i C.1 D.﹣1【解答】解:复数==1﹣i的虚部为﹣1.故选:D.2.(5分)已知集合,则A∩B=()A.(1,+∞)B.[1,+∞)C.(﹣∞,0]∪(1,+∞) D.[0,1]【解答】解:∵集合,∴A={x|x≤0或x>1},B={y|y≥1},∴A∩B=(1,+∞).故选:A.3.(5分)已知函数f(x)是奇函数,且当x>0时,f(x)=x2+,则f(﹣1)=()A.﹣2 B.0 C.1 D.2【解答】解:∵f(x)是定义在R上的奇函数,∴f(﹣x)=﹣f(x),f(﹣1)=﹣f(1),又当x>0时,f(x)=x2+,∴f(1)=12+1=2,∴f(﹣1)=﹣2,故选:A.4.(5分)在区间[0,π]上随机取一个数x,使的概率为()A.B.C.D.【解答】解:∵0≤x≤π,,∴≤x≤π,区间长度为,则对应的概率P==,故选:B.5.(5分)若|+|=|﹣|=2||,则向量+与的夹角为()A.B.C. D.【解答】解:作,,以OA,OB为邻边作平行四边形OACB,则=.∵|+|=|﹣|=2||,∴四边形OACB为矩形,∴==,∴向量+与的夹角为.故选:A.6.(5分)如果对于任意实数x,[x]表示不超过x的最大整数.例如[3.27]=3,[0.6]=0.那么“[x]=[y]”是“|x﹣y|<1”的()A.充分而不必要条件B.必要而不充分条件C.充分必要条件D.既不充分也不必要条件【解答】解:[x]=[y]⇒﹣1<x﹣y<1即|x﹣y|<1而取x=1.9,y=2.1,此时|x﹣y|=0.2<1,而[x]=1,[y]=2,[x]≠[y]∴“[x]=[y]”是“|x﹣y|<1”的充分而不必要条件故选:A.7.(5分)二项式(x2﹣)11的展开式中,系数最大的项为()A.第五项B.第六项C.第七项D.第六和第七项【解答】解:二项式(x2﹣)11的展式的通项公式为T r=•x22﹣2r•(﹣1)r•x+1﹣r =•x22﹣3r,故当r=6时,展开式的系数=最大,故选:C.8.(5分)根据如图所示程序框图,若输入m=42,n=30,则输出m的值为()A.0 B.3 C.6 D.12【解答】解:第一次执行循环体后,r=12,m=30,n=12,不满足退出循环的条件;第二次执行循环体后,r=6,m=12,n=6,不满足退出循环的条件;第三次执行循环体后,r=0,m=6,n=0,满足退出循环的条件;故输出的m值为6,故选:C.9.(5分)数列{a n}的前n项和为S n,若a1=1,a n+1=3S n(n≥1),则a6=()A.3×44B.3×44+1 C.44D.44+1【解答】解:由a n=3S n,得到a n=3S n﹣1(n≥2),+1﹣a n=3(S n﹣S n﹣1)=3a n,两式相减得:a n+1=4a n(n≥2),又a1=1,a2=3S1=3a1=3,则a n+1得到此数列除去第一项后,为首项是3,公比为4的等比数列,所以a n=a2q n﹣2=3×4n﹣2(n≥2)则a6=3×44.故选:A.10.(5分)若α∈(,π)且3cos2α=4sin(﹣α),则sin2α的值为()A.B.﹣ C.﹣ D.【解答】解:∵α∈(,π),且3cos2α=4sin(﹣α),∴3(cos2α﹣sin2α)=4(cosα﹣sinα),化简可得:3(cosα+sinα)=2,平方可得1+sin2α=,解得:sin2α=﹣,故选:C.11.(5分)身穿红、黄两种颜色衣服的各有两人,身穿蓝颜色衣服的有一人,现将这五人排成一行,要求穿相同颜色衣服的人不能相邻,则不同的排法共有()A.24种B.48种C.36种D.28种【解答】解:由题意知先使五个人的全排列,共有A55=120种结果.穿红色相邻或穿黄色相邻两种情况,有2A22A44=96种,穿红色相邻且穿黄色也相邻情况,有A22A22A33=24种,故:穿相同颜色衣服的人不能相邻的排法是120﹣96+24=48,故选:B.12.(5分)已知函数f(x)的导函数f′(x)=2+sinx,且f(0)=﹣1,数列{a n}是以为公差的等差数列,若f(a2)+f(a3)+f(a4)=3π,则=()A.2016 B.2015 C.2014 D.2013【解答】解:∵函数f(x)的导函数f′(x)=2+sinx,可设f(x)=2x﹣cosx+c,∵f(0)=﹣1,∴﹣1+c=﹣1,可得c=0.∴f(x)=2x﹣cosx.∵数列{a n}是以为公差的等差数列,∴a n=a1+(n﹣1)×,∵f(a2)+f(a3)+f(a4)=3π,∴2(a2+a3+a4)﹣(cosa2+cosa3+cosa4)=3π,∴6a2+﹣cosa2﹣﹣=3π,∴6a2﹣=.令g(x)=6x﹣cos﹣,则g′(x)=6+sin在R上单调递增,又=0.∴a2=.则==2015.故选:B.二、填空题(本大题共4小题,每小题5分,共20分.将答案填在答题卡相应的位置上)13.(5分)将高三(1)班参加体检的36名学生,编号为:1,2,3, (36)若采用系统抽样的方法抽取一个容量为4的样本,已知样本中含有编号为6号、24号、33号的学生,则样本中剩余一名学生的编号是15.【解答】解:样本间距为36÷4=9,则另外一个编号为6+9=15,故答案为:15.14.(5分)已知,则|a0|+|a1|+|a2|+…+|a9|= 512.【解答】解:已知,则|a0|+|a1|+|a2|+…+|a9|,即(1+x)9展开式的各项系数和,令x=1,可得(1+x)9展开式的各项系数和为|a0|+|a1|+|a2|+…+|a9|=29=512,故答案为:512.15.(5分)袋子中装有大小相同的6个小球,2红4白,现从中有放回的随机摸球3次,每次摸出1个小球,则至少有2次摸出白球的概率为.【解答】解:∵袋子中装有大小相同的6个小球,2红4白,现从中有放回的随机摸球3次,每次摸出1个小球,∴每次摸到红球的概率都是,摸到白球的概率都是,∴至少有2次摸出白球的概率为:p=()()2+()3=,故选答案为:.16.