高三限时训练(二轮复习)

一、词汇语法巩固练习A)1.The researchers who study jokes want to know __________ people from differentnations and cultures find very funny.2.Just like a voyage at sea, our life journey, __________ days are limited, is full ofdifficulties as well as opportunities and therefore needs to be cherished.3.I had just nodded off to sleep ___________ I was woken up by an almighty crashfrom their house.4.It is known that water is not an endless resource, ___________ can it be madeonce more, so we should save and protect it.ter in this chapter the case will be introduced to readers _________ consumers’support became a stream of motivation for improvement of the company.6.Mr. Zhang gave me many valuable presents, __________ that I had never seen.7.In spite __________ what has recently been done to provide more buses forpeople, a shortage of public vehicles remains a serious problem.8.__________ great the difficulty may be, we will not lose heart because we knowwe are supported by you all.9.Having a mix of male and female nurses also helps create a fun atmosphere,__________ helps patients recover faster.10.If you come across an 85-year-old woman __________ (walk) slowly with awalker in the Children's Hospital of Soochow University, don't mistake her _____a patient. Instead, she is a doctor, _______ still insists on working in the hospital. B)1.Male nurses are difficult to hire as many men reject this ______ out of thelong-existing discrimination.A. conceptB. assessmentC. identityD. personality2.An artist who was recently travelling on a ferry to the southern island discovered______ a long lost antique Greek vase.A. at randomB. by chanceC. on occasionD. in turn3.______ honesty, I know a lot of people who don’t pay bus fares.A. Concentrating onB. Speaking ofC. Leading toD. Living up to4.For many years, we have been trying to ______ to such people the fact thatlanguage is slightly different in expressions.A. wind upB. get acrossC. go all outD. look forward5.My mission is to remind you of the ______ power within your own being and toencourage you to recognize and use it.A. ultimateB. energeticC. incredibleD. dramatic6.Angela and Ajani will pick up the marriage______ after they meet online severaltimes, which is not sensible.A. acquisitionB. identificationC. certificateD. qualification7.I went to the ticket office, but the tickets had been sold out already. I could have______ myself an unnecessary trip by phoning in advance.A. preventedB. guaranteedC. enjoyedD. spared8.They stared at me, their ____ faces giving no clue as to what they would say next.A. dizzyB. funnyC. mildD. blank9.Mr. Brown thought it would be a good idea to have ______ from his workers toimprove his plan.A. summariesB. admissionsC. commentsD. connections10.Gradually I became ______ to having her follow me around the house andwatching me with those dark almond shaped eyes.A. accustomedB. tolerantC. consistentD. concerned11.Nowadays, basic health care services are ______ to almost all the Chinese people.This accounts for the fact that the average life expectancy of all the Chinese has already risen to 75.A. accessibleB. abundantC. accurateD. attractive12.Japanese media report the trend of wearing adult diapers is becoming ______popular, especially for women looking to save time.A. frequentlyB. increasinglyC. individuallyD. deliberately13.When kids are struggling with subjects like reading or math, schools may provideteaching ______, a specific program to address academic needs.A. comparisonB. accumulationC. interventionD. conservation14.Mary was greatly inspired though she made ______ improvements in her Englishwriting.A. brilliantB. modestC. substantialD. considerable15.That Peter has spent too much time on novels recently may ______ his poorgrade in the exam.A. answer forB. stand forC. account forD. apply for二、七选五Mind-wandering can help boost our moodPrior research suggests a wandering mind is an unhappy mind: We tend to be less happy when we're not focused on what we're doing. That's likely true. _____1_____ For example, as one 2013 study showed, when people found their wandering thoughts more interesting, their moods actually improved while mind-wandering. Similarly, other studies have found that thinking about people you love or thinking more about your potential future than about what happened in the past produces positive results.How you use mind-wandering may also be important. ___2___ It has been mostly unexplored in previous research, but likely has distinct effects. As one 2017 study found, people who use daydreaming for self-reflection typically have more pleasant thoughts than people who simply think about unpleasant experiences.There is even some evidence that mind-wandering may be more of an antidote (缓解方法) to depression than a cause. People who are depressed may simply replay events from their past to better understand what happened to cause their dark mood and avoid future problems. Also, when researchers studied whether a negative mood preceded or allowed a mind-wandering episode, they found poor moods led to more mind-wandering but not vice versa. _____3_____Now, findings from a 2021 study suggest that mind-wandering that is more freely moving can actually improve your mood. In this study, participants were prompted randomly via cell phone over three days to report how they were feeling and how much their thoughts were freely moving and related to what they were doing. After analyzing the data, the researchers found that when people's thoughts were of-task, they generally felt more negative—similar to what earlier findings showed. _____4_____“Our findings suggest there might be positive aspects of mind-wandering," the researchers conclude.____5____ If I simply put myself in a space that lets my mind move freely, I don’t get depressed. On the contrary, I'm happier because of it.A. In some cases, people intentionally mind-wander.B. Mind-wandering is part of our human inheritance.C. Again, I find that science supports my own experience.D. There may be a right and a wrong way to mind-wander.E. But if their thoughts were free-moving, it had the opposite effect.F. In fact, the content of wandering thoughts makes a big difference.G. It suggests that mind-wandering may be helping people feel better.三、完形填空Growing up, the relationship between my father and I is that I asked him questions and he told me the answers. In my teen years, he taught me things I’d need to know to ___1___ in the real world. When I moved out on my own, I called him at least once a week, usually when something broke in my apartment and I needed to know how to ___2___ it.But then, ___3___ I needed him less. I got married, and my husband had most of the knowledge I ___4___. For everything else, we had Google. I don’t know when it ___5___ but our conversations when I ___6___ are just six words. Me: “Hi, Dad.” Him: “Hi, sweets. Here’s Mom.” I loved my dad, of course, but I ___7___ at times if maybe he had already shared everything I ___8___ to know.Then, this past summer, my family moved to my parents’ ___9___ our house was being renovated. Dad asked me to help him rebuild a lake house with wood. Without hesitation, I agreed. After all, I could