小题满分限时练(一)~(四)
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高考数学艺术生冲刺精准训练(限时:40分钟)一、选择题(本大题共12个小题,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A ={x |x (x -2)=0},B ={x ∈Z |4x 2-9≤0},则A ∪B 等于( ) A.{-2,-1,0,1} B.{-1,0,1,2} C.[-2,2] D.{0,2}【答案】 B【解析】 A ={x |x (x -2)=0}={0,2},B ={x ∈Z |4x 2-9≤0}={-1,0,1},则A ∪B ={-1,0,1,2}. 2.已知(1-i)z =2+4i ,则复数z 在复平面内对应的点所在的象限是( ) A.第一象限 B.第二象限 C.第三象限 D.第四象限【答案】 C【解析】 ∵(1-i)z =2+4i ,∴z =2+4i 1-i =(2+4i )(1+i )(1+i )(1-i )=-2+6i2=-1+3i ,则z =-1-3i ,其在复平面内所对应的点位于第三象限.3.已知向量a =(1,2),b =(m ,-4),若|a ||b |+a ·b =0,则实数m 等于( ) A.-4 B.4 C.-2D.【答案】 C【解析】 向量a =(1,2),b =(m ,-4),且|a ||b |+a ·b =0,∴|a ||b |+|a ||b |cos θ=0,∴cos θ=-1,∴a ,b 的方向相反,∴b =-2a ,∴m =-2.4.设a =60.4,b =log 0.40.5,c =log 80.4,则a ,b ,c 的大小关系是( ) A.a <b <c B.c <b <a C.c <a <bD.b <c <a【答案】 B【解析】 ∵a =60.4>1,b =log 0.40.5∈(0,1),c =log 80.4<0,∴a >b >c .5.在一次化学测试中,高一某班50名学生成绩的平均分为82分,方差为8.2,则下列四个数中不可能是该班化学成绩的是( ) A.60 B.70 C.80D.100【答案】 A【解析 】∵∑50i =1 150(x i -82)2=8.2,150(60-82)2=9.68,8.2<9.68,因此化学成绩不可能为60.6.执行如图所示的程序框图,输出的s 值为( )A.2B.32C.53D.85【答案】 C7.“m >2”是“不等式|x -3m|+|x -3|>23对∀x ∈R 恒成立”的( ) A.充分不必要条件 B.必要不充分条件 C.充要条件 D.既不充分也不必要条件【答案】 A【解析】 ∵|x -3m|+|x -3|≥|3m-3|,又不等式|x -3m|+|x -3|>23对∀x ∈R 恒成立,只需3m>33,则m >32.故“m >2”是“|x -3m |+|x -3|>23对∀x ∈R 恒成立”的充分不必要条件.8.已知实数x ,y 满足约束条件⎩⎪⎨⎪⎧x -y +1≥0,2x +y -a ≥0,2x -y -4≤0,若z =y +1x +1的最小值为-14,则正数a 的值为( )A.76 B.1C.34D.89【答案】 D【解析】 满足约束条件的可行域如图中阴影部分所示,∵z =y +1x +1表示过可行域内的点(x ,y )与(-1,-1)连线的斜率,由题意知a >0,所以作出可行域,可知可行域内的点A 与(-1,-1)连线的斜率最小,由⎩⎪⎨⎪⎧2x +y -a =0,2x -y -4=0,解得A ⎝ ⎛⎭⎪⎫1+a 4,a 2-2,又z =y +1x +1的最小值为-14,则⎝ ⎛⎭⎪⎫y +1x +1min =a2-2+1a4+1+1=2a -4a +8=-14⇒a =89. 9. 把不超过实数x 的最大整数记作[x],则函数f (x )=[x]称作取整函数,又叫高斯函数.在[1,4]上任取x ,则[x]=[]的概率为( ) A .B .C .D .【答案】D10.已知函数f (x )=sin(2x +φ),其中0<φ<2π,若f (x )≤⎪⎪⎪⎪⎪⎪f ⎝ ⎛⎭⎪⎫π6对x ∈R 恒成立,且f ⎝ ⎛⎭⎪⎫π2>f (π),则φ等于( ) A.π6B.5π6C.7π6D.11π6【答案】 C【解析 】若f (x )≤⎪⎪⎪⎪⎪⎪f ⎝ ⎛⎭⎪⎫π6对x ∈R 恒成立,则f ⎝ ⎛⎭⎪⎫π6为函数的最大值或最小值,即2×π6+φ=k π+π2,k ∈Z ,则φ=k π+π6,k ∈Z ,又f ⎝ ⎛⎭⎪⎫π2>f (π),即sin φ<0,又0<φ<2π,故当k =1时,此时φ=7π6,满足条件.11.已知双曲线C :x 2a 2-y 2b2=1(a >0,b >0)的左顶点为A ,右焦点为F (c ,0).直线x =c 与双曲线C 在第一象限的交点为P .过F 的直线l 与双曲线C 过二、四象限的渐近线平行,且与直线AP 交于点B .若△ABF 与△PB F 的面积的比值为2,则双曲线C 的离心率为( ) A.53 B.322C. 2D. 3【答案】 A【解析】 ∵△ABF 与△PBF 的面积的比值为2,∴|AB ||BP |=2.∵A (-a ,0),P ⎝ ⎛⎭⎪⎫c ,b 2a ,∴点B 的坐标为⎝⎛⎭⎪⎫2c -a 3,2b 23a ,代入直线l 的方程y =-b a (x -c )得2b =a +c ,即3c 2-2ac -5a 2=0,解得3c =5a 或a =-c (舍去).∴双曲线C 的离心率为53.12.设min{m ,n }表示m ,n 二者中较小的一个,已知函数f (x )=x 2+8x +14,g (x )=min ⎩⎪⎨⎪⎧⎭⎪⎬⎪⎫⎝ ⎛⎭⎪⎫12x -2,log 24x (x >0).若∀x 1∈[-5,a ](a ≥-4),∃x 2∈(0,+∞),使得f (x 1)=g (x 2)成立,则a 的最大值为( ) A.-4 B.-3C.-2D.0【答案】 C【解析】 由题意得g (x )=⎩⎪⎨⎪⎧log 24x ,0<x <1,⎝ ⎛⎭⎪⎫12x -2,x ≥1,则g (x )max =g (1)=2.在同一坐标系作出函数f (x )和g (x )的图象,如图所示.由f (x )=2得x =-6或-2,∵∀x 1∈[-5,a ],∃x 2∈(0,+∞),使得f (x 1)=g (x 2)成立,∴a ≤-2.二、填空题(本大题共4个小题,每小题5分,共20分.请把正确的答案填写在各小题的横线上.) 13.已知一个正方体的所有顶点在一个球面上,若这个正方体的表面积为18,则这个球的体积为________. 【答案 】92π【解析】 设正方体棱长为a ,则6a 2=18,∴a 2=3,a = 3.外接球直径为2R =3a =3,∴R =32,∴V =43πR 3=43π×278=92π.14.在△ABC 中,角A ,B ,C 所对的边分别为a ,b ,c ,且b 2+c 2=a 2+bc ,若sin B ·sin C =sin 2A ,则△ABC 的形状是________三角形.学_科网 【答案】 等边15.如图,抛物线y 2=4x 的一条弦AB 经过焦点F ,取线段OB 的中点D ,延长OA 至点C ,使|AC |=|OA |,过点C ,D 作y 轴的垂线,垂足分别为E ,G ,则|EG |的最小值为________.【答案】 4【解析】 设直线AB 的方程为x =my +1,代入抛物线y 2=4x ,得y 2-4my -4=0.设A (x 1,y 1),B (x 2,y 2),则y 1+y 2=4m ,y 1y 2=-4,∴|EG |=12y 2-2y 1=12y 2+8y 2≥4,当且仅当y 2=4时取等号,即|EG |的最小值为4.16.在数列{a n }中,a 1=1,a n =n 2n 2-1a n -1(n ≥2,n ∈N *),则数列⎩⎨⎧⎭⎬⎫a n n 2的前n 项和T n =________. 【答案】2nn +1【解析】 在数列{a n }中,a 1=1,a n =n 2n 2-1a n -1(n ≥2,n ∈N *),得a n n =n n +1·a n -1n -1. 令b n =a nn,可得b n =nn +1·b n -1,则b n =b 1·b 2b 1·b 3b 2·…·b n -1b n -2·b n b n -1=1·23·34·…·n n +1=2n +1.∴a n =2n n +1,因此a n n 2=2n (n +1)=2⎝ ⎛⎭⎪⎫1n -1n +1,所以T n =2⎝ ⎛⎭⎪⎫1-12+12-13+…+1n -1n +1=2n n +1.。
50分阅读限时满分练(五)(限时35分钟)Ⅰ.阅读理解(共15小题; 每小题2.5分, 满分37.5分)A(2021·安徽合肥二模)CVCC’s High School ProgrammeCareer-technical programmes at Cuyahoga Valley Career Centre allow students flexible career paths with advanced knowledge to prepare them for college and work.Electrical SystemsEach student is provided with trade-related classroom training that produces competency and pride that lead to true craftsmanship. Learn how to use many of the latest tools and technologies with hands-on training in our fully equipped lab. The programme’s diversified coursework gives students a strong foundation in electrical systems installation and repair, in addition to receiving skill training, working and earning a paycheque on the job.Engineering TechnologyYou’ll learn engineering concepts and technology such as applied logic, digital electronics, computer-aided design, robotics and computer-integrated manufacturing. The project-based approach lets you apply your skill to real situations. In practice, you’ll pick your own real-world problems to solve and get ready for a college engineering programme.Sports Medicine Exercise ScienceThis science-based programme combines lectures, hands-on labs and critical thinking activities with all of the latest fitness technology. Students have the unique opportunity to learn in exercise and sports environments alongside exercise and sports professionals using advanced medical techniques to prevent and heal injuries. Upon completion of the programme, students may earn college credits at various Ohio universities. Applicants must have a minimum 3.0 GPA(Grade Point Average) and “C” or better in Biology.Education ProfessionEducation Profession introduces college-minded students to the field of education. This foundational course benefits all education majors regardless of specialty. Observation and field experiences provide the first-hand experience you need to decide if a career in education is for you. Admission is with high school counselor(顾问) recommendation only.1.Which programme offers students pay?A.Electrical Systems.B.Engineering Technology.C.Sports Medicine Exercise Science.cation Profession.2.What is special about Sports Medicine Exercise Science?A.It is targeted at college students in Ohio.B.It is aimed at training professional athletes.C.It only admits applicants with recommendation.D.It has academic requirements for applicants.3.What can we infer about the programmes mentioned in the text?A.They are science-related.B.They are technology-centred.C.They stress practical training.D.They guarantee admission to college.B(2021·宁夏银川高三教学质量检测)Chinese cuisine is widely known and enjoyed all around the world. Who doesn’t long for a favourite Chinese dish? But there is one interesting concept concerning Chinese food which is almost unheard of in the West, and which is becoming increasingly ignored by the youth of the East—the ancient custom of “tonic food”.Tonic food is food which is consumed to improve one’s well-being or avoid sickness. For instance, it was once the