(5分)已知x,y∈R,满足x2+2xy+4y2=6,则z=x2+4y2的取值范围为[4,12] .【解答】解:x2+2xy+4y2=6变形为=6,设,,θ∈[0,2π).∴y=sinθ,x=,∴z=x2+4y2==+6=2×(1﹣cos2θ)﹣+6=,∵∈[﹣1,1].∴z∈[4,12].故答案为:[4,12].三、解答题(本大题共5小题,共70分.解答应写出文字说明、证明过程或演算步骤)17.(12分)△ABC的内角A,B,C所对的边分别为a,b,c,向量=(a,c),=(1﹣2cosA,2cosC﹣1),(Ⅰ)若b=5,求a+c值;(Ⅱ)若,且角A是△ABC中最大内角,求角A的大小.【解答】(本大题满分12分)解:(Ⅰ)因为:,所以,2sinAcosC﹣sinA=sinC﹣2sinCcosA,可得:2sinAcosC+2sinCcosA=2sin(A+C)=sinC+sinA,所以,sinA+sinC=2sinB,由正弦定理得2b=a+c=10.….6分(Ⅱ),又因为sinA+sinC=2sinB=sinA+sin(π﹣A﹣B),则,2sinA+cosA=2,又sin2A+cos2A=1,所以,解得,由于A是最大角,所以,.….12分18.(12分)中国乒乓球队备战里约奥运会热身赛暨选拨赛于2016年7月14日在山东威海开赛,种子选手A与非种子选手B1,B2,B3分别进行一场对抗赛,按以往多次比赛的统计,A获胜的概率分别为,且各场比赛互不影响.(Ⅰ)若A至少获胜两场的概率大于,则A入选征战里约奥运会的最终名单,否则不予入选,问A是否会入选最终的名单?(Ⅱ)求A获胜场数X的分布列和数学期望.【解答】解:(Ⅰ)记“种子A与非种子B1、B2、B3比赛获胜”分别为事件A1、A2、A3=所以,A入选最终名单 (6)(Ⅱ)X的可能值为0、1、2、3所以,X的分布列为所以,数学期望: (12)19.(12分)已知各项为正数的数列{a n}的前n项和为S n,且满足(Ⅰ)求证:{a n}为等差数列,并求数列{a n}的通项公式;(Ⅱ)设,求证:.【解答】证明:(1)∵满足,当n=1时,a1=2.当n≥2时,由(1)﹣(2)得(a n+a n﹣1)(a n﹣a n﹣1﹣4)=0(a n>0)则a n﹣a n﹣1=4,∴{a n}是以4为公差的等差数列.a n=4n﹣2.(2)证明:设,则f(n+1)﹣f(n)<0所以,{f(n)}递减,即:…12.20.(12分)已知函数f(x)=x﹣2sinx.(Ⅰ)求函数f(x)在上的最值;(Ⅱ)若存在,使得不等式f(x)<ax成立,求实数a的取值范围.【解答】(本大题满分12分)(1)f'(x)=1﹣2cosx,…(2分)…(6分)(2)f(x)<ax,∴2sinx﹣(1﹣a)x>0设g(x)=2sinx﹣(1﹣a)x,则g'(x)=2cosx﹣(1﹣a)…(7分)由①1﹣a≥2即a≤﹣1,此时g'(x)<0得出g(x)在单调递减,g(x)<g(0)=0不成立…(8分)②1﹣a≤0即a≥1,此时g'(x)>0得出g(x)在单调递增,g(x)>g(0)=0成立…(9分)③0<1﹣a<2即﹣1<a<1,令,存在唯一,使得.当x∈(0,x0)时,g'(x)>0得出g(x)>g(0)=0,∴存在,有g(x)>0成立…(11分)综上可知:a>﹣1…(12分)21.(12分)已知函数,其中a,b,c∈R.(Ⅰ)若a=b=1,求函数f(x)的单调区间;(Ⅱ)若a=0,且当x≥0时,f(x)≥1总成立,求实数b的取值范围;(Ⅲ)若a>0,b=0,若f(x)存在两个极值点x1,x2,求证;f(x1)+f(x2)<e.【解答】解:(Ⅰ),f'(x)>0⇒x>1或x<0,f'(x)<0⇒0<x<1,∴f(x)增区间为(﹣∞,0),(1,+∞),减区间为(0,1).…(4分)(Ⅱ)在[0,+∞)恒成立⇒b≥0…(5分)当b≥0时,f(x)≥1⇔e x﹣bx﹣1≥0.设g(x)=e x﹣bx﹣1,g'(x)=e x﹣b①当0≤b≤1时,g'(x)≥0⇒g(x)在[0,+∞)单调递增,⇒g(x)≥g(0)=0成立②当b>1时,g'(x)=0⇔x=lnb,当x∈(0,lnb)时,g'(x)<0⇒g(x)在(0,lnb)单调递减,⇒g(x)<g(0)=0,不成立综上,0≤b≤1…(8分)(Ⅲ)有条件知x1,x2为ax2﹣2ax+1=0两根,,且,由成立,作差得:,得∴f(x1)+f(x2)<e (12)或由x1+x2=2,,(可不妨设0<x1<1)设(0<x<1),在(0,1)单调递增,h(x)<h(1)=e,∴f(x1)+f(x2)<e成立.[选作题]22.(10分)已知函数f(x)=|x﹣a|﹣2.(Ⅰ)若a=1,求不等式f(x)+|2x﹣3|>0的解集;(Ⅱ)若关于x的不等式f(x)<|x﹣3|恒成立,求实数a的取值范围.【解答】(本大题满分10分)解:(Ⅰ)函数f(x)=|x﹣a|﹣2.若a=1,不等式f(x)+|2x﹣3|>0,化为:|x﹣1|+|2x﹣3|>2.当x≥时,3x>6.解得x>2,当x∈(1,)时,可得﹣x+2>2,不等式无解;当x≤1时,不等式化为:4﹣3x>2,解得x.不等式的解集为: (5)(Ⅱ)关于x的不等式f(x)<|x﹣3|恒成立,可得|x﹣a|﹣2<|x﹣3|设f(x)=|x﹣a|﹣|x﹣3|,因为|x﹣a|﹣|x﹣3|≤|a﹣3|,所以,f(x)max=|a﹣3|即:|a﹣3|<2所以,a的取值范围为(1,5) (10)[选作题]23.(Ⅰ)已知x2+y2=1,求2x+3y的取值范围;(Ⅱ)已知a2+b2+c2﹣2a﹣2b﹣2c=0,求证:.【解答】(Ⅰ)解:由柯西公式(x2+y2)(4+9)≥(2x+3y)2,则|2x+3y|,∴﹣≤2x+3y≤.(Ⅱ)证明:由a2+b2+c2﹣2a﹣2b﹣2c=0,得(a﹣1)2+(1﹣b)2+(1﹣c)2=3,由柯西公式[(a﹣1)2+(1﹣b)2+(1﹣c)2](4+1+1)≥[2(a+1)+(1﹣b)+(1﹣c)]2得证:18≥(2a﹣b﹣c)2,所以.。