do for free rent. As we put the new wood together piece by piece, my dad knowing ___10___ what went where, I looked at him. “How do you know how to build a(n) ___11___?” “I spent a summer in college building it on the Jersey Shore.” “You did?”I thought I knew ___12___ about my dad. I knew the summer when he burned his hands raw at the horseradish (辣根酱) manufacturing plant and the diner line-cook (流水线厨师) position that ___13___ him how to make the best egg roll. But I ___14___ knew this. I realized that maybe it’s not that there’s nothing left to say ___15__ it’s just that I’ve spent my life asking him the wrong questions.1. A. survive B. exist C. last D. live2. A. work B. fix C. store D. purchase3. A. consequently B. gradually C. actually D. frequently4. A. gained B. requested C. lacked D. grasped5. A. appeared B. happened C. changed D. challenged6. A. visited B. chatted C. called D. asked7. A. wondered B. considered C. ordered D. wandered8. A. desired B. seemed C. suspected D. needed9. A. though B. while C. unless D. if10. A. nearly B. exactly C. originally D. merely11. A. apartment B. garden C. cabin D. house12. A. nothing B. something C. everything D. anything13. A. constructed B. suggested C. caught D. taught14. A. ever B. never C. hardly D. almost15. A. but B. and C. or D. so四、语法填空A)As a general rule, all forms of activity lead to boredom when they are performed _____1_____ a routine basis. As a matter of fact, we can see this principle at work in people of all _____2_____ (age). For example, on Christmas morning, children are excited about playing with their new toys. But their interest soon _____3_____ (wear) off and by January those same toys can be found put away in the basement. The world is full of half­filled stamp albums and _____4_____ (finish) models, each _____5_____ (stand) as a monument to someone's passing interest. When parents bring home a pet, their child gladly bathes it and brushes its fur. Within a short time, _____6_____, the burden of caring for the animal is handed over to the parents. Adolescents enter high school with great _____7_____ (excite) but are soon looking forward to graduation. The same is true of the young adults going to college. And then, how many adults, who now complain about the long drives to work, eagerly drove for hours at _____8_____ time when they first obtained their driver's licence? Before people retire, they usually plan to do a lot of great things, _____9_____ they never had time to do while working. But soon after retirement, the golfing, the fishing, the reading and all of the other pastimes become as _____10_____ (bore) as the jobs they left. And, like the child in January, they go searching for new toys.B)White Cane Safety Day is an observance celebrated on October 14 of each year since 1964. It is a day of the White Cane, a tool that allows the visually impaired (障碍) people to travel independently and get ___1___ (identify) easily.The White Cane ___2___ (introduce) by an American man called George A. Bonham. One day, he watched a man who was blind ___3___ (attempt) to cross a street. The man’s cane was black and motorists couldn’t see it, so Bonham proposed painting the cane white with a red stripe to make it more ___4___ (notice). The ideaquickly ___5___ (catch) on around the U.S., making the White Cane an important tool for blind people to move ___6___(safe).Today, there are about 17,000,000 people with visual impairment in China, over 8,000,000 of ___7___ are blind. It is important that people raise awareness of their ___8___ (exist) and help improve their living conditions.Modern White Canes are becoming more and more lightweight and high-tech, making it ___9___ (easy) for them to travel independently. But it usually takes more effort for them to receive education and pursue careers. Therefore, it is important that the whole society help create ____10____ barrier-free environment for them.