custom for new mothers to eat a sesame oil(麻油) hot pot every day for the first month after giving birth. It was believed that this dish would benefit the muscles, reduce pain, improve circulation, stimulate sweating, and warm the body. Some foods, such as mutton and spinach(菠菜), are seen as “hot”, while others, such as Chinese cabbage and radish(樱桃萝卜), are seen as “cold”. One should be careful not to eat too much of either “hot” or “cold” food. However, how much “hot” or “cold” food one should eat depends on the time of the year, how the food is prepared and what it is prepared with, and the individual’s health.The custom of tonic food for a healthier life also influences the catering industry. Chinese herbal medicines, such as wolfberry(枸杞), can be found on many restaurant menus, either added to fruit tea or as a beneficial addition to a dish. These herbs attract customers, such as over-worked office staff, in need of a modest pick-me-up.So, whether you need to boost your strength with a large helping of chicken soup, or increase your mental powers with a serving of pig’s brain soup, you may find that this ancient Chinese custom could be just the tonic you are looking for.4.What is the present situation of tonic food?A.It is catching less attention.B.It is enjoyed by many young people.C.It is well-known worldwide.D.It is becoming increasingly popular.5.What is believed to benefit new mothers?A.Chinese cabbage.B.Fruit tea.C.Sesame oil hot pot.D.Pig’s brain soup.6.What does the underlined word “catering” mean in paragraph 3?A.Medicine.B.Restaurant.cation.D.Fashion.7.What can be a suitable title for the text?A.The Charm of Chinese CuisineB.The Popularity of Tonic FoodC.An Introduction to Chinese Tonic FoodD.Differences between Hot Food and Cold FoodC(2021·黑龙江齐齐哈尔二模)A new study looks at how pets provide important support through petting, cuddles(拥抱), and a frequent physical presence. This is especially key when people feel lonely.The researchers interviewed 32 people aged 59 to 83. Pets included dogs, cats, birds, etc. More than 90% of the people spoke about touch in the relationship with their pets. “Participants often described touch-based interactions with their pets as being comforting or relaxing in a way that contributed to their overall well-being,” the researchers say. “For our participants, comfort is the sense of being somehow cared for by another being.”Many people in the study spoke about how their pets just seemed to “know” when they weren’t feeling well and they’d move to get physically near them. Different pets were better at providing comfort, some participants insisted. Many said cats were more relaxing than dogs, while others said dogs could be relaxing as long as they were “the right kind of dog”. But almost all pets provided some sort of comfort to their owners when it came to touch.Janette Young, lecturer in health sciences at the University of South Australia, and her team uncovered an interesting thing about the pet-human relationship. People in the study often spoke about how their animals demanded to be petted or seemed to take joy in the interaction. This, in turn, made the humans feel good.The researchers suggest that pets can be “helpful in reducing touch deprivation(匮乏)” and the benefits can come from all kinds of pets. This connection can be particularly important in health care and senior care settings where patients and workers are less likely to be able to see friends and family; yet touch is important and can improve health and happiness.Pets can also offer benefits that in some cases people can’t offer. “Relationships with pets are different to those with humans,” Young says. “Animals don’t judge and are always with us.”8.About what did some participants hold different opinions?A.What pets were better at providing comfort.B.Whether pets could make them feel relaxed.C.How they felt when their pets touched them.D.When pets could notice their owners’ discomfort.9.What did Janette Young find about the pet-human relationship?A.It is full of judgments.B.It turns out temporary.C.It benefits both parties.D.It is determined by pets.10.What may researchers suggest hospitals do?A.Encourage pet connection programmes.B.Raise pets to reduce human touch.C.Select the best pet companions.D.Order workers’ friends to pay visits.11.What is the main idea of the text?A.Humans need spiritual support.B.Pet-human interactions are common.C.Lonely people like living with pets.D.Pets offer humans touch-based comfort.D(2021·山东日照一模)Looking for a place to take a rest, sip a cup of coffee and have a furry companion? Welcome to animal-themed cafes.In recent years, the combination of pets and catering has become incredibly popular in many cities throughout China. While many cafes started out offering the company of cats and dogs, others are expanding their range. Some cafes include animals like squirrels, raccoons(浣熊) and even capybaras(水豚).Many Internet users have commented that their main reason for visiting these businesses is to interact closely with animals they have only seen from afar.“The customers love animals. They are curious about these pets and want to raise one at home but circumstances don’t allow it,” said pet cafe owner Tian Rui. Liu Jing, a Beijing resident who works in finance, loves pigs, but her parents would not endorse her getting one as a pet. The 24-year-old had to admire the animals through videos and images. But now, she can have close contact with an adorable pig every weekend in a pet cafe.Also, these pet cafes serve as a shelter for those who find comfort in being around animals. “In this place, you can get close to the cute little ones in a very comfortable way without having to worry about safety and hygiene(卫生)—overall, it is an extremely comfortable experience,” said Zhao Ming from Tianjin, a frequent pet cafe visitor.Animal lovers also expect the pet cafes to change people’s attitudes towards pets. Animals like ducks, pigs and sheep have been served as food for many years. People hardly see them as pets. But the pet cafes help shift their attitude from seeing them as playthings to regarding them as family members or close friends.“Whether at home or in stores, these little animals are there for companionship and they bring warmth to our hearts. They care and understand just like family members. They give us much more than the love we give to them,” Zhao said.12.Why do people visit pet cafes?A.To pursue inner peace.B.To appreciate animal images.C.To experience a clean environment.D.To have close contact with animals.13.What does the underlined word “endorse” in paragraph 4 mean?A.Believe in.B.Approve of.C.Prevent.D.Suspect.14.We can conclude from paragraph 6 that pet cafes may.A.change our views of animalsB.teach us how to look after pets betterC.show us the importance of having petsD.remind us to value our family and friends15.Which of the following can be the best title for the text?A.Cafes Enjoy Great PopularityB.Ideal Shelters for Lovely AnimalsC.Animals Turn into CompanionsD.Animal Lovers from Different CitiesⅡ.