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1分所以,两者共同减速 1分错误!未找到引用源。
m 1分 错误!未找到引用源。
m 1分25(1)错误!未找到引用源。
1分错误!未找到引用源。
1分错误!未找到引用源。
错误!未找到引用源。
由几何关系可知错误!未找到引用源。
2分错误!未找到引用源。
M (-3m,0) (2)由几何关系可得错误!未找到引用源。
2解得: 错误!未找到引用源。
由错误!未找到引用源。
可得错误!未找到引用源。
错误!未找到引用源。
错误!未找到引用源。
2分由错误!未找到引用源。
可得: 1分错误!未找到引用源。
1分错误!未找到引用源。
1分错误!未找到引用源。
1分错误!未找到引用源。
1分33.(1)CDE(2)①根据图像可以读出,h=1600m时,P=70cmHg 2分②对气球内气体,有错误!未找到引用源。
2分由图像可得:P0=76cmHg 1分T0=27+273=300K 1分解得:错误!未找到引用源。
3分34.(1)ACD(2)解:①错误!未找到引用源。
2分错误!未找到引用源。
1分错误!未找到引用源。
1分错误!未找到引用源。
1分②因为AP=4错误!未找到引用源。
AP=5错误!未找到引用源。
哈三中2015-2016学年度上学期高三学年期末测试英语答案听力: 1-5 CBCBB 6-10 ACCAB 11-15 CAABA 16-20 BABCC阅读: 21-25 CCADA 26-29 ADCC 30-32 DCA 33-35 ADB七选五: 36-40 BFDGA完形填空:41-45 BCDCB 46-50 ADDBC 51-55 ABDAC 56-60 ABDAC语法填空:61. the 62.when 63.closer 64. long 65. weeks 66.was walking67. checked 68.nothing 69.from/ after 70.Hopefully短文改错:Nowadays parents arrange everything for their children and try to pave the way for theirsuccess, which is a common phenomena. For me, therefore, that is a great conc ern. I’mphenomenon howeverstrongly opposed of this kind of a love. For one thing, children become very dependent on their to 去掉a soparents that they failed to have independent thoughts or creative ideas. For another,failoverprotecting the children or giving out too much care is extremely bad for our development去掉out theirin the future. I think only by encouraging the children to overcome difficulties and developingdevelopby themselves ^they help their children to grow up to be independent and become trulycansuccessful.作文(One possible version):For the three years of my high school life, there is a person that I believe has the greatest influence on me. It’s my grandpa. I can still remember when I was a child, I got used to living in his house, not my home. Strict and serious as he was, he did love me. Now though I have left for high school, I still enjoy my holiday with him. Once I watched his flowers, he started telling me about how he cared for them. At that time, I could recognize that grandpa was really fond of gardening. He was always so patient with his flowers and as a result, they grew better and better. Grandpa isn’t a talkative man, but I have learned a lot from him --- if you do love something, try to do the best. The same is true of learning. How wise my grandpa is!I believe the lesson I’ve learned from him will accompany me forever. I will respect him forever.。
哈师大附中2016-2017高二学年上学期期末英语试题2017.1本试卷分为第I卷(选择题)和第II卷(非选择题)两部分。