一、词汇语法巩固练习A) 1. what 2. whose 3. when 4. nor 5. where6. ones7. of8. However9. which 10. walking; for; whoB) CBBBC CDDCA ABCBC二、FAGEC三、21-25 ABBCB 26-30 CADBB 31-35 CCDBA四、A) 1. on 2. ages 3. wears 4. unfinished 5. standing6. however7. excitement8. a9. which 10. boringB) 1. identified 2. was introduced 3. attempting 4. noticeable 5. caught6. safely7. whom8. existence9. easier 10. a本文是说明文。

xzOy v 高三物理第二轮复习专题限时训练10一．选择题1．水平地面上有一个倾角为θ的斜面，其表面绝缘。

另一个带正电的滑块放在斜面上，两物体均处于静止状态，如图所示。

当加上水平向右的匀强电场后，滑块与斜面仍相对地面静止（ ） A ．滑块与斜面间的摩擦力一定变大 B ．斜面体与地面间的摩擦力可能不变 C ．滑块对斜面的压力一定变大 D. 斜面体对地面的压力一定变大2．图中平行金属板中带电质点P 处于静止状态，不考虑电流表和电压表对电路的影响，当滑动变阻器R 4的滑片向b 端移动时 ( ) A ．电流表读数减小 B ．电压表读数增大 C ．R 4上消耗的功率可能先增加后减小 D ．质点P 将向上运动3．如图所示，在同时存在匀强电场、匀强磁场的空间中取正交坐标系Oxyz （z 轴正方向竖直向上），一质量为m 、电荷量为q 的带正电粒子（重力不能忽略）从原点O 以速度 v 沿x 轴正方向出发。

下列说法正确的是（ ）A ．若电场、磁场分别沿z 轴正方向和x 轴正方向，粒子只能做曲线运动B ．若电场、磁场均沿z 轴正方向, 粒子有可能做匀速圆周运动C ．若电场、磁场分别沿y 轴负方向和z 轴正方向，粒子有可能做平抛运动D ．若电场、磁场分别沿z 轴正方向和y 轴负方向，粒子有可能做匀速直线运动4．右图所示，一个由绝缘材料制成的轻弹簧水平放置，一端固定于竖直墙上，另一端与一带负电的小球相连，小球置于光滑的绝缘水平面上。

当整个装置处于水平向左的匀强电场中时，振子在O 点处于平衡状态，在B 、C 两点间振动。

假定在振动过程中小球的电荷量保持不变，则 ( )A ．小球在由B 到O 的过程中，弹性势能和电势能减少，动能增加 B ．小球在由O 到C 的过程中，弹性势能增加，电势能和动能减少C ．小球在由B 经O 到C ，电势能的变化量和弹性势能的变化量大小相等D ．小球在由C 到O 的过程中，电势能的变化量和弹性势能的变化量大小相等5．如图所示，绝缘的斜劈A 置于水平地面上，滑块B 带正电，在竖直平面内存在匀强电场。

高考二轮复习限时训练（一）（时间：60分钟）班级 姓名 得分一、填空题：本大题共12小题，每小题5分，共60分。

1、复数ii4321+-在复平面上对应的点位于第__ 象限. 2、命题“2,220x R x x ∃∈++≤”的否定是3、设{}{}=⋂+==∈==B A x y y x B R x x y y A 则,2|),(,,|2 4、已知x 、y 的取值如下表：从散点图分析，y 与x 线性相关，且回归方程为 0.95y x a =+，则a = 5、若椭圆的焦距长等于它的短轴长，则椭圆的离心率等于____ ____。

6、如果执行右面的程序框图，那么输出的S = 7、把函数4cos()3y x π=++1的图象向左平移ϕ个单位，所得的图象对应的函数为偶函数，则ϕ的最小正值为8、如果实数x ,y 满足x 2+y 2=1,则（1+xy ）(1－xy )的最小值为9、已知实数x ，y 满足条件5003x y x y x -+⎧⎪+⎨⎪⎩≥≥≤，i z x y =+（i 为虚数单位），则|12i |z -+的最小值是 ．10、一枚半径为1的硬币随机落在边长为3的正方形所在平面内，且硬币一定落在正方形内部或与正方形有公共点，则硬币与正方形没有公共点的概率是11、若函数f (x )=e x -2x-a 在R 上有两个零点，则实数a 的取值范围是12、设函数)0](,[,321)1ln()(2>-∈+-+=t t t x x e x x f x ，若函数的最大值是M ，最小值是m ，则M+m=二、解答题：本大题共2小题，共30分。

13、（本小题满分15分）在ABC ∆中，角A 、B 、C 的对边分别为,,a b c ，已知向量33(cos ,sin ),22A A m = (cos ,sin ),22A An =且满足m n +=（Ⅰ）求角A 的大小；（Ⅱ）若,b c +=试判断ABC ∆的形状。

14、（本小题满分15分）已知圆C ：224x y +=.（1）直线l 过点()1,2P ，且与圆C 交于A 、B 两点，若||AB =l 的方程;（2）过圆C 上一动点M 作平行于x 轴的直线m ，设m 与y 轴的交点为N ，若向量OQ OM ON =+，求动点Q 的轨迹方程，并说明此轨迹是什么曲线.高考二轮复习限时训练（一）1、三2、2,220.x R x x ∀∈++>3、∅4、2.6 5 6、25507、32π 8、43 9 10、 π+21111、()+∞-,2ln 2212、6 二、解答题： 13、（1）3A π=———————————————————7分（2）ABC ∆为直角三角形。

2024届高三二轮复习“8+3+3”小题强化训练(1)一、选择题：本题共8小题，每小题5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对两个具有线性相关关系的变量x 和y 进行统计时，得到一组数据1,0.3 ,2,4.7 ,3,m ,4,8 ，通过这组数据求得回归直线方程为y=2.4x -2，则m 的值为()A.3B.5C.5.2D.6【答案】A【解析】易知x =1+2+3+44=52,y =13+m4，代入y =2.4x -2得13+m 4=2.4×52-2⇒m =3.故选：A2已知m ，n 表示两条不同直线，α表示平面，下列说法正确的是()A.若m ⎳α,n ⎳α,则m ⎳nB.若m ⊥α，n ⊂α，则m ⊥nC.若m ⊥α，m ⊥n ，则n ⎳αD.若m ⎳α，m ⊥n ，则n ⊥α【答案】B【解析】线面垂直，则有该直线和平面内所有的直线都垂直，故B 正确.故选：B3已知向量a ，b 满足a =3，b =23，且a ⊥a +b，则b 在a 方向上的投影向量为()A.3B.-3C.-3aD.-a【答案】D【解析】a ⊥a +b ，则a ⋅a +b =a 2+a ⋅b =9+a ⋅b =0，故a ⋅b=-9，b 在a 方向上的投影向量a ⋅b a 2⋅a =-99⋅a =-a.故选：D .4若n 为一组从小到大排列的数1，2，4，8，9，10的第六十百分位数，则二项式3x +12xn的展开式的常数项是()A.7B.8C.9D.10【答案】A【解析】因为n 为一组从小到大排列的数1，2，4，8，9，10的第六十百分位数，6×60%=3.6，所以n =8，二项式3x +12x8的通项公式为T r +1=C r 8⋅3x 8-r ⋅12x r =C r 8⋅12 r⋅x8-r 3-r，令8-r 3-r =0⇒r =2，所以常数项为C 28×12 2=8×72×14=7，故选：A5折扇是我国古老文化的延续，在我国已有四千年左右的历史，“扇”与“善”谐音，折扇也寓意“善良”“善行”.它常以字画的形式体现我国的传统文化，也是运筹帷幄、决胜千里、大智大勇的象征(如图1).