七选五(共5小题; 每小题2.5分, 满分12.5分)(2021·江苏淮安高三适应性联考)If you’re looking to maximise the amount of fat burning in your next workout, think about having a coffee half an hour before you get started. 16Researchers found that 3 milligrams(毫克) of caffeine per kilogram of body weight can raise the rate of fat burning during aerobic(有氧的) exercise. 17 The coffee dose was shown to increase maximal fat oxidation rate(脂肪氧化率) by an average of 10.7 percent in the morning and 29 percent in the afternoon.The recommendation to exercise on an empty stomach in the morning to increase fat oxidation is common. 18 It is still unknown whether this increase is due to exercising in the morning or due to going without food for a longer period of time.19 Caffeine is often associated with improved athletic performance, though the science behind this link isn’t as comprehensive as it could be. In later cycling tests after the caffeine had been given, thefat-burning capability of the body was measured, showing the impact that caffeine had made on MFO. Maximum oxygen uptake was also raised, as was the intensity(强度) of exercise.The results of the study showed that acute caffeine intake 30 minutes before performing an aerobic exercise test increased MFO during exercise regardless of the time of day.20 Overall, these results suggest that a combination of acute caffeine intake and exercise at moderate intensity in the afternoon provides the best solution for individuals seeking to increase whole-body fat oxidation during aerobic exercise.A.It is based on the results gathered from 15 male volunteers.B.However, this recommendation may lack a scientific basis.C.Evidence is mounting that morning exercise is good for your health.D.A new study suggests it can make a significant difference to fat burning.E.They are clear enough to suggest that there is some kind of association there.F.Researchers have found that if cyclists take more caffeine, they will ride much faster.G.Researchers were keen to look in detail at the relationship between caffeine and exercise.50分阅读限时满分练(五)【语篇导读】本文是一篇应用文。
高中数学专题复习
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得分 一、选择题
1.在锐角中ABC ∆,角,A B 所对的边长分别为,a b .若2sin 3,a B b A =则角等于
A.12π
B.6π
C.4π
D.3
π (汇编年高考湖南卷(理)) 2.若钝角三角形三内角的度数成等差数列,且最大边长与最小边长的比值为m ,则m 的范围是( )
A .(1,2)
B .(2,+∞)
C .[3,+∞)
D .(3,+∞)(汇编辽宁)
3.在△ABC 中,角A ,B ,C 所对的边长分别为a,b,c ,若∠C=120°,2c a =,则( ) A 、a>b B 、a<b C 、a=b D 、a 与b 的大小关系不能确定(汇编湖南理6)
4.在直角ABC ∆中,CD 是斜边AB 上的高,则下列等式不成立的是( ) A.2AC AC AB =⋅ B.2BC BA BC =⋅ C.2AB AC CD =⋅。
小题限时练 选择题专练一(数与代数) 姓名:________ 班级:________ 限时:______分钟1.|-13|的相反数是( ) A.13 B .-13C .3D .-3 2.81的平方根是( )A .±3B .3C .±9D .93.下列各数中,小于-3的数是( )A .2B .-13C .-52D .-44.据国家统计局统计,2018年全国居民人均可支配收入28 228元,比上年名义增长8.7%,扣除价格因素,实际增长6.5%.将28 228用科学记数法表示为( )A .28 228×105B .2 822.8×102C .2.822 8×104D .0.282 28×1055.目前我国能制造芯片的最小工艺水平已经达到7纳米.已知1纳米=0.00 000 0001米,用科学记数法将7纳米表示为( )A .0.7×10-8米B .7×10-9米C .0.7×10-10米D .7×10-10米6.学校为创建“书香校园”,购买了一批图书.已知购买科普类图书花费10 000元,购买文学类图书花费9 000元,其中科普类图书平均每本的价格比文学类图书平均每本的价格贵5元,且购买科普书的数量比购买文学书的数量少100本.求科普类图书平均每本的价格是多少元?若设科普类图书平均每本的价格是x 元,则可列方程为( )A.10 000x -9 000x -5=100 B.9 000x -5-10 000x=100 C.10 000x -5-9 000x=100 D.9 000x -10 000x -5=100 7.有理数a ,b 在数轴上的位置如图所示,则下列结论中,不正确的是( )A .a +b <0B .a -b <0C.a b<0 D .|a|<|b| 8.下列运算正确的是( )A .a 5+a 3=a 8B .(3a 3)2=9a 9C .a 3·a 3=a 6D .2a -a =29.已知一元二次方程x 2+k -3=0有一个根为1,则k 的值为( )A .-2B .2C .-4D .410.关于x 的分式方程2x +a x +1=1的解为负数,则a 的取值范围是( )A .a >1B .a <1C .a <1且a≠-2D .a >1且a≠211.若3a -2b =2,则代数式2b -3a +1的值等于( )A .-1B .-3C .3D .512.在平面直角坐标系的第二象限内有一点M ,点M 到x 轴的距离为3,到y 轴的距离为4,则点M 的坐标是( )A .(3,-4)B .(4,-3)C .(-4,3)D .(-3,4)13.如图,在平面直角坐标系中,菱形OABC 的边OA 在x 轴的正半轴上,反比例函数y =k x(x >0)的图象经过对角线OB 的中点D 和顶点C.若菱形OABC 的面积为12,则k 的值为( )A.6 B.5 C.4 D.314.如图,菱形ABCD的边长是4 cm,∠B =60°,动点P以1 cm/s 的速度自A点出发沿AB方向运动至B点停止,动点Q以2 cm/s的速度自B点出发沿折线BCD运动至D点停止.若点P,Q同时出发运动了t s,记△BPQ的面积为S cm2.下面图象中能表示S与t之间的函数关系的是( )参考答案1.B 2.A 3.D 4.C 5.B 6.B 7.D 8.C 9.B 10.D 11.A 12.C 13.C 14.D。
小题满分限时练一一、填空题(本题共6小题,每空1分,共14分)1.人们要认识事物,就要对事物进行描述,从而引入相应的物理量.如:为了描述物体所含物质的多少,我们引入了物理量“质量”;为了描述,我们引入了物理量“”.2.在抗击新冠肺炎疫情期间,医务人员在进入隔离病房前必须要穿戴好防护服、口罩、面罩等.医务人员戴着面罩工作一段时间后,面罩会模糊不清,这是由于人呼出的水蒸气遇冷(填写物态变化名称)形成的小水珠附着在了面罩的(选填“内侧”或“外侧”).3.我国最大的核电站——秦山核电站是利用(选填“核裂变”或“核聚变”)释放的核能来发电的.核能属于(选填“可再生”或“不可再生”)能源.4.“唯有牡丹真国色,花开时节动京城”,每年的牡丹花开时节都吸引了大量游客到洛阳赏牡丹.我们能从不同方向看到同一朵牡丹花,这是由于光在花的表面发生了.阳光透过牡丹花和叶子的缝隙,在地面上形成了许多圆形的光斑,这是光形成的.5.在科技制作比赛中,某同学设计了如图甲所示的喷泉控制电路,EF两端提供的电压随时间周期性变化的情况如图乙所示.R为5 Ω的定值电阻,A为特殊电子元件,其两端的电压恒为电源电压.当R的功率大于或等于5 W时,水泵启动喷水.则喷水时,EF两端提供的电压至少为的13V;喷水过程中,R的最大功率是W;该电路持续工作10 min,喷水时间共为min.6.如图所示,弹簧测力计左端固定,右端与木块A相连,木块A放在上表面水平的小车上,弹簧测力计保持水平.现用大小为10 N的拉力F水平向右拉动水平地面上的小车,木块A处于静止状态,弹簧测力计的示数为5 N.木块A所受摩擦力的大小为N,方向.在上述过程中,要想使弹簧测力计的示数减小,可以采取的措施是.二、选择题(本题共8小题,每小题2分,共16分.第7~12题每小题只有一个选项符合题目要求,第13~14题每小题有两个选项符合题目要求,全部选对得2分,选对但不全的得1分,有选错的得0分)7.下列有关古筝演奏的说法错误..的是( )A.拨动古筝的弦时,弦因振动而发出声音B.拨动粗细不同的弦时,发出声音的音色不同C.拨动同一根弦的力度不同时,发出声音的响度不同D.优美的古筝声也可能是噪声8.如图所示,消毒车在水平路面上沿直线匀速行驶并对街道喷洒消毒液.下列说法不正确...的是( )A.以消毒车为参照物,道路两旁的大楼是运动的B.消毒车的动能减小C.停止喷液后,消毒车受到的重力和水平路面对它的支持力是一对平衡力D.刹车后,消毒车不能立即停下来,是因为它受到惯性的作用9.据统计,2020年1~3月份,在郑州市发生的火灾事故中,由各类电器引发的火灾数量最多.下列关于安全用电的描述正确的是( )A.控制灯泡的开关应接在灯泡与零线之间B.用电器失火时,先用水灭火,再切断电源C.为了节约插线板,应将多个大功率用电器插在同一个插线板上使用D.不可以用湿布擦洗正在工作的用电器10.下列关于粒子和宇宙的说法正确的是( )A.马路上尘土飞扬,说明分子在不停地运动B.原子的核式结构模型与太阳系的结构很相似C.登月成功标志着人类完成了对宇宙的探索D.人类能对微观世界研究的最小尺度是纳米尺度11.如图所示,某同学欲用焦距不变的照相机拍摄一棵高大树木的全景,但从取景框中只能看到树的一部分.要想拍摄这棵树的全景,该同学可以采取的方法是( )A.原地不动,将相机向上举高些B.远离大树,镜头向后缩C.靠近大树,镜头向后缩D.远离大树,镜头向前伸12.如图所示,滑轮组上挂有a、b两物体,两物体处于静止状态.不计绳重、滑轮重和摩擦,若在a、b下各挂一个质量相等的物体,则( )A.a下降B.a上升C.a、b仍静止D.以上情况都有可能13.(双选)如图所示的电路中,电源电压保持不变,R1为定值电阻,R2为滑动变阻器.当开关S闭合,滑片P向右移动时,下列说法正确的是( )A.电压表V1的示数变大B.电压表V2的示数变大C.电流表A的示数变大D.电压表V1与电流表A的示数之比变大14.(双选)如图所示,两个完全相同的盛水容器放在水平地面上,用细线悬挂体积相同的实心铅球和铝球(ρ铅>ρ铝),将它们浸没在容器内的水中,此时两容器中的水面高度相同,则下列说法正确的是( )A.如果铅球和铝球所受的浮力分别为F浮1和F浮2,则F浮1>F浮2B.如果铅球和铝球所受绳的拉力分别为F拉1和F拉2,则F拉1>F拉2C.如果图甲和图乙中水对容器底的压力分别为F'1和F'2,则F'1<F'2D.如果图甲和图乙中容器对水平地面的压强分别为p1和p2,则p1=p2三、作图题(本题共2小题,每小题2分,共4分)15.如图所示,杠杆OBA可绕O点在竖直平面内转动,在B点用细绳悬挂一重物,在A点对杠杆施加动力,使杠杆在图示位置平衡.试画出作用在A点的最小动力F1及阻力F2的示意图.16.如图所示,一个盛有水的水缸底部铺有一块平面镜,一束单色光AB沿图示方向从空气入射到水面上.请画出这束光射入水中后折射光的大致传播方向,并画出这束光经平面镜反射后从水中射出的光路图.小题满分限时练一1.压力的作用效果压强(本题答案不唯一,对应且合理即可)【解析】为了描述物体运动的快慢,引入了物理量“速度”;为了描述物体的冷热程度,引入了物理量“温度”;为了描述物体所占空间的大小,引入了物理量“体积”.2.液化内侧【解析】人呼出的水蒸气遇到较冷的面罩后液化成小水珠附着在面罩的内侧.3.核裂变不可再生【解析】利用核能的方式主要有核裂变和核聚变.目前核电站均是利用可控核裂变释放的核能来发电的.消耗后短期内不能从自然界得到补充的能源是不可再生能源,常见的不可再生能源有:煤、石油、天然气、核能等.4.