满分150分。
考试时间为120分钟。
第I卷(选择题,共100分)第一部分听力(共两节,满分20分)第一节(共5小题;每小题l分,满分5分)听下面5段对话。
每段对话仅读一遍。
1. Where will the woman go?A. To a hotel.B. To the man’s house.C. To a supermarket.2. What does the man mean?A. The price tag of the tennis shoes is missing.B. The woman should buy a pair of hiking shoes.C. The tennis shoes are not so expensive as the woman thinks.3. How does the man suggest the woman go to Bali?A. By air.B. By ship.C. By train.4. What does the man suggest the woman do?A. Do some exercise.B. Take a short shower.C. Have a rest.5. What are the speakers mainly talking about?A. A robbery.B. A newspaper.C. A supermarket.第二节(共15小题;每小题1分,满分15分)听下面5段对话或独白。
每段对话或独白读两遍。
听第6段材料,回答第6、7题。
6. Where does the conversation probably take place?A.At a hotel.B. At t he customs.C. At a travel agency.7. Why does the woman want to keep the man’s passport?A. To follow the practice.B. To record the details later.C. To make a copy of it.听第7段材料,回答第8至10题。
1 哈师大附中高三上学期十二月月考题(英语) 第一部分: 听力 (共两节, 满分30分 每小题1.5分) 第一节(共5小题;每小题1.5分,满分7.5分) 听下面5段对话。每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。每段对话仅读一遍。 1. Where are the two speakers? A. In a store. B. In a hotel. C. At home. 2. Who is Christ Paine? A. A writer. B. A book seller. C. A computer engineer. 3. What’s the relationship between the speakers? A. Neighbors. B. Doctor and patient. C. Boss and employee. 4. How many ties does the man have? A. Nine. B. Five. C. Seven. 5. When is the man checking in? A. Tuesday. B. Thursday. C. Friday. 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,各小题将给出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、7 题。 6. What does the woman want to do? A. To buy an MP4. B. To replace her MP4. C. To repair her MP4. 7. Who is the man? A. A repairman. B. A manager. C. A sales clerk. 听第7段材料,回答第8至10 题。 8. What does the woman intend to do? A. Introduce the Workers’ Club. B. Invite the man to see a film. C. Invite the man to do sports. 9. What can we learn about the club? A. It’s open until 12:00 at night. B. People can skate there. C. Three buses pass the club. 10. Where does the man work? A. At the Workers’ Club. B. At a computer center. C. At a high shool. 听第8段材料,回答第11至13题。 11. How long is it since Mary and John last talked to each other? A. More than one year. B. Less than one year. C. Three months. 12. Where does the man work? A. In a computer factory. B. At a university. C. At the National Bank. 13. Which of the following is true? A. The woman speaks German better than Spanish. B. Tom, John’s son, is in Grade Three. C. The man has two children. 2