图2是一个圆台的侧面展开图(扇形的一部分)，若两个圆弧DE ，AC 所在圆的半径分别是3和6，且∠ABC =120°，则该圆台的体积为()A.5023π B.9π C.7π D.1423π【答案】D【解析】设圆台上下底面的半径分别为r 1,r 2，由题意可知13×2π×3=2πr 1，解得r 1=1，13×2π×6=2πr 2，解得：r 2=2，作出圆台的轴截面，如图所示：图中OD =r 1=1,O A =r 2=2，AD =6-3=3，过点D 向AP 作垂线，垂足为T ，则AT =r 2-r 1=1，所以圆台的高h =AD 2-AT 2=32-1=22，则上底面面积S 1=π×12=π，S 2=π×22=4π，由圆台的体积计算公式可得：V =13×(S 1+S 2+S 1⋅S 2)×h =13×7π×22=142π3，故选：D .6已知函数f x =x 2-bx +c (b >0,c >0)的两个零点分别为x 1,x 2，若x 1,x 2,-1三个数适当调整顺序后可为等差数列，也可为等比数列，则不等式x -bx -c≤0的解集为()A.1,52B.1,52C.-∞,1 ∪52,+∞D.-∞,1 ∪52,+∞ 【答案】A【解析】由函数f x =x 2-bx +c (b >0,c >0)的两个零点分别为x 1,x 2，即x 1,x 2是x 2-bx +c =0的两个实数根据，则x 1+x 2=b ,x 1x 2=c 因为b >0,c >0，可得x 1>0,x 2>0，又因为x 1,x 2,-1适当调整可以是等差数列和等比数列，不妨设x 1<x 2，可得x 1x 2=-1 2=1-1+x 2=2x 1 ，解得x 1=12,x 2=2，所以x 1+x 2=52,x 1x 2=1，所以b =52,c =1，则不等式x -b x -c ≤0，即为x -52x -1≤0，解得1<x ≤52，所以不等式的解集为1,52.故选：A .7已知双曲线C ：x 2a 2-y 2b2=1a >0,b >0 的左、右焦点分别为F 1，F 2，M ，N 为双曲线一条渐近线上的两点，A 为双曲线的右顶点，若四边形MF 1NF 2为矩形，且∠MAN =2π3，则双曲线C 的离心率为()A.3B.7C.213D.13【答案】C【解析】如图，因为四边形MF 1NF 2为矩形，所以MN =F 1F 2 =2c (矩形的对角线相等)，所以以MN 为直径的圆的方程为x 2+y 2=c 2.直线MN 为双曲线的一条渐近线，不妨设其方程为y =bax ，由y =b a x ,x 2+y 2=c 2,解得x =a y =b ，或x =-a ,y =-b , 所以N a ,b ，M -a ,-b 或N -a ,-b ，M a ,b .不妨设N a ,b ，M -a , -b ，又A a ,0 ，所以AM =a +a 2+b 2=4a 2+b 2，AN =a -a 2+b 2=b .在△AMN 中，∠MAN =2π3，由余弦定理得MN 2=AM 2+AN 2-2AM AN ⋅cos 2π3，即4c 2=4a 2+b 2+b 2+4a 2+b 2×b ，则2b =4a 2+b 2，所以4b 2=4a 2+b 2，则b 2=43a 2，所以e =1+b 2a2=213.故选:C .8已知a =ln 1.2e ,b =e 0.2,c =1.2e 0.2，则有()A.a <b <cB.a <c <bC.c <a <bD.c <b <a【答案】C【解析】令f x =e x -ln x +1 -1,x >0，则f x =e x -1x +1.当x >0时，有e x >1,1x +1<1，所以1x +1<1，所以，f (x )>0在0,+∞ 上恒成立，所以，f (x )在0,+∞ 上单调递增，所以，f (x )>f (0)=1-1=0，所以，f (0.2)>0，即e 0.2-ln1.2-1>0，所以a <b令g x =e x -x +1 ,x >0，则g x =e x -1在x >0时恒大于零，故g x 为增函数，所以x +1ex <1,x >0，而a =ln 1.2e =1+ln1.2>1，所以c <a ，所以c <a <b ，故选：C二、选择题：本题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项符合题目要求.全部选对的得6分，部分选对的得部分分，有选错的得0分.9已知函数f x =sin 2x +3π4 +cos 2x +3π4，则()A.函数f x -π4 为偶函数 B.曲线y =f x 对称轴为x =k π,k ∈ZC.f x 在区间π3,π2单调递增D.f x 的最小值为-2【答案】AC【解析】f x =sin 2x +3π4 +cos 2x +3π4=sin2x cos 3π4+sin 3π4cos2x +cos2x cos 3π4-sin2x sin3π4=-22sin2x +22cos2x -22cos2x -22sin2x =-2sin2x ，即f x =-2sin2x ，对于A ，f x -π4 =-2sin 2x -π2=2cos2x ，易知为偶函数，所以A 正确；对于B ，f x =-2sin2x 对称轴为2x =π2+k π,k ∈Z ⇒x =π4+k π2,k ∈Z ，故B 错误；对于C ，x ∈π3,π2 ,2x ∈2π3,π ，y =sin2x 单调递减，则f x =-2sin2x 单调递增，故C 正确；对于D ，f x =-2sin2x ，则sin2x ∈-1,1 ，所以f x ∈-2,2 ，故D 错误；故选：AC10设z 为复数，则下列命题中正确的是()A.z 2=zz B.若z =(1-2i )2，则复平面内z对应的点位于第二象限C.z 2=z 2D.若z =1，则z +i 的最大值为2【答案】ABD【解析】对于A ，设z =a +bi ，故z =a -bi ，则z 2=a 2+b 2，zz =(a +bi )(a -bi )=a 2+b 2，故z 2=zz成立，故A 正确，对于B ，z =(1-2i )2=-4i -3，z =4i -3，显然复平面内z对应的点位于第二象限，故B 正确，对于C ，易知z 2=a 2+b 2，z 2=a 2+b 2+2abi ，当ab ≠0时，z 2≠z 2，故C 错误，对于D ，若z =1，则a 2+b 2=1，而z +i =a 2+(b +1)2=2b +2，易得当b =1时，z +i 最大，此时z +i =2，故D 正确.故选：ABD11已知菱形ABCD 的边长为2，∠ABC =π3．将△DAC 沿着对角线AC 折起至△D AC ，连结BD ．设二面角D -AC -B 的大小为θ，则下列说法正确的是()A.若四面体D ABC 为正四面体，则θ=π3B.四面体D ABC 的体积最大值为1C.四面体D ABC 的表面积最大值为23+2D.当θ=2π3时，四面体D ABC 的外接球的半径为213【答案】BCD【解析】如图，取AC 中点O ，连接OB ,OD ，则OB =OD ，OB ⊥AC ,OD ⊥AC ，∠BOC 为二面角D AC -B 的平面角，即∠BOC =θ．若D ABC 是正四面体，则BD =BC ≠BO ，△OBD 不是正三角形，θ≠π3，A 错；四面体D ABC 的体积最大时，BO ⊥平面ACD ，此时B 到平面ACD 的距离最大为BO =3，而S △ACD=34×22=3，所以V =13×3×3=1，B 正确；S △ABC =S △DAC =3，易得△BAD ≅△BCD ，S △BAD=S △BCD=12×22sin ∠BCD =2sin ∠BCD ，未折叠时BD =BD =23，折叠到B ,D 重合时，BD =0，中间存在一个位置，使得BD =22，则BC 2+D C 2=BD 2，∠BCD =π2，此时S △BAD=S △BCD=2sin ∠BCD 取得最大值2，所以四面体D ABC 的表面积最大值为23+2 ，C 正确；当θ=2π3时，如图，设M ,N 分别是△ACD 和△BAC 的外心，在平面AOD 内作PM ⊥OD ，作PN ⊥OB ，PM ∩PN =P ，则P 是三棱锥外接球的球心，由上面证明过程知平面OBD 与平面ABC 、平面D AC 垂直，即P ,N ,O ,M 四点共面，θ=2π3，则∠PON =π3，ON =13×32×2=33，PN =ON tan π3=33×3=1，PB =PN 2+BN 2=12+233 2=213为球半径，D 正确．