漫反射沿直线传播【解析】我们能从不同方向看到同一朵牡丹花是由于光在花的表面发生了漫反射;牡丹花和叶子间有缝隙,这些缝隙可看作一个个小孔,地面上的圆形光斑是太阳经小孔形成的实像,是光沿直线传播形成的.5.7.520 5【解析】依题意,当R的功率达到5 W时,水泵启动喷水,此时,R两端的电压U R=√P R R=√5W×5Ω=5 V.因电子元件A两端的电压恒为电源电压的13,由串、并联电路的电压规律可知,R两端的电压为电源电压的23,故电源电压U EF=32U R=32×5 V=7.5 V.当EF两端的电压为15 V时,定值电阻R两端的电压最大,此时R两端的电压U'R=23U'EF=23×15 V=10 V,此时电阻R的功率最大,为P=U'2RR =(10V)25Ω=20 W.由题图乙可知,0~3 s内电压均匀增加,大于或等于7.5 V的时间为1.5s ,由对称性可知,3~6 s 内电压大于或等于 7.5 V 的时间也为1.5 s ,所以每个周期内喷水时间与总时间的比值是 3s 6s =12,该电路持续工作10 min ,喷水时间共为12×10 min=5 min. 6.5 水平向右 减小木块A 的质量(或减小接触面的粗糙程度)【解析】 木块A 在水平方向上受到弹簧测力计对它水平向左的拉力和小车对它水平向右的摩擦力的作用,因为木块A 处于静止状态,所以这两个力是一对平衡力,因此木块A 受到的摩擦力的大小等于弹簧测力计的示数,为5 N.因为小车与木块A 之间发生了相对运动,故木块A 所受的摩擦力是滑动摩擦力.影响滑动摩擦力大小的因素是压力的大小和接触面的粗糙程度,故要想使弹簧测力计的示数减小,可以减小木块A 对小车的压力,即减小木块A 的重力(或质量),也可以减小接触面的粗糙程度.7.B 【解析】 拨动粗细不同的弦时,弦振动的频率不同,发出声音的音调不同,B 错误.8.D 【解析】 消毒车做匀速直线运动时,消毒车与道路两旁的大楼的相对位置发生了变化,所以以消毒车为参照物,道路两旁的大楼是运动的,A 正确;消毒车在对街道消毒的过程中,速度不变,消毒车的质量减小,故消毒车的动能减小,B 正确;停止喷液后,消毒车在竖直方向上处于平衡状态,故其受到的重力和地面对它的支持力是一对平衡力,C 正确;刹车后,消毒车不能立即停下来,是因为消毒车具有惯性,惯性不是力,不能说“受到惯性的作用”,D 错误.9.D 【解析】 为保证用电安全,在安装照明电路时,控制灯泡的开关一定要接在火线和灯泡之间,A 错误;用电器失火时,应先切断电源,再灭火,B 错误;将多个大功率用电器插在同一个插线板上使用,可能会导致插线板中的电流过大,烧坏插线板甚至引发火灾,C 错误.10.B 【解析】 尘土是宏观物质,不是微观分子,A 错误;卢瑟福提出了原子的核式结构模型,原子的核式结构模型与太阳系的结构很相似,B 正确;宇宙很大,人类登月成功标志着人类对宇宙的探索又向前迈了一步,但人类探索宇宙的脚步不会停止,C 错误;纳米是长度单位,人类研究的比纳米更小的长度单位还有皮米、飞米等,D 错误.11.B 【解析】 由题意可知,该同学拍照时,取景框内只能看到树的一部分,说明树经凸透镜所成的像太大,要想拍到这棵树的全景,则应使像变小.根据“凸透镜成实像时,物远像近像变小”的规律可知,凸透镜应远离大树,同时减小像距,即镜头向后缩,应选B.12.B 【解析】 由于不计绳重、滑轮重和摩擦,与物体a 相连的滑轮为动滑轮,使用动滑轮能省一半的力,与物体b 相连的滑轮为定滑轮,使用定滑轮不省力,则有2m b g=m a g ;设在物体a 、b 下所挂物体的质量均为m ,由于2(m b +m )g>(m a +m )g ,所以在a 、b 下各挂一个质量相等的物体后,a 上升,b 下降.故选B.13.AD 【解析】 分析电路图可知,闭合开关S ,R 1与R 2串联,V 1测滑动变阻器R 2两端的电压,V 2测电源电压,因电源电压不变,所以在移动滑动变阻器滑片的过程中,电压表V 2的示数不变,故B 错误;当滑片P 向右移动时,滑动变阻器接入电路的阻值变大,电路的总电阻变大,根据I=U R可知,电路中的电流变小,即电流表A 的示数变小,故C 错误;根据U=IR 可知,R 1两端的电压变小,根据串联电路的电压规律可知,滑动变阻器两端的电压增大,即V 1的示数变大,故A 正确;电压表V 1与电流表A 的示数的比值等于滑动变阻器连入电路的阻值,当滑片P 向右移动时,滑动变阻器连入电路的阻值变大,即电压表V 1与电流表A 的示数之比变大,故D 正确.14.BD【解析】因为铅球和铝球的体积相等,且两球均浸没在水中,根据阿基米德原理可知,它们所受的浮力相等,即F浮1=F浮2,A错;由于ρ铅>ρ铝,根据G=mg=ρVg可得,G铅>G铝,悬吊在水中的球受重力、浮力和细线的拉力作用,且在这三个力的作用下处于平衡状态,即F拉=G-F浮,因此F拉1>F拉2,B正确;由p=ρgh可知,容器中的水面高度相同,故容器底受到水的压强相等,根据F=pS可知,容器底受到水的压力相等,即F'1=F'2,C错误;两容器的重力相同,两容器中的水的体积相等,则水的重力也相等,两容器中的水受到金属球的“压力”(即浮力的反作用力)也相等,故两容器对水平地面的压力相等,根据p=FS可得,p1=p2,D正确.15.如图所示.【解析】由杠杆平衡条件可知,在阻力和阻力臂一定时,动力臂越长,动力越小;由题图可知,以线段OA作为动力臂时,动力臂最长;为使杠杆在图示的位置平衡,动力F1的方向应垂直于OA斜向上.该杠杆的阻力F2为细绳对杠杆的拉力,阻力的作用点在B点,方向竖直向下,且F2>F1.16.如图所示.【解析】光从空气斜射入水中时,折射角小于入射角,所以此时的折射光线应靠近法线.根据光的反射定律(反射角等于入射角),可作出这束折射光经平面镜反射后的反射光线.光从水中斜射入空气中时,折射角大于入射角,所以此时的折射光线应远离法线.同时,根据光路的可逆性,光最终射出水面时的折射角和最初入射时的入射角相等.。
新沂三中2023届高三英语限时练(4)限时:40分钟一、阅读理解AAllowing women to give birth in hospital with their pet dog by their side could make them less anxious, one of the first women to use a “birth dog” has said.Charlotte Beard, 24, who suffers from non-epileptic seizures, was supported through her 50-hour labour by her aid pet Flump, who is trained to detect seizures before they happen.Flump was prepared for the birth by being played audio tracks of babies crying at home and being introduced to staff at the maternity ward (产房) in Poole hospital. Beard joked. “He was visiting the maternity hospital on a very regular basis, getting to know staff, getting to know the different rooms and the setups.”It is a legal requirement that assistance dogs are allowed into hospitals, but Beard said a lot of hospitals were not aware of this.“It would have caused high levels of anxiety if Flump was not there,” she said. “I would have certainly been more concerned about my safety.”“I am not suggesting that if I were to have a seizure, staff would not have been able to look after me. But no staff member can detect beforehand whether you are going to have a seizure or not. That is something that Flump can do that nobody else can, and it gives them such valuable time to make sure that I am safe and have the right support in place before it happens. Prevention is always much better than treatment.”Beard got Flump at 10 weeks and has trained him since then. “From a human standpoint I cannot help but think how lovely it is that Flump was there. He was one of the first to see Alfie,” Beard said. “Being present there from the birth and meeting Alfie within moments of him being born, whether it has increased the bond or not, it was such a lovely experience for everyone. Flump was absolutely besotted (痴恋) with Alfie from the moment he first saw him.”1. Why did Beard have an assistance pet present during birth?A. To cure her disease.B. To monitor the doctors.C. To ease her anxiety.D. To keep the door.2. What was the assistance dog trained to do before doing the job?A. Understand human voices.B. Get familiarized with the hospital.C. Play music for babies.D. Learn about medical devices.3. What does the underlined word “it” in paragraph 6 refer to?A. Detecting the disease.B. Medical treatment.C. Looking after the patient.D. A seizure.4. What does Beard think of future bond between the dog and her son?A. It will be close.B. It will be tense.C. It needs improving.D. It is hard to say.BEven tree-planting can increase health risks to local human populations if it focuses too narrowly on a small number of species, as is often the case in commercial forests. Outbreaks of infectious diseases are more likely in areas of monoculture plantations, according to a new study.The researchers of the study said this was because diseases are filtered (过滤) and blocked by a range of predators (食肉动物) and habitats in a healthy, biodiverse forest. When this is replaced by a palm oil plantation or soy fields, the specialist species die off, leaving generalists such as rats and mosquitoes to thrive (繁殖) and spread pathogens across human and non-human habitats. The net result is a loss of natural disease regulation.The researchers examined the correlation between trends for forest cover, plantations, population and disease around the globe using statistics from international institutions such as the World Health Organization and the Food and Agricultural Organization. Over the period of study from 1990 to 2016, this covered 3,884 outbreaks of 116 diseases that crossed the species barrier and 1,996 outbreaks of 69 vector-borne infectious diseases, mostly carried by mosquitoes, ticks or flies.The new study adds to a growing body of evidence that viruses are more likely to transfer to humans or animals if they live in or near human-disturbed ecosystems. This is shaped by trade patterns and consumer behaviour. A quarter of global forest loss is driven by the production of commodities such as beef, soy, palm oil and wood fibre.Morand, lead-author of the study, said his study showed that disease risks need to be added to risk-benefit analysis of new projects. “We should take the costs of public health into account when considering new plantations. The