听第9段材料,回答第14至16题。 14. What are the speakers talking about? A. American TV. B. American movie. C. American newspaper. 15. What does the woman feel about the people he met? A. They are not interesting. B. They are violent. C. They are kind. 16. Why doesn’t the woman like American TV? A. It’s not interesting. B. It’s full of violence. C. It’s always about the West. 听第10段材料,回答第17至20 题。 17. Whose dog is Chevy? A. Mine. B. My friend’s. C. My husband’s. 18. Why do people go to the cliff? A. To swim. B. To fish. C. To go cliff jumping. 19. Why did Chevy jump off after my friend? A. He wanted to protect my friend. B. He wanted to surprise my friend. C. He wanted to take a swim. 20. Why was my friend happy? A. She was saved by Chevy. B. She saved Chevy. C. She had a brave dog. 第二部分: 阅读理解(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。 A Last Thursday, Michael and Linda stood behind large food trucks distributing meals to 4,000 homeless people for their wedding reception on the border town of Kilis. The couple had decided that instead of hosting their friends and family for a traditional feast reception, they would feed the victims from an earthquake-stricken area. The idea came from the bridegroom's father, Ted, who volunteers for a Turkish relief organization. For the past few years, the organization has distributed daily meals to thousands of people who've suffered from natural disasters. He approached a representative of the organization and suggested that the family cover part of the costs of feeding them for the day. Then he told his son, who was surprised by the suggestion, but soon won over. When he told that to the bride, she was really shocked but finally accepted because in southeastern Turkey there is a real culture of sharing with people in need. They love to share their food, their table and everything they have. And afterwards she was quite amazed about it. So, they arrived at the distribution center on Thursday to spend the day serving food and taking photographs with their grateful recipients(接受者). On Tuesday evening, the newly married couple were still pleased with their decision to quit a personal celebration for one with a greater good. "It's like sharing a dinner with your friends and family who have this kind of thing on a daily basis or sharing something with people who don't even have the most basic things," Michael said. "Hopefully, this will also give the start for other wedding dinners to be held here with our brothers and sisters