故选：BCD ．三、填空题：本题共3小题，每小题5分，共15分.12设集合M =x log 2x <1 ，N =x 2x -1<0 ，则M ∩N =．【答案】x 0<x <12【解析】因为log 2x <1=log 22，所以0<x <2，即M =x log 2x <1 =x 0<x <2 ，因为2x -1<0，解得x <12，所以N =x 2x -1<0 =x x <12，所以，M ∩N =x 0<x <12 .故答案为：x 0<x <12 13已知正项等比数列a n 的前n 项和为S n ，且S 8-2S 4=6，则a 9+a 10+a 11+a 12的最小值为.【答案】24【解析】设正项等比数列a n 的公比为q ，则q >0，所以，S 8=a 1+a 2+a 3+a 4+a 5+a 6+a 7+a 8=a 1+a 2+a 3+a 4+q 4a 1+a 2+a 3+a 4 =S 41+q 4 ，则S 8-2S 4=S 4q 4-1 =6，则q 4>1，可得q >1，则S 4=6q 4-1，所以，a 9+a 10+a 11+a 12=q 8a 1+a 2+a 3+a 4 =S 4q 8=6q 8q 4-1=6q 4-1+1 2q 4-1=6q 4-1 2+1+2q 4-1 q 4+1=6q 4-1 +1q 4-1+2 ≥62q 4-1 ⋅1q 4-1+2 =24，当且仅当q 4-1=1q 4-1q >1 时，即当q =42时，等号成立，故a 9+a 10+a 11+a 12的最小值为24.故答案为：2414已知F 为拋物线C :y =14x 2的焦点，过点F 的直线l 与拋物线C 交于不同的两点A ，B ，拋物线在点A ,B 处的切线分别为l 1和l 2，若l 1和l 2交于点P ，则|PF |2+25AB的最小值为.【答案】10【解析】C :x 2=4y 的焦点为0,1 ，设直线AB 方程为y =kx +1，A x 1,y 1 ,B x 2,y 2 .联立直线与抛物线方程有x 2-4kx -4=0，则AB =y 1+y 2+2=k x 1+x 2 +4=4k 2+4.又y =14x 2求导可得y =12x ，故直线AP 方程为y -y 1=12x 1x -x 1 .又y 1=14x 21，故AP :y =12x 1x -14x 21，同理BP :y =12x 2x -14x 22.联立y =12x 1x -14x 21y =12x 2x -14x 22可得12x 1-x 2 x =14x 21-x 22 ，解得x =x 1+x 22，代入可得P x 1+x 22,x 1x 24 ，代入韦达定理可得P 2k ,-1 ，故PF =4k 2+4.故|PF |2+25AB=4k 2+4+254k 2+4≥24k 2+4 ×254k 2+4=10，当且仅当4k 2+4=254k 2+4，即k =±12时取等号.故答案为：102024届高三二轮复习“8+3+3”小题强化训练(2)一、选择题：本题共8小题，每小题5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抛物线y =12x 2的焦点坐标为()A.18,0B.12,0 C.0,18D.0,12【答案】D 【解析】由y =12x 2可得抛物线标准方程为：x 2=2y ，∴其焦点坐标为0,12 .故选：D .2二项式3x 2-1x 47的展开式中常数项为()A.-7B.-21C.7D.21【答案】A 【解析】二项式3x 2-1x47的通项公式为Tr +1=C r 7⋅3x 27-r⋅-1x4r=Cr 7⋅-1 r⋅x14-14r 3，令14-14r 3=0⇒r =1，所以常数项为C 17⋅-1 =-7，故选：A3已知集合A =x log 2x ≤1 ，B =y y =2x ,x ≤2 ，则()A.A ∪B =BB.A ∪B =AC.A ∩B =BD.A ∪(C R B )=R【答案】A【解析】由log 2x ≤1，则log 2x ≤log 22，所以0<x ≤2，所以A =x log 2x ≤1 =x 0<x ≤2 ，又B =y y =2x ,x ≤2 =y 0<y ≤4 ，所以A ⊆B ，则A ∪B =B ，A ∩B =A .故选：A ．4若古典概型的样本空间Ω=1,2,3,4 ，事件A =1,2 ，甲：事件B =Ω，乙：事件A ,B 相互独立，则甲是乙的()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【答案】A【解析】若B =Ω，A ∩B =1,2 ，则P A ∩B =24=12，而P A =24=12，P B =1，所以P A P B =P A ∩B ，所以事件A ,B 相互独立，反过来，当B =1,3 ，A ∩B =1 ，此时P A ∩B =14，P A =P B =12，满足P A P B =P A ∩B ，事件A ,B 相互独立，所以不一定B =Ω，所以甲是乙的充分不必要条件.故选：A5若函数f x =ln e x -1 -mx 为偶函数，则实数m =()A.1B.-1C.12D.-12【答案】C【解析】由函数f x =ln e x -1 -mx 为偶函数，可得f -1 =f 1 ，即ln e -1-1 +m =ln e -1 -m ，解之得m =12，则f x =ln e x -1 -12x (x ≠0)，f -x =ln e -x -1 +12x =ln e x -1 -x +12x =ln e x -1 -12x =f x故f x =ln e x -1 -12x 为偶函数，符合题意.故选：C6已知函数y =f (x )的图象恰为椭圆C :x 2a 2+y 2b2=1(a >b >0)x 轴上方的部分，若f (s -t )，f (s )，f (s +t )成等比数列，则平面上点(s ，t )的轨迹是()A.线段(不包含端点) B.椭圆一部分C.双曲线一部分D.线段(不包含端点)和双曲线一部分【答案】A【解析】因为函数y =f (x )的图象恰为椭圆C :x 2a 2+y 2b2=1(a >b >0)x 轴上方的部分，所以y =f (x )=b ⋅1-x 2a2(-a <x <a )，因为f (s -t )，f (s )，f (s +t )成等比数列，所以有f 2(s )=f (s -t )⋅f (s +t )，且有-a <s <a ,-a <s -t <a ,-a <s +t <a 成立，即-a <s <a ,-a <t <a 成立，由f 2(s )=f (s -t )⋅f (s +t )⇒b ⋅1-s 2a 22=b ⋅1-(s -t )2a 2⋅b ⋅1-(s +t )2a 2，化简得：t 4=2a 2t 2+2s 2t 2⇒t 2(t 2-2a 2-2s 2)=0⇒t 2=0，或t 2-2a 2-2s 2=0，当t 2=0时，即t =0，因为-a <s <a ，所以平面上点(s ，t )的轨迹是线段(不包含端点)；当t 2-2a 2-2s 2=0时，即t 2=2a 2+2s 2，因为-a <t <a ，所以t 2<a 2，而2a 2+2s 2>a 2，所以t 2=2a 2+2s 2不成立，故选：A7若tan α+π4=-2，则sin α1-sin2α cos α-sin α=()A.65B.35C.-35D.-65【答案】C【解析】因为tan α+π4 =tan α+tan π41-tan αtan π4=tan α+11-tan α=-2，解得tan α=3，所以，sin α1-sin2αcos α-sin α=sin αsin 2α+cos 2α-2sin αcos α cos α-sin α=sin αcos α-sin α 2cos α-sin α=sin αcos α-sin 2α=sin αcos α-sin 2αcos 2α+sin 2α=tan α-tan 2α1+tan 2α=3-91+9=-35.