risks are first to local people, but thenworldwide because we have seen with Covid how quickly diseases can spread.”Morand is now working on a more detailed study that will use satellite analysis of forest cover to examine links with disease. With more information, he believes it may be possible to predict future outbreaks and to work with local communities to build ecologically diverse and economically productive landscapes that reduce the risks.5. Why are outbreaks of infectious diseases more likely in areas of monoculture plantations?A. The specialist species are wiped out by their predators.B. Biodiversity decline destroys natural disease regulation.C. Local human populations have no knowledge of health risks.D. Commercial forests provide food for disease-carrying insects.6. How did the researchers get their findings?A. By conducting data analysis.B. By referring to another study.C. By making field investigation.D. By consulting authoritative agencies.7. What does Morand intend to do by his words in paragraph 5?A. Criticize policy-makers.B. Offer a solution to Covid.C. Make a suggestion.D. Support evidence for her findings.8. What does the last paragraph talk about concerning Morand-led study?A. Its theoretical basis.B. Its appeal to the public.C. Expectations for future studies.D. Researchers with new perspectives.第二节(共5小题;每小题分,满分分)If you’re going travelling after university or as a career break, an extended period of time of unemployment can leave a gaping hole in your CV. ____1____ As long as you plan your trip carefully and think about what you can do, travelling can actually boost your CV and career.Recharge you batteriesIf you’ve just spent three or four years at university, or working without a break, then you’ll need to recharge your batteries. If you take time out to travel you’ll have the opportunity to think about what’s important in your life and refocus on your career. ____2____ Learn a new skillIf you’re going on a round-the-world trip then you could be travelling from anywhere between a month to a year, why don’t youlearn a new skill while you’re on the road? ___3____ All businesses and companies need candidates with a broad skill set, so look at learning another language or getting further qualifications to boost your CV.Make contactsWhen you’re backpacking and travelling, try to make contacts with people who might be useful for your career. With the rise of social media, making contacts is as simple as saying “just tweet me” or “I’ll add you on Facebook”. The contacts you make while travelling could be your avenue into a new career. ____4____Volunteer____5____ Not only will you make a difference to the community and society, but it’ll look great on your CV. With volunteering you’ll learn what it means to live and work in another country, and employers are actively encouraging it. In a survey, 63% of HR professionals said that those who have taken a constructive gap year of volunteering or working abroad had an application that stood out from the rest.A. But don’t worry.B. Consider the simplest way you can meet their needs.C. It’s important to network wherever you are in the world.D. More often than not, you’ll come home feeling refreshed and energetic.E. Think about what employers are looking for in your industry and cater to that.F One of the hardest things to overcome when applying for a job is work experience.G. Volunteering is one of the most popular activities for gap years and round-the-world trips.第三部分语言运用(共两节,满分30分)When I was pregnant with Paige, I read about the donor milk bank and was instantly____1____. I learned that nearly 4,000 babies are born ____2____ every year; because their organs and immune systems have had less time to mature, they’re at greater risk of____3____, which can be fatal (致命的) in their____4____ early days. Breast milk is often the only thing they can ____5____, but mothers with a ____6____ pregnancy often find that their milk has not yet come in. The thought of helping the tiny babies ____7____ a lot to me.Paige was born a month early and in special care for a couple of days. That’s ____8____ we saw those tiny babies who could____9____ in your hand, only managing to take about 4 ml of milk at atime — and the reality of milk ______10______ hit home. Watching those other parents desperately hoping for their babies to______11______, and realizing I could help, was when the feeling that I should become a milk donor changed to wanting to ______12______ it.I contacted the coordinator of the donor milk bank. She came to my house to do my blood tests and ______13______ me through the process, so I was good to go. I now spend about six hours a day expressing milk. The milk is frozen and then _____14______ every week by one of the volunteer drivers. I’ve been told I’m donating record-breaking amounts, and I’ll______15______ donating for as long as I can.1. A. hesitant B. embarrassed C. interestedD. grateful2. A. prematurely B. safely C. purposelyD. secretly3. A. depression B. infection C. injuryD. conflict4. A. remaining B. boring C. criticalD. merry5. A. tolerate B. expect C. affordD. order6. A. lengthened B. shortened C. worsenedD. strengthened7. A. troubled B. amused C. offendedD. meant8. A. how B. when C. whetherD. what9. A. fit B. feed C. drinkD. grow10. A. delivery B. donation C. productionD. consumption11. A. recover B. survive C. laughD. communicate12. A. do B. forget C. useD. escape13. A. warned B. talked C. blamedD. teased14. A. thrown out B. sold out C. picked upD. drunk up15. A. give up B. carry on C. put offD. bring down第二节(共10小题;每小题分,满分15分)Shangsi Festival is an ancient Chinese festival ____1____ (celebrate) on the third day of the third lunar month. It is afestival____2____ (main) observed by Chinese ethnic groups to drive away evil spirits and natural disasters.There are many ____3____ (story) about the festival’s true origins. Some say it stems from a dinner party held on the banks of the Qushui River during the Zhou Dynasty. Others say it____4____ (come) from the custom of getting rid of evil by bathing in the river.Shangsi Festival activities have changed with the times. The feast and praying for later generations by the river____5____ (add) in the Han Dynasty. It was after the Wei and Jin dynasties ___6____ the event developed into the Double Third Day.After the Ming and Qing dynasties, the festival developed into____7____ spring outing featuring lively activities like hiking and listening to choir music.To the Zhuang ethnic group, March 3 is a festival for young people ____8____ (sing) songs and find their true love. Since 1983, singing festivals____9____ (hold) annually on this day throughout Guangxi.Five-colored sticky rice and painted eggs are traditional food for the Shangsi Festival. The Zhuang people believe rice is good for health and painted eggs are seen ____10____ a symbol of love.参考答案:【答案】CBD31~35ABACC36~40ADECG41~45CABCA46~50BDBAB51~55BABCBaddedbeen【解析】C本篇是一篇报道,介绍了英国一位妇女让宠物狗陪她在医院产子的事情。
2020学年七年级(上)英语限时一考生须知:1. 全卷共八大题,66小题,满分为100分。
2. 全卷分为卷Ⅰ(选择题) 和卷Ⅱ(非选择题) 两部分,全部在“答题纸”上作答。
卷Ⅰ的答案必须用2B铅笔填涂;卷Ⅱ的答案必须用黑色字迹签字笔写在“答题纸”的相应位置上。
3. 请用黑色字迹签字笔在“答题纸”上先填写姓名和准考证号。
卷Ⅰ说明:本卷共有五大题,45小题, 满分65分。
请用2B铅笔在“答题纸”上将你认为正确的选项所对应的小方框涂黑、涂满。
第一部分听力部分一、听力(本题有15小题,第一节每小题1分,第二、三节每小题2分,共计25分)第一节:听小对话,回答问题。
1. What’s Henry’s last name?A. GreenB. BrownC. Miller2. Who is Mary?A. Eric’s auntB. Eric’s sisterC. Eric’s mother3. Which school is Bill in?A. No, 3 Middle SchoolB. No, 4 Middle SchoolC. No, 5 Middle School4. What color is Steve’s phone?A. RedB. BlueC. Yellow5. How old is Molly?A. 7B. 8C. 9第二节:听较长对话,回答问题。
听下面一段对话,回答第6-7小题。
6. Who’s Shirley?A. Robert’s sisterB. Robert’s friendC. Robert’s cousin7. Where is Shirley now?A. In ChinaB. In the UKC. In the USA听下面一段对话,回答第8-10小题。
8. Who has a blue jacket?A. BenB. HillC. Rose9. What size(尺码) is the girl’s jacket?A. SB. MC. L10. What’s on the girl’s jacket?A. A CDB. A TVC. A UFO第三节:听独白,回答问题。
第四单元综合素质评价限时:150分钟满分:120分一、积累与运用(10分)1. 阅读下面的语段,完成(1)~(4)题。
(6分)然后,我们来到了历史的转折关头。
渴求进步但又常常因夸大某种正确思想而②的人类精神,开始致力于将青少年从平衡状态中挣脱出来。
于是,青少年开始为呆板而复杂的教育枷锁所套牢,被在愚蠢的放纵和不明智的严厉交互作用下的道德说教以及拙.劣肤浅的世界观所束缚。
这就是我们要重启奥林匹克时代,并为体格训练的复兴隆重庆祝。
我们不断推动盎格鲁-撒克逊人的运动功利思想向古希腊遗留下来的一呼百应的体育观靠拢,两者逐渐①合为一体。
(1)语段中加点字“拙”的正确读音是( )(1分)A. zhuōB. zhuóC. chùD. chū(2)在语段中的①处填入汉字,正确的是( )(1分)A. 溶B. 熔C. 融D. 镕(3)在语段中的②处填入的词语,恰当的一项是( )(2分)A. 迷途知返B. 另辟蹊径C. 万劫不复D. 误入歧途(4)语段中画波浪线的句子有语病,下列修改正确的一项是( )(2分)A. 这就是为何我们要重启奥林匹克时代,并为体格训练的复兴隆重庆祝。
B. 这就是为何我们要重启奥林匹克时代,并为体格训练的复兴隆重。
C. 这就是我们要重启奥林匹克时代,并为体格训练的复兴隆重。
D. 这就是因为我们要重启奥林匹克时代,并为体格训练的复兴隆重庆祝。
2. 【2022·徐州模拟】下列句子的语序排列正确的一项是( )(2分)①因为每个人都有自己希望实现的梦想,而个人梦想的实现又是建立在国家繁荣的基础上的。
②其实,换个角度想,“中国梦”与我们每个人息息相关。
③它使人联想到民族奋进的滚滚大潮,气势磅礴的宏伟蓝图。
④所以,我们每一天都向着自己的梦想出发,其实也在缔造中华民族的复兴之梦。
⑤“中国梦”即中华民族的复兴之梦,是十四亿中华儿女共同拥有的伟大梦想。
A. ⑤④②③①B. ①④⑤③②C. ①②④⑤③D. ⑤③②①④3. 下面是小明同学演讲稿的开头部分,针对这段话的说法有误的一项是( )(2分)同学们:大家好!我是来自八年级(5)班的小明。
限时练(一)(限时:40分钟)一、选择题(本大题共10小题,每小题4分,共40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M ={x |x 2-4x <0},N ={x |m <x <5},若M ∩N ={x |3<x <n },则m +n 等于( ) A.9 B.8 C.7D.6解析 ∵M ={x |x 2-4x <0}={x |0<x <4}, N ={x |m <x <5},且M ∩N ={x |3<x <n },∴m =3,n =4,∴m +n =3+4=7.故选C. 答案 C2.《张丘建算经》卷上第22题——“女子织布”问题:某女子善于织布,一天比一天织得快,而且每天增加的数量相同.已知第一天织布5尺,30天共织布390尺,则该女子织布每天增加( ) A.47尺 B.1629尺 C.815尺D.1631尺解析 依题意知,每天的织布数组成等差数列,设公差为d ,则5×30+30×292d=390,解得d =1629.故选B. 答案 B3.已知直线l :x +y +m =0与圆C :x 2+y 2-4x +2y +1=0相交于A ,B 两点,若△ABC 为等腰直角三角形,则m =( ) A.1 B.2 C.-5D.1或-3解析 △ABC 为等腰直角三角形,等价于圆心到直线的距离等于圆的半径的22.圆C 的标准方程是(x -2)2+(y +1)2=4,圆心到直线l 的距离d =|1+m |2,依题意得|1+m |2=2,解得m =1或-3.故选D.答案 D4.多面体MN -ABCD 的底面ABCD 为矩形,其正视图和侧视图如图,其中正视图为等腰梯形,侧视图为等腰三角形,则该多面体的体积是( )A.16+33B.8+632C.163D.203解析 将多面体分割成一个三棱柱和一个四棱锥,如图所示,∵正视图为等腰梯形,侧视图为等腰三角形,∴四棱锥底面BCFE 为正方形,S BCFE =2×2=4,四棱锥的高为2,∴V N -BCFE =13×4×2=83.可将三棱柱补成直三棱柱,则 V ADM -EFN =12×2×2×2=4,∴多面体的体积为203.故选D. 答案 D5.若函数f (x )=sin ⎝ ⎛⎭⎪⎫ωx +π6(ω>0)的图象的相邻两条对称轴之间的距离为π2,且该函数图象关于点(x 0,0)成中心对称,x 0∈⎣⎢⎡⎦⎥⎤0,π2,则x 0=( )A.5π12 B.π4 C.π3D.π6解析 由题意得T 2=π2,T =π,ω=2,又2x 0+π6=k π(k ∈Z ),x 0=k π2-π12(k ∈Z ),而x 0∈⎣⎢⎡⎦⎥⎤0,π2,∴x 0=5π12.答案 A6.已知向量a ,b 的模都是2,其夹角是60°,又OP →=3a +2b ,OQ →=a +3b ,则P ,Q 两点间的距离为( ) A.2 2 B. 3 C.2 3D. 2解析 ∵a ·b =|a |·|b |·cos 60°=2×2×12=2,PQ →=OQ →-OP →=-2a +b ,∴|PQ →|2=4a 2-4a ·b +b 2=12,∴|PQ →|=2 3. 答案 C7.在(1+x )6(1+y )4的展开式中,记x m y n 项的系数为f (m ,n ),则f (3,0)+f (2,1)+f (1,2)+f (0,3)=( ) A.45 B.60 C.120D.210解析 在(1+x )6的展开式中,x m 的系数为C m 6,在(1+y )4的展开式中,y n 的系数为C n 4,故f (m ,n )=C m 6·C n 4.从而f (3,0)=C 36·C 04=20,f (2,1)=C 26·C 14=60, f (1,2)=C 16·C 24=36,f (0,3)=C 06·C 34=4,所以f (3,0)+f (2,1)+f (1,2)+f (0,3)=120. 答案 C8.从装有除颜色外完全相同的3个白球和m 个黑球的布袋中随机摸取一球,有放回地摸取5次,设摸得白球数为X ,已知E (X )=3,则D (X )=( ) A.85 B.65 C.45D.25解析 由题意,X ~B ⎝ ⎛⎭⎪⎫5,3m +3,又E (X )=5×3m +3=3,∴m =2, 则X ~B ⎝ ⎛⎭⎪⎫5,35,故D (X )=5×35×⎝ ⎛⎭⎪⎫1-35=65.答案 B9.设双曲线x 24-y 23=1的左、右焦点分别为F 1、F 2,过F 1的直线l 交双曲线左支于A 、B 两点,则|BF 2|+|AF 2|的最小值为( ) A.192 B.11 C.12D.16解析 由双曲线定义可得|AF 2|-|AF 1|=2a =4,|BF 2|-|BF 1|=2a =4,两式相加可得|AF 2|+|BF 2|=|AB |+8,由于AB 为经过双曲线的左焦点与左支相交的弦,而|AB |min =2b 2a =3,∴|AF 2|+|BF 2|=|AB |+8≥3+8=11. 答案 B10.已知函数f (x )=x 3+ax 2+bx +c ,且0<f (-1)=f (-2)=f (-3)≤3,则( ) A.c ≤3 B.3<c ≤6 C.6<c ≤9D.c >9解析 由题意,不妨设g (x )=x 3+ax 2+bx +c -m ,m ∈(0,3],则g (x )的三个零点分别为x 1=-3,x 2=-2,x 3=-1,因此有(x +1)(x +2)(x +3)=x 3+ax 2+bx +c -m ,则c -m =6,因此c =m +6∈(6,9]. 答案 C二、填空题(本大题共7小题,多空题每小题6分,单空题每小题4分,共36分.把答案填在题中的横线上)11.若x 、y 满足约束条件⎩⎨⎧x +y ≥1,x -y ≥-1,2x -y ≤2,若目标函数z =ax +3y 仅在点(1,0)处取得最小值,则实数a 的取值范围为________.解析 画出关于x 、y 约束条件的平面区域如图中阴影部分所示,当a =0时,显然成立.当a >0时,直线ax +3y -z =0的斜率k =-a3>k AC =-1,∴0<a <3.当a <0时,k =-a3<k AB =2,∴-6<a <0.综上所得,实数a 的取值范围是(-6,3).答案 (-6,3)12.已知{a n }为等差数列,若a 1+a 5+a 9=8π,则{a n }前9项的和S 9=________,cos(a 3+a 7)的值为________.解析 由{a n }为等差数列得a 1+a 5+a 9=3a 5=8π,解得a 5=8π3,所以{a n }前9项的和S 9=9(a 1+a 9)2=9a 5=9×8π3=24π.cos(a 3+a 7)=cos 2a 5=cos 16π3=cos 4π3=-12. 答案 24π -1213.函数f (x )=4sin x cos x +2cos 2x -1的最小正周期为________,最大值为________.解析 f (x )=2sin 2x +cos 2x =5sin(2x +φ),其中tan φ=12,所以最小正周期T=2π2=π,最大值为 5. 答案 π514.设函数f (x )=⎩⎪⎨⎪⎧|log 3(x +1)|,-1<x ≤0,tan ⎝ ⎛⎭⎪⎫π2x ,0<x <1,则f ⎣⎢⎡⎦⎥⎤f ⎝ ⎛⎭⎪⎫33-1=________,若f (a )<f ⎝ ⎛⎭⎪⎫12,则实数a 的取值范围是________.解析 由题意可得f ⎝ ⎛⎭⎪⎫33-1=⎪⎪⎪⎪⎪⎪log 333=12,则f ⎣⎢⎡⎦⎥⎤f ⎝ ⎛⎭⎪⎫33-1=f ⎝ ⎛⎭⎪⎫12=tan π4=1.-1<a ≤0时,f (a )=|log 3(a +1)|<1,-1<log 3(a +1)<1,解得-23<a <2,所以-23<a ≤0;当0<a <1时,f (a )=tan ⎝ ⎛⎭⎪⎫π2a <1,0<π2a <π4,0<a <12,综上可得实数a 的取值范围是⎝ ⎛⎭⎪⎫-23,12.答案 1 ⎝ ⎛⎭⎪⎫-23,1215.已知圆O :x 2+y 2=r 2与圆C :(x -2)2+y 2=r 2(r >0)在第一象限的一个公共点为P ,过点P 作与x 轴平行的直线分别交两圆于不同两点A ,B (异于P 点),且OA ⊥OB ,则直线OP 的斜率k =________,r =________.解析 两圆的方程相减可得点P 的横坐标为 1.易知P 为AB 的中点,因为OA ⊥OB ,所以|OP |=|AP |=|PB |,又|AO |=|OP |,所以△OAP 为等边三角形,同理可得△CBP 为等边三角形,所以∠OPC =60°.又|OP |=|OC |,所以△OCP 为等边三角形,所以∠POC =60°,所以直线OP 的斜率为 3.设P (1,y 1),则y 1=3,所以P (1,3),代入圆O ,解得r =2.答案3 216.已知偶函数f (x )满足f (x +2)=f (x ),且当x ∈[0,1]时,f (x )=x ,若区间[-1,3]上,函数g (x )=f (x )-kx -k 有3个零点,则实数k 的取值范围是________.解析 根据已知条件知函数f (x )为周期为2的周期函数;且x ∈[-1,1]时,f (x )=|x |;而函数g (x )的零点个数便是函数f (x )和函数y =kx +k 的交点个数. ①若k >0,如图所示,当y =kx +k 经过点(1,1)时,k =12;当经过点(3,1)时,k =14,∴14<k <12.②若k <0,即函数y =kx +k 在y 轴上的截距小于0,显然此时该直线与f (x )的图象不可能有三个交点,即这种情况不存在.③若k =0,得到直线y =0,显然与f (x )图象只有两个交点.综上所得,实数k 的取值范围是⎝ ⎛⎭⎪⎫14,12.答案 ⎝ ⎛⎭⎪⎫14,1217.已知数列{a n }满足a 1=-1,a 2>a 1,|a n +1-a n |=2n ,若数列{a 2n -1}单调递减,数列{a 2n }单调递增,则数列{a n }的通项公式为a n =________.解析 由题意得a 1=-1,a 2=1,a 3=-3,a 4=5,a 5=-11,a 6=21,……,然后从数字的变化上找规律,得a n +1-a n =(-1)n +12n ,则利用累加法即得a n =a 1+(a 2-a 1)+(a 3-a 2)+…+(a n -a n -1)=-1+2-22+…+(-1)n 2n -1=(-1)[1-(-2)n ]1-(-2)=(-2)n -13.答案 (-2)n -13限时练(二)(限时:40分钟)一、选择题(本大题共10小题,每小题4分,共40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.若f (x )=sin(2x +θ),则“f (x )的图象关于x =π3对称”是“θ=-π6”的( ) A.充分不必要条件 B.必要不充分条件 C.充要条件D.既不充分又不必要条件解析 若f (x )的图象关于x =π3对称,则2π3+θ=π2+k π,k ∈Z ,即θ=-π6+k π,k ∈Z ,此时θ的值不一定为-π6;若θ=-π6时,f (x )=sin ⎝ ⎛⎭⎪⎫2x -π6,2x -π6=π2+k π,k ∈Z ,∴x =π3+k π2,k ∈Z ,当k =0时,f (x )的图象关于x =π3对称.即“f (x )的图象关于x =π3对称”是“θ=-π6”的必要不充分条件. 答案 B2.若1a <1b <0,则下列四个不等式恒成立的是( ) A.|a |>|b | B.a <b C.a 3<b 3D.a +b <ab解析 由1a <1b <0可得b <a <0,从而|a |<|b |,b 3<a 3,即A ,B ,C 项均不正确;a +b <0,ab >0,则a +b <ab ,即D 项正确. 答案 D3.如图,AB 是⊙O 的直径,点C 、D 是半圆弧AB 上的两个三等分点,AB →=a ,AC→=b ,则AD →=( )A.12a +b B.12a -b C.a +12b D.a -12b解析 连接CD 、OD ,∵点C 、D 是半圆弧AB 的两个三等分点,∴AC ︵=BD ︵=CD ︵,∴CD ∥AB ,∠CAD =∠DAB =13×90°=30°,∵OA =OD ,∴∠ADO =∠DAO =30°,由此可得∠CAD =∠DAO =30°,∴AC ∥DO ,∴四边形ACDO 为平行四边形,∴AD →=AO →+AC →=12AB →+AC →=12a +b .答案 A4.在△ABC 中,角A 、B 、C 的对边分别是a 、b 、c ,若a =5b sin C ,且cos A =5cos B cos C ,则tan A 的值为( ) A.5 B.6 C.-4D.-6解析 由正弦定理得sin A =5sin B sin C ①,又cos A =5cos B cos C ②,②-①得,cos A -sin A =5(cos B cos C -sin B sin C )=5cos(B +C )=-5cos A ,∴sin A =6cos A ,∴tan A =6. 答案 B5.已知S n 表示数列{a n }的前n 项和,若对任意n ∈N *满足a n +1=a n +a 2,且a 3=2,则S 2 014=( ) A.1 006×2 013 B.1 006×2 014 C.1 007×2 013D.1 007×2 014解析 在a n +1=a n +a 2中,令n =1,则a 2=a 1+a 2,∴a 1=0,令n =2,则a 3=2a 2=2,∴a 2=1,于是a n +1-a n =1,∴数列{a n }是首项为0,公差为1的等差数列,∴S 2 014=2 014×2 0132=1 007×2 013.答案 C6.北京某大学为第十八届四中全会招募了30名志愿者(编号分别是1,2,…,30号),现从中任意选取6人按编号大小分成两组分配到江西厅、广电厅工作,其中三个编号较小的人在一组,三个编号较大的在另一组,那么确保6号、15号与24号同时入选并被分配到同一厅的选取种数是( ) A.25 B.32 C.60D.100解析 要“确保6号、15号与24号入选并分配到同一厅”,则另外三人的编号或都小于6或都大于24,于是根据分类加法计数原理,得选取种数是(C 35+C 36)A 22=60. 答案 C7.已知C 0n +2C 1n +22C 2n +23C 3n +…+2n C n n =729,则C 1n +C 2n +C 3n +…+C n n 等于( ) A.63 B.64 C.31D.32解析 逆用二项式定理得C 0n +2C 1n +22C 2n +23C 3n +…+2n C n n =(1+2)n =3n =729,即3n =36,所以n =6,所以C 1n +C 2n +C 3n +…+C n n =26-C 0n =64-1=63.答案 A8.口袋中有5只球,编号分别为1,2,3,4,5,从中任取3只球,以X 表示取出的球的最大号码,则X 的数学期望E (X )的值是( ) A.4 B.4.5 C.4.75D.5解析 由题意知,X 可以取3,4,5,P (X =3)=1C 35=110,P (X =4)=C 23C 35=310,P (X =5)=C 24C 35=610=35,所以E (X )=3×110+4×310+5×35=4.5. 答案 B9.椭圆ax 2+by 2=1(a >0,b >0)与直线y =1-x 交于A 、B 两点,过原点与线段AB 中点的直线的斜率为32,则ba =( ) A.32 B.233 C.932D.2327解析 A ,B 两点坐标分别为(x 1,y 1)、(x 2,y 2),AB 的中点为(x 中,y 中),代入椭圆方程得ax 21+by 21=1,ax 22+by 22=1,由两式相减整理得:b a ·y 1-y 2x 1-x 2·y 1+y 2x 1+x 2= -1,即b a ·y 1-y 2x 1-x 2·y 中x 中=-1,又y 中x 中=y 中-0x 中-0=32,可得b a ·(-1)·32=-1,即ba =233. 答案 B10.如图,在棱长为a 的正方体ABCD -A 1B 1C 1D 1中,P 是A 1D 1的中点,Q 是A 1B 1上任意一点,E 、F 是CD 上任意两点,且EF 长为定值,现有下列结论:①异面直线PQ 与EF 所成的角为定值; ②点P 到平面QEF 的距离为定值; ③直线PQ 与平面PEF 所成的角为定值; ④三棱锥P -QEF 的体积为定值. 其中正确结论的个数为( ) A.0B.1C.2D.3解析 当点Q 与A 1重合时,异面直线PQ 与EF 所成的角为π2;当点Q 与B 1重合时,异面直线PQ 与EF 所成的角不为π2,即①错误;平面QEF 也是平面A 1B 1CD ,既然P 和A 1B 1CD 都是定的,所以P 到平面QEF 的距离是定值,即②正确;因为EF 是定长,Q 到EF 的距离就是Q 到CD 的距离也为定长,④也正确;当点Q 在A 1B 1上运动时,直线QP 与平面PEF 所成的角随点Q 的变化而变化,即③错误. 答案 C二、填空题(本大题共7小题,多空题每小题6分,单空题每小题4分,共36分.把答案填在题中的横线上)11.α,β是两个平面,m ,n 是两条直线,有下列四个命题: ①如果m ⊥n ,m ⊥α,n ∥β,那么α⊥β. ②如果m ⊥α,n ∥α,那么m ⊥n . ③如果α∥β,m ⊂α,那么m ∥β.④如果m ∥n ,α∥β,那么m 与α所成的角和n 与β所成的角相等. 其中正确的命题有________(填写所有正确命题的编号).解析 当m ⊥n ,m ⊥α,n ∥β时,两个平面的位置关系不确定,故①错误,经判断知②③④均正确. 答案 ②③④12.以椭圆x 24+y 2=1的焦点为顶点、长轴顶点为焦点的双曲线的渐近线方程是________,离心率为________.解析 设双曲线的方程为x 2a 2-y 2b 2=1(a >0,b >0),由题意得双曲线的顶点为(±3,0),焦点为(±2,0),所以a =3,c =2,所以b =1,所以双曲线的渐近线方程为y =±b a x =±33x ,离心率为e =c a =233. 答案 y =±33x 23313.函数f (x )=2sin(ωx +φ)⎝ ⎛⎭⎪⎫ω>0,|φ|<π2的图象如图所示,则ω=________,φ=________.解析 由图象知函数f (x )的周期为π,所以ω=2πT =2,所以f (x )=2sin(2x +φ).把点(π,1)代入得2sin(2π+φ)=1,即sin φ=12.因为|φ|<π2,所以φ=π6. 答案 2 π614.某几何体的三视图如图所示(单位:cm),则该几何体的体积为________cm 3,表面积为________cm 2.解析 由三视图知该几何体为一个半球被割去14后剩下的部分,其球半径为1,所以该几何体的体积为12×34×43π×13=π2,表面积为12×34×4π×12+34×π×12+2×14×π×12=11π4. 答案π2 11π415.已知x ,y ∈R 且满足不等式组⎩⎨⎧x ≥1,2x +y -5≤0,kx -y -k -1≤0.当k =1时,不等式组所表示的平面区域的面积为________; 若目标函数z =3x +y 的最大值为7,则k 的值为________.解析当k =1时,不等式组为⎩⎨⎧x ≥1,2x +y -5≤0,x -y -2≤0,作出不等式组满足的平面区域如图中△ABC ,易求得A (1,3),B (1,-1),C ⎝ ⎛⎭⎪⎫73,13,所以S △ABC =12×4×43=83;由目标函数z =3x +y 的最大值为7知⎩⎨⎧3x +y =7,2x +y -5=0,解得⎩⎨⎧x =2,y =1,则点(2,1)在kx -y -k -1=0上,即2k -1-k -1=0,解得k =2.答案 83 216.在实数集R 中定义一种运算“*”,对任意a 、b ∈R ,a *b 为唯一确定的实数,且具有性质:(1)对任意a ∈R ,a *0=a ;(2)对任意a 、b ∈R ,a *b =ab +(a *0)+(b *0). 关于函数f (x )=(e x )*1e x 的性质,有如下说法:①函数f (x )的最小值为3;②函数f (x )为偶函数;③函数f (x )的单调递增区间为 (-∞,0].其中所有正确说法的序号为________.解析 依题意得f (x )=(e x )*1e x =e x ·1e x +[(e x )*0]+⎣⎢⎡⎦⎥⎤⎝ ⎛⎭⎪⎫1e x *0=1+e x +1e x ,其中x ∈R .∴f ′(x )=e x -1e x ,令f ′(x )=0,则x =0,∴函数f (x )在(-∞,0)上单调递减,在(0,+∞)上单调递增,∴当x =0,f (0)min =3,即①正确,③错误.又f (-x )=1+e -x +1e-x =1+e x+1e x =f (x ),∴函数f (x )为偶函数,即②正确. 答案 ①② 17.若关于x 的方程|x |x +2=kx 2有四个不同的实根,则实数k 的取值范围是________.解析由于关于x的方程|x|x+2=kx2有四个不同的实根,x=0是此方程的一个根,故关于x的方程|x|x+2=kx2有3个不同的非零的实数解.∴方程1k=⎩⎨⎧x(x+2),x>0,-x(x+2),x<0有3个不同的非零的实数解,即函数y=1k的图象和函数g(x)=⎩⎨⎧x(x+2),x>0,-x(x+2),x<0的图象有3个交点,画出函数g(x)图象,如图所示,故0<1k<1,解得k>1.答案(1,+∞)限时练(三)(限时:40分钟)一、选择题(本大题共10小题,每小题4分,共40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.设平面α与平面β相交于直线m,直线a在平面α内,直线b在平面β内,且b⊥m,则“a⊥b”是“α⊥β”的()A.充分不必要条件B.必要不充分条件C.充分必要条件D.既不充分也不必要条件解析因为α⊥β,b⊥m,α∩β=m,b⊂β,所以b⊥α,又直线a在平面α内,所以a⊥b;但直线a,m不一定相交,所以“a⊥b”是“α⊥β”的必要不充分条件,故选B.答案 B2.已知a=413,b=log1413,c=log314,则()A.a>b>cB.b>c>aC.c>b>aD.b>a>c解析因为a=413>1,0<b=log1413=log43<1,c=log314<0,所以a>b>c,故选A.答案 A3.已知函数f(x)=sin x-cos x,且f′(x)=12f(x),则tan 2x的值是()A.-23 B.-43C.-34 D.34解析因为f′(x)=cos x+sin x=12sin x-12cos x,所以tan x=-3,所以tan 2x=2tan x1-tan2x=-61-9=34,故选D.答案 D4.若某几何体的三视图(单位:cm)如图所示,则此几何体的体积是()A.36 cm3B.48 cm3C.60 cm3D.72 cm3解析由三视图可知,上面是个长为4,宽为2,高为2的长方体,下面是一个放倒的四棱柱,高为4,底面是个梯形,上、下底分别为2,6,高为2.所以长方体的体积为4×2×2=16,四棱柱的体积为4×2+62×2=32,所以该几何体的体积为32+16=48,选B.答案 B5.已知x,y满足约束条件⎩⎨⎧x≥2,x+y≤4,-2x+y+c≥0,目标函数z=6x+2y的最小值是10,则z 的最大值是( ) A.20 B.22 C.24D.26解析 由⎩⎨⎧6x +2y =10,x =2,解得⎩⎨⎧x =2,y =-1.代入直线-2x +y +c =0得c =5,即直线方程为-2x +y +5=0.平移直线3x +y =0,由⎩⎨⎧-2x +y +5=0,x +y =4,得⎩⎨⎧x =3,y =1,即D (3,1),当直线经过点D 时,直线的纵截距最大,此时z 取最大值,代入直线z =6x +2y 得z =6×3+2=20,故选A. 答案 A6.等差数列{a n }中的a 4,a 2 016是函数f (x )=x 3-6x 2+4x -1的极值点,则log 14a 1 010=( ) A.12 B.2 C.-2D.-12解析 因为f ′(x )=3x 2-12x +4,而a 4和a 2 016为函数f (x )=x 3-6x 2+4x -1的极值点,所以a 4和a 2 016为f ′(x )=3x 2-12x +4=0的根,所以a 4+a 2 016=4,又a 4、a 1 010和a 2 016为等差数列,所以2a 1 010=a 4+a 2 016,即a 1 010=2,所以 log 14a 1 010=-12,故选D. 答案 D7.将标号为1,2,3,4的四个篮球分给三位小朋友,每位小朋友至少分到一个篮球,且标号1、2的两个篮球不能分给同一个小朋友,则不同的分法种数为( ) A.15 B.20 C.30D.42解析 四个篮球两个分到一组有C 24种,3个篮球进行全排列有A 33种,标号1、2的两个篮球分给一个小朋友有A 33种,所以有C 24A 33-A 33=36-6=30,故选C.答案 C8.二项式⎝ ⎛⎭⎪⎫x +2x 2n 的展开式中只有第6项的二项式系数最大,则展开式中的常数项是( ) A.180 B.90 C.45D.360解析 依题意n =10,则⎝ ⎛⎭⎪⎫x +2x 210的通项公式T r +1=C r 10(x )10-r ⎝ ⎛⎭⎪⎫2x 2r = 2r C r 10x 5-52r . 令5-52r =0,得r =2.∴展开式中的常数项T 3=22C 210=180. 答案 A9.已知函数f (x )=2x 2-4ax +2b 2,若a ∈{4,6,8},b ∈{3,5,7},则该函数有两个零点的概率为( ) A.13 B.23 C.59D.79解析 要使函数f (x )=2x 2-4ax +2b 2有两个零点,即方程x 2-2ax +b 2=0要有两个不等实根,则Δ=16a 2-16b 2>0,即a >b ,又a ∈{4,6,8},b ∈{3,5,7},故a 、b 的取法共有3×3=9种,其中满足a >b 的取法有(4,3),(6,3),(6,5),(8,3),(8,5),(8,7)6种,所以所求的概率为69=23. 答案 B10.已知点A 是抛物线y 2=4x 的对称轴与准线的交点,点B 是其焦点,点P 在该抛物线上,且满足|P A |=m |PB |,当m 取得最大值时,点P 恰在以A ,B 为焦点的双曲线上,则双曲线的离心率为( ) A.2-1 B.22-2 C.2+1D.22+2解析 设P (x ,y ),可知A (-1,0),B (1,0), 所以m =|P A ||PB |=(x +1)2+y 2(x -1)2+y 2=(x +1)2+4x(x -1)2+4x=1+4xx 2+2x +1,当x=0时,m =1;当x >0时,m =1+4xx 2+2x +1=1+4x +1x +2≤ 2.当且仅当x =1x ,即x =1时取等号,所以P (1,±2),所以|P A |=22,|PB |=2,又点P 在以A ,B 为焦点的双曲线上,所以由双曲线的定义知2a =|P A |-|PB |=22-2,即a =2-1,c =1,所以e =12-1=2+1,故选C.答案 C二、填空题(本大题共7小题,多空题每小题6分,单空题每小题4分,共36分.把答案填在题中的横线上)11.△ABC 中,点M 是边BC 的中点,|AB →|=4,|AC →|=3,则AM →·BC →=________.解析 AM →·BC →=12(AB →+AC →)·(AC →-AB →)=12(|AC →|2-|AB |2)=12(9-16)=-72. 答案 -7212.已知φ∈[0,π),函数f (x )=cos 2x +cos(x +φ)是偶函数,则φ=________,f (x )的最小值为________.解析 因为函数f (x )为偶函数,所以cos 2x +cos(x +φ)=cos(-2x )+cos(-x +φ),即cos(x +φ)=cos(x -φ).因为φ∈[0,π),所以x +φ=x -φ,所以φ=0,所以f (x )=cos 2x +cos x =2cos 2x -1+cos x =2⎝ ⎛⎭⎪⎫cos x +142-98,所以当cos x =-14时,f (x )取得最小值-98. 答案 0 -9813.已知函数f (x )=⎩⎨⎧log 2 x ,x >0,x 2+x ,x ≤0,则f ⎝ ⎛⎭⎪⎫f ⎝ ⎛⎭⎪⎫12=________,方程f (x )=2的解为________.解析 因为f ⎝ ⎛⎭⎪⎫12=log 212=-1,所以f ⎝ ⎛⎭⎪⎫f ⎝ ⎛⎭⎪⎫12=f (-1)=(-1)2-1=0;当x ≤0时,由x 2+x =2,解得x =-2,当x >0时,由log 2x =2,解得x =4.答案 0 -2或414.在数列{a n }中,如果对任意n ∈N *都有a n +2-a n +1a n +1-a n=k (k 为常数),则称{a n }为等差比数列,k 称为公差比.现给出下列命题: ①等差比数列的公差比一定不为0; ②等差数列一定是等差比数列;③若a n =-3n +2,则数列{a n }是等差比数列; ④若等比数列是等差比数列,则其公比等于公差比. 其中正确的命题的序号为________.解析 若k =0,{a n }为常数列,分母无意义,①正确;公差为0的等差数列不是等差比数列,②错误;a n +2-a n +1a n +1-a n=3,满足定义,③正确;设a n =a 1q n -1,则a n +2-a n +1a n +1-a n =a 1q n +1-a 1q na 1q n -a 1q n -1=q ,④正确.答案 ①③④15.已知向量a ,b ,且|b |=3,b ·(2a -b )=0,则|a |的最小值为________;|t b +(1-2t )a |(t ∈R )的最小值为________.解析 设向量a ,b 的夹角为θ,则b ·(2a -b )=2a ·b -b 2=2|a ||b |cos θ-|b |2=6|a |cos θ-9=0,所以|a |cos θ=32,当cos θ取得最大值1时,|a |取得最小值32;又由b ·(2a -b )=0,得2a ·b =b 2=9,所以|t b +(1-2t )a |2=t 2b 2+2a ·b (1-2t )t +(1-2t )2a 2=9t 2+9(1-2t )t +(1-2t )2a 2=(4|a |2-9)t 2+(9-4|a |2)t +|a |2=(4|a |2-9)⎝ ⎛⎭⎪⎫t -122+94,因为|a |≥32,所以4|a |2-9≥0,所以当t =12时,|t b +(1-2t )a |2取得最小值94,所以|t b +(1-2t )a |的最小值为32. 答案 32 3216.罐中有6个红球,4个白球,从中任取1球,记住颜色后再放回,连续摸取4次,设X 为取得红球的次数,则E (X )=________,D (X )=________.解析 因为是有放回地摸球,所以每次摸球(试验)摸得红球(成功)的概率均为35,连续摸4次(做4次试验),X 为取得红球(成功)的次数,则X ~B ⎝ ⎛⎭⎪⎫4,35,∴E (X )=4×35=125,D (X )=4×35⎝ ⎛⎭⎪⎫1-35=2425.答案 125 242517.若函数f (x )满足f (x -1)=1f (x )-1,当x ∈[-1,0]时,f (x )=x ,若在区间[-1,1)上,g (x )=f (x )-mx +m 有两个零点,则实数m 的取值范围是________. 解析 因为当x ∈[-1,0]时,f (x )=x ,所以当x ∈(0,1)时,x -1∈(-1,0),由f (x -1)=1f (x )-1可得,x -1=1f (x )-1,所以f (x )=1x -1+1,作出函数f (x )在[-1,1)上的图象如图所示,因为g (x )=f (x )-mx +m 有两个零点,所以y =f (x )的图象与直线y =mx -m 有两个交点,由图可得m ∈⎝ ⎛⎦⎥⎤0,12.答案 ⎝ ⎛⎦⎥⎤0,12限时练(四)(限时:40分钟)一、选择题(本大题共10小题,每小题4分,共40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P ={x |x 2-2x ≥3},Q ={x |2<x <4},则P ∩Q =( ) A.[3,4) B.(2,3] C.(-1.2) D.(-1,3]答案 A2.已知双曲线C :x 2a 2-y 2b 2=1(a >0,b >0)的离心率为52,则C 的渐近线方程为( ) A.y =±14x B.y =±13x C.y =±12x D.y =±x答案 C3.在平行四边形ABCD 中,AC 与BD 交于点O ,E 是线段OD 的中点,AE 的延长线与CD 交于点F .若AC →=a ,BD →=b ,则AF →=( ) A.14a +12b B.12a +14b C.23a +13bD.12a +23b解析 ∵AC →=a ,BD →=b ,∴AD →=AO →+OD →=12AC →+12BD →=12a +12b ,因为E 是OD 的中点,∴|DE ||EB |=13, ∴|DF |=13|AB |,∴DF →=13AB →=13(OB →-OA →) =13×⎝ ⎛⎭⎪⎫-12BD →-⎝ ⎛⎭⎪⎫-12AC → =16AC →-16BD →=16a -16b ,AF →=AD →+DF →=12a +12b +16a -16b =23a +13b .答案 C4.将函数y =cos 2x 的图象向左平移π4个单位,得到函数y =f (x )·cos x 的图象,则f (x )的表达式可以是( ) A.f (x )=-2sin xB.f (x )=2sin xC.f (x )=22sin 2x D.f (x )=22(sin 2x +cos 2x )解析 将函数y =cos 2x 的图象向左平移π4个单位,得到函数y =cos ⎣⎢⎡⎦⎥⎤2⎝ ⎛⎭⎪⎫x +π4=cos ⎝ ⎛⎭⎪⎫2x +π2=-sin 2x 的图象,因为-sin 2x =-2sin x cos x ,所以f (x )=-2sin x .答案 A5.设{a n }是等差数列,下列结论中正确的是( ) A.若a 1+a 2>0,则a 2+a 3>0 B.若a 1+a 3<0,则a 1+a 2<0 C.若0<a 1<a 2,则a 2>a 1a 3 D.若a 1<0,则(a 2-a 1)(a 2-a 3)>0解析 A ,B 选项易举反例,C 中若0<a 1<a 2,∴a 3>a 2>a 1>0,∵a 1+a 3>2a 1a 3,又2a 2=a 1+a 3,∴2a 2>2a 1a 3,即a 2>a 1a 3成立. 答案 C6.在直角坐标系中,P 点的坐标为⎝ ⎛⎭⎪⎫35,45,Q 是第三象限内一点,|OQ |=1且∠POQ=3π4,则Q 点的横坐标为( ) A.-7210 B.-325 C.-7212D.-8213解析 设∠xOP =α,则cos α=35,sin α=45,x Q =cos ⎝ ⎛⎭⎪⎫α+3π4=35·⎝ ⎛⎭⎪⎫-22-45×22=-7210. 答案 A7.某几何体的三视图如图所示,则该几何体的体积为( )A.13+π B.23+π C.13+2πD.23+2π解析 这是一个三棱锥与半个圆柱的组合体,V =12π×12×2+13×⎝ ⎛⎭⎪⎫12×1×2×1=π+13. 答案 A8.口袋里放有大小相等的两个红球和一个白球,有放回地每次摸取一个球,定义数列{a n }:a n =⎩⎨⎧-1,第n 次摸取红球,1,第n 次摸取白球,如果S n 为数列{a n }的前n 项和,那么S 7=3的概率为( ) A.C 57⎝⎛⎭⎪⎫132·⎝ ⎛⎭⎪⎫235B.C 27⎝⎛⎭⎪⎫232·⎝ ⎛⎭⎪⎫135C.C 57⎝ ⎛⎭⎪⎫132·⎝ ⎛⎭⎪⎫135D.C 37⎝ ⎛⎭⎪⎫132·⎝ ⎛⎭⎪⎫235 解析 S 7=3即为7次摸球中,有5次摸到白球,2次摸到红球,又摸到红球的概率为23,摸到白球的概率为13.故所求概率为P =C 27⎝ ⎛⎭⎪⎫232⎝ ⎛⎭⎪⎫135.答案 B9.设双曲线x 2a 2-y 2b 2=1(a >0,b >0)的右焦点是F ,左、右顶点分别是A 1,A 2,过F 作A 1A 2的垂线与双曲线交于B ,C 两点,若A 1B ⊥A 2C ,则该双曲线的渐近线的斜率为( ) A.±12 B.±22 C.±1D.± 2解析 不妨令B 在x 轴上方,因为BC 过右焦点F (c ,0),且垂直于x 轴,所以可求得B ,C 两点的坐标分别为⎝ ⎛⎭⎪⎫c ,b 2a ,⎝ ⎛⎭⎪⎫c ,-b 2a ,又A 1,A 2的坐标分别为(-a ,0),(a ,0),所以A 1B →=⎝ ⎛⎭⎪⎫c +a ,b 2a ,A 2C →=⎝ ⎛⎭⎪⎫c -a ,-b 2a ,因为A 1B ⊥A 2C ,所以A 1B →·A 2C →=0, 即(c +a )(c -a )-b 2a ·b 2a =0,即c 2-a 2-b 4a 2=0,所以b 2-b 4a 2=0,故b 2a 2=1,即b a =1,又双曲线的渐近线的斜率为±ba ,故该双曲线的渐近线的斜率为±1.答案 C10.现定义e i θ=cos θ+isin θ,其中i 为虚数单位,e 为自然对数的底,θ∈R ,且实数指数幂的运算性质对e i θ都适用,若a =C 05cos 5θ-C 25cos 3θsin 2θ+ C 45cos θsin 4θ,b =C 15cos 4θsin θ-C 35cos 2θsin 3θ+C 55sin 5θ,那么复数a +b i等于( )A.cos 5θ+isin 5θB.cos 5θ-isin 5θC.sin 5θ+icos 5θD.sin 5θ-icos 5θ解析 (e i θ=cos θ+isin θ其实为欧拉公式)a +b i =C 05cos 5θ+C 15cos 4θ(isin θ)-C 25cos 3θsin 2θ- C 35cos 2θ(isin 3θ)+C 45cos θsin 4θ+C 55(isin 5θ) =C 05cos 5θ+C 15cos 4θ(isin θ)+C 25cos 3θ(i 2sin 2θ)+ C 35cos 2θ(i 3sin 3θ)+C 45cos θ(i 4sin 4θ)+C 55(i 5sin 5θ)=(cos θ+isin θ)5=(e i θ)5=e i ×5θ=cos 5θ+isin 5θ. 答案 A二、填空题(本大题共7小题,多空题每小题6分,单空题每小题4分,共36分.把答案填在题中的横线上)11.若抛物线y 2=2px (p >0)的准线经过双曲线x 2-y 2=1的一个焦点,则p =________.解析 抛物线y 2=2px (p >0)的准线方程是x =-p2,双曲线x 2-y 2=1的一个焦点F 1(-2,0),因为抛物线y 2=2px (p >0)的准线经过双曲线x 2-y 2=1的一个焦点,所以-p2=-2,解得p =2 2. 答案 2 212.计算:log 222=________,2log 2 3+log 4 3=________.解析 log 222=log 22-12=-12,2log 23+log 43=232log 2 3=2log 2332=27=3 3. 答案 -12 3 313.已知{a n }是等差数列,公差d 不为零.若a 2,a 3,a 7成等比数列,且2a 1+a 2=1,则a 1=________,d =________.解析 由a 2,a 3,a 7成等比数列,得a 23=a 2a 7,则2d 2=-3a 1d ,则d =-32a 1.又2a 1+a 2=1,所以a 1=23,d =-1. 答案 23 -114.函数f (x )=sin 2x +sin x cos x +1的最小正周期是________,最小值是________. 解析 由题可得f (x )=22sin ⎝ ⎛⎭⎪⎫2x -π4+32 ,所以最小正周期T =π,最小值为3-22. 答案 π3-2215.设函数f (x )=-ln(-x +1),g (x )=⎩⎨⎧x 2 (x ≥0),f (x ) (x <0),则g (-2)=________;函数y =g (x )+1的零点是________.解析 由题意知g (-2)=f (-2)=-ln 3,当x ≥0时,x 2+1=0没有零点,当x <0时,由-ln(-x +1)+1=0,得x =1-e. 答案 -ln 3 1-e16.已知实数x 、y 满足⎩⎨⎧y ≤2,3x -y -3≤0,2x +y -2≥0,则目标函数z =3x +y 的最大值为________.解析 作出可行域如图所示:作直线l 0:3x +y =0,再作一组平行于l 0的直线l :3x +y =z ,当直线l 经过点M 时,z =3x +y 取得最大值, 由⎩⎨⎧3x -y -3=0,y =2,得⎩⎪⎨⎪⎧x =53,y =2,所以点M 的坐标为⎝ ⎛⎭⎪⎫53,2,所以z max =3×53+2=7.答案 717.已知平面四边形ABCD 为凸四边形(凸四边形即任取平面四边形一边所在直线,其余各边均在此直线的同侧),且AB =2,BC =4,CD =5,DA =3,则平面四边形ABCD 面积的最大值为________.解析 设AC =x ,在△ABC 中,由余弦定理有: x 2=22+42-2×2×4cos B =20-16cos B , 同理,在△ADC 中,由余弦定理有: x 2=32+52-2×3×5cos D =34-30cos D , 即15cos D -8cos B =7,①又平面四边形ABCD 面积为S =12×2×4sin B +12×3×5sin D =12(8sin B +15sin D ),即8sin B +15sin D =2S ,② ①②平方相加得64+225+240(sin B sin D -cos B cos D )=49+4S 2, 即S 2=240[1-cos (B +D )]4,当B +D =π时,S 取最大值230. 答案 230。