故选：C .8函数f x =2ln xx,x >0sin ωx +π6,-π≤x ≤0，若2f 2(x )-3f (x )+1=0恰有6个不同实数解，正实数ω的范围为()A.103,4B.103,4 C.2,103D.2,103【答案】D【解析】由题知，2f 2x -3f x +1=0的实数解可转化为f (x )=12或f (x )=1的实数解，即y =f (x )与y =1或y =12的交点，当x >0时，f x =2ln xx ⇒f (x )=21-ln x x 2所以x ∈0,e 时，f (x )>0，f x 单调递增，x ∈e ,+∞ 时，f (x )<0，f x 单调递减，如图所示：所以x =e 时f x 有最大值：12<f (x )max =2e<1所以x >0时，由图可知y =f (x )与y =1无交点，即方程f (x )=1无解，y =f (x )与y =12有两个不同交点，即方程f (x )=12有2解当x <0时，因为ω>0，-π≤x ≤0，所以-ωπ+π6≤ωx +π6≤π6，令t =ωx +π6，则t ∈-ωπ+π6,π6则有y =sin t 且t ∈-ωπ+π6,π6，如图所示：因为x >0时，已有两个交点，所以只需保证y =sin t 与y =12及与y =1有四个交点即可，所以只需-19π6<-ωπ+π6≤-11π6，解得2≤ω<103．故选：D二、选择题：本题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项符合题目要求.全部选对的得6分，部分选对的得部分分，有选错的得0分.9已知复数z 1,z 2是关于x 的方程x 2+bx +1=0(-2<b <2,b ∈R )的两根，则下列说法中正确的是()A.z 1=z 2B.z 1z 2∈R C.z 1 =z 2 =1D.若b =1，则z 31=z 32=1【答案】ACD【解析】Δ=b 2-4<0，∴x =-b ±4-b 2i 2，不妨设z 1=-b 2+4-b 22i ，z 2=-b2-4-b 22i ，z 1=z 2，A 正确；z 1 =z 2 =-b 22+4-b 222=1，C 正确；z 1z 2=1，∴z 1z 2=z 21z 1z 2=z 21=b 2-22-b 4-b 22i ，b ≠0时，z 1z 2∉R ，B 错；b =1时，z 1=-12+32i ，z 2=-12-32i ，计算得z 21=-12-32i =z 2=z 1 ，z 22=z 1=z 2 ，z 31=z 1z 2=1，同理z 32=1，D 正确．故选：ACD ．10四棱锥P -ABCD 的底面为正方形，P A 与底面垂直，P A =2，AB =1，动点M 在线段PC 上，则()A.不存在点M ，使得AC ⊥BMB.MB +MD 的最小值为303C.四棱锥P -ABCD 的外接球表面积为5πD.点M 到直线AB 的距离的最小值为255【答案】BD【解析】对于A ：连接BD ，且AC ∩BD =O ，如图所示，当M 在PC 中点时，因为点O 为AC 的中点，所以OM ⎳P A ，因为P A ⊥平面ABCD ，所以OM ⊥平面ABCD ，又因为AC ⊂平面ABCD ，所以OM ⊥AC ，因为ABCD 为正方形，所以AC ⊥BD .又因为BD ∩OM =O ，且BD ，OM ⊂平面BDM ，所以AC ⊥平面BDM ，因为BM ⊂平面BDM ，所以AC ⊥BM ，所以A 错误；对于B ：将△PBC 和△PCD 所在的平面沿着PC 展开在一个平面上，如图所示，则MB +MD 的最小值为BD ，直角△PBC 斜边PC 上高为1×56，即306，直角△PCD 斜边PC 上高也为1×56，所以MB +MD 的最小值为303，所以B 正确；对于C ：易知四棱锥P -ABCD 的外接球直径为PC ，半径R =12PC =1222+12+12=62，表面积S =4πR 2=6π，所以C 错误；对于D ：点M 到直线AB 距离的最小值即为异面直线PC 与AB 的距离，因为AB ⎳CD ，且AB ⊄平面PCD ，CD ⊂平面PCD ，所以AB ⎳平面PCD ，所以直线AB 到平面PCD 的距离等于点A 到平面PCD 的距离，过点A 作AF ⊥PD ，因为P A ⊥平面ABCD ，所以P A ⊥CD ,又AD ⊥CD ,且P A ∩AD =A ,故CD ⊥平面P AD ，AF ⊂平面P AD ，所以AF ⊥CD ，因为PD ∩CD =D ，且PD ，CD ⊂平面PCD ，所以AF ⊥平面PCD ，所以点A 到平面PCD 的距离，即为AF 的长，如图所示，在Rt △P AD 中，P A =2，AD =1，可得PD =5，所以由等面积得AF =255，即直线AB 到平面PCD 的距离等于255，所以D 正确，故选：BCD .11今年是共建“一带一路”倡议提出十周年.某校进行“一带一路”知识了解情况的问卷调查，为调动学生参与的积极性，凡参与者均有机会获得奖品.设置3个不同颜色的抽奖箱，每个箱子中的小球大小相同质地均匀，其中红色箱子中放有红球3个，黄球2个，绿球2个；黄色箱子中放有红球4个，绿球2个；绿色箱子中放有红球3个，黄球2个，要求参与者先从红色箱子中随机抽取一个小球，将其放入与小球颜色相同的箱子中，再从放入小球的箱子中随机抽取一个小球，抽奖结束.若第二次抽取的是红色小球，则获得奖品，否则不能获得奖品，已知甲同学参与了问卷调查，则()A.在甲先抽取的是黄球的条件下，甲获得奖品的概率为47B.在甲先抽取的不是红球的条件下，甲没有获得奖品的概率为1314C.甲获得奖品的概率为2449D.若甲获得奖品，则甲先抽取绿球的机会最小【答案】ACD【解析】设A 红，A 黄，A 绿，分别表示先抽到的小球的颜色分别是红、黄、绿的事件，设B 红表示再抽到的小球的颜色是红的事件，在甲先抽取的是黄球的条件下，甲获得奖品的概率为：P B 红∣A 黄 =P B 红A 黄 P A 黄=27×4727=47，故A 正确；在甲先抽取的不是红球的条件下，甲没有获得奖品的概率为：P B 红 ∣A 红 =P A 红 B 红 P A 红 =P A 黄B 红 +P A 绿B 红 P A 红 =27×37+27×1247=1328，故B 错误；由题意可知，P A 红 =37,P A 黄 =27,P A 绿 =27,P B 红∣A 红 =37,P B 红∣A 黄 =47，P B 红∣A 绿 =12，由全概率公式可知，甲获得奖品的概率为：P =P A 红 P B 红∣A 红 +P A 黄 ⋅P B 红∣A 黄 +P A 绿 ⋅P B 红∣A 绿 =37×37+27×47+27×12=2449，故C 正确；因为甲获奖时红球取自哪个箱子的颜色与先抽取小球的颜色相同，则P A 红∣B 红 =P A 红 ⋅P B 红∣A 红 P B 红=37×37×4924=38，P A 黄∣B 红 =P A 黄 ⋅P B 红∣A 黄P B 红=27×47×4924=13，P A 绿∣B 红 =P A 绿 ⋅P B 红∣A 绿 P B 红 =27×12×4924=724，所以甲获得奖品时，甲先抽取绿球机会最小，故D 正确.故选：ACD .三、填空题：本题共3小题，每小题5分，共15分.12已知△ABC 的边BC 的中点为D ，点E 在△ABC 所在平面内，且CD =3CE -2CA ，若AC =xAB +yBE，则x +y =．【答案】11【解析】因为CD =3CE -2CA ，边BC 的中点为D ，所以12CB=3BE -BC +2AC ，因为12CB =3BE -3BC +2AC ，所以52BC =3BE +2AC ，所以52BC =52AC -AB =3BE +2AC ，所以5AC -5AB =6BE +4AC ，即5AB +6BE =AC ，因为AC =xAB +yBE ，所以x =5，y =6，故x +y =11.故答案为：1113已知圆锥母线长为2，则当圆锥的母线与底面所成的角的余弦值为时，圆锥的体积最大，最大值为．【答案】①.63②.16327π【解析】设圆锥的底面半径为r ，圆锥的母线与底面所成的角为θ,θ∈0,π2 ，易知cos θ=r 2.圆锥的体积为V =13πr 2⋅4-r 2=43πcos 2θ⋅2sin θ=8π3cos 2θ⋅sin θ=8π31-sin 2θ sin θ令x =sin θ,x ∈0,1 ，则y =1-sin 2θ sin θ=-x 3+x ，y =-3x 2+1当y >0时，x ∈0,33，当y<0时，x ∈33,1 ，即函数y =-x 3+x 在0,33 上单调递增，在33,1上单调递减，即V max =8π333-33 3 =163π27，此时cos θ=1-323 =62.故答案为：62；163π2714已知双曲线C ：x 2-y 23=1的左、右焦点分别为F 1，F 2，右顶点为E ，过F 2的直线交双曲线C 的右支于A ，B 两点(其中点A 在第一象限内)，设M ，N 分别为△AF 1F 2，△BF 1F 2的内心，则当F 1A ⊥AB 时，AF 1=；△ABF 1内切圆的半径为．【答案】①.7+1##1+7②.7-1##-1+7【解析】由双曲线方程知a =1,b =3,c =2，如下图所示：由F 1A ⊥AB ，则AF 1 2+AF 2 2=F 1F 2 2=16，故AF 1 -AF 2 2+2AF 1 AF 2 =16，而AF 1 -AF 2 =2a =2，所以AF 1 AF 2 =6，故AF 2 2+2AF 2 -6=0，解得AF 2 =7-1，所以AF 1 =7+1，若G 为△ABF 1内切圆圆心且F 1A ⊥AB 可知，以直角边切点和G ,A 为顶点的四边形为正方形，结合双曲线定义内切圆半径r =12AF 1 +AB -BF 1 =12AF 1 +AF 2 +BF 2 -BF 1所以r =1227+BF 2 -BF 1 =1227-2 =7-1；故答案为：7+1，7-1；2024届高三二轮复习“8+3+3”小题强化训练(3)一、选择题：本题共8小题，每小题5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有一组按从小到大顺序排列数据：3，5，x ，8，9，10，若其极差与平均数相等，则这组数据的中位数为()A.7B.7.5C.8D.6.5【答案】B【解析】依题意可得极差为10-3=7，平均数为163+5+x +8+9+10 =1635+x ，所以1635+x =7，解得x =7，所以中位线为7+82=7.5.故选：B .2已知集合A =x x -1 >2 ，B =x log 4x <1 ，则A ∩B =()A.3,4B.-∞,-1 ∪3,4C.1,4D.-∞,4【答案】A【解析】由x -1 >2，得x <-1或x >3，所以A =x x <-1或x >3 ，由log 4x <1，得0<x <4，所以B =x 0<x <4 ，所以A ∩B =x 3<x <4 .故选：A .3已知向量a =(2,0)，b =sin α,32，若向量b 在向量a 上的投影向量c =12,0 ，则|a +b |=()A.3B.7C.3D.7【答案】B【解析】由已知可得，b 在a 上的投影向量为a ⋅b |a |⋅a |a |=2sin α2×2(2,0)=(sin α,0)，又b 在a 上的投影向量c =12,0 ，所以sin α=12，所以b =12,32，所以a +b =52,32 ，所以|a +b |=52 2+322=7.故选：B .4如图是两个底面半径都为1的圆锥底面重合在一起构成的几何体，上面圆锥的侧面积是下面圆锥侧面积的2倍，AP ⊥AQ ，则PQ =()A.74B.262C.52D.3【答案】C【解析】设两圆锥的高OP =x ，OQ =y ，则AP =x 2+1，AQ =y 2+1，由AP ⊥AQ ，有AP 2+AQ 2=PQ 2，可得x 2+1+y 2+1=x +y 2，可得xy =1，又由上下圆锥侧面积之比为2:1，即π×1×P A =2×π×1×QA ，可得P A =2QA ，则有x 2+1=2y 2+1，即x 2=4y 2+3，代入y =1x整理为x 4-3x 2-4=0，解得x =2(负值舍)，可得y =12，OP =x +y =2+12=52．故选：C ．5已知Q 为直线l :x +2y +1=0上的动点，点P 满足QP=1,-3 ，记P 的轨迹为E ，则()A.E 是一个半径为5的圆B.E 是一条与l 相交的直线C.E 上的点到l 的距离均为5D.E 是两条平行直线【答案】C【解析】设P x ,y ，由QP=1,-3 ，则Q x -1,y +3 ，由Q 在直线l :x +2y +1=0上，故x -1+2y +3 +1=0，化简得x +2y +6=0，即P 轨迹为E 为直线且与直线l 平行，E 上的点到l 的距离d =6-112+22=5，故A 、B 、D 错误，C 正确.故选：C .6已知x +1 x -1 5=a 0+a 1x +a 2x 2+a 3x 3+a 4x 4+a 5x 5+a 6x 6，则a 1+a 3的值为()A.-1B.1C.4D.-2【答案】C【解析】在x +1 x -1 5=a 0+a 1x +a 2x 2+a 3x 3+a 4x 4+a 5x 5+a 6x 6中，而x +1 x -1 5=x x -1 5+x -1 5，由二项式定理知x -1 5展开式的通项为T r +1=C r 5x 5-r (-1)r ，令5-r =2，解得r =3，令5-r =3，r =2，故a 3=C 35(-1)3+C 25(-1)2=0，同理令5-r =1，解得r =4，令5-r =0，解得r =5，故a 1=C 45(-1)4+C 55(-1)5=4，故a 1+a 3=4.故选：C7已知P 为抛物线x 2=4y 上一点，过P 作圆x 2+(y -3)2=1的两条切线，切点分别为A ，B ，则cos ∠APB 的最小值为()A.12B.23C.34D.78【答案】C【解析】如图所示：因为∠APB =2∠APC ，sin ∠APC =AC PC=1PC，设P t ,t 24，则PC 2=t 2+t 24-3 2=t 416-t 22+9=116t 2-4 2+8，当t 2=4时，PC 取得最小值22，此时∠APB 最大，cos ∠APB 最小，且cos ∠APB min =1-2sin 2∠APC =1-21222=34，故C 正确.故选：C8已知函数f x ,g x 的定义域为R ,g x 为g x 的导函数且f x +g x =3,f x -g 4-x =3，若g x 为偶函数，则下列结论一定成立的是()A.f -1 =f -3B.f 1 +f 3 =65C.g 2 =3D.f 4 =3【答案】D【解析】对于D ，∵g x 为偶函数，则g x =g -x ，两边求导可得g x =-g -x ，则g x 为奇函数，则g 0 =0，令x =4，则f 4 -g 0 =3，f 4 =3，D 对；对于C ，令x =2，可得f 2 +g 2 =3f 2 -g 2 =3 ，则f 2 =3g 2 =0 ，C 错；对于B ，∵f x +g x =3，可得f 2+x +g 2+x =3，f x -g 4-x =3可得f 2-x -g 2+x =3，两式相加可得f 2+x +f 2-x =6，令x =1，即可得f 1 +f 3 =6，B 错；又∵f x +g x =3，则f x -4 +g x -4 =f x -4 -g 4-x =3，f x -g 4-x =3，可得f x =f x -4 ，所以f x 是以4为周期的函数，所以根据以上性质不能推出f -1 =f -3 ，A 不一定成立.故选：D二、选择题：本题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项符合题目要求.全部选对的得6分，部分选对的得部分分，有选错的得0分.9下列结论正确的是()A.若a <b <0，则a 2>ab >b 2B.若x ∈R ，则x 2+2+1x 2+2的最小值为2C.若a +b =2，则a 2+b 2的最大值为2D.若x ∈(0,2)，则1x +12-x ≥2【答案】AD【解析】因为a 2-ab =a (a -b )>0，所以a 2>ab ，因为ab -b 2=b (a -b )>0，所以ab >b 2，所以a 2>ab >b 2，故A 正确；因为x 2+2+1x 2+2≥2的等号成立条件x 2+2=1x 2+2不成立，所以B 错误；因为a 2+b 22≥a +b 2 2=1，所以a 2+b 2≥2，故C 错误；因为1x +12-x =12(x +2-x )1x +12-x =122+2-x x +x 2-x ≥12(2+2)=2，当且仅当1x =12-x，即x =1时，等号成立，所以D 正确.故选：AD10若函数f x =2sin 2x ⋅log 2sin x +2cos 2x ⋅log 2cos x ，则()A.f x 的最小正周期为πB.f x 的图像关于直线x =π4对称C.f x 的最小值为-1D.f x 的单调递减区间为2k π,π4+2k π ,k ∈Z【答案】BCD【解析】由sin x >0,cos x >0得f x 的定义域为2k π,π2+2k π ,k ∈Z .对于A ：当x ∈0,π2时，x +π∈π,32π 不在定义域内，故f x +π =f x 不成立，易知f x 的最小正周期为2π，故选项A 错误；对于B ：又f π2-x =2cos 2x ⋅log 2cos x +2sin 2x ⋅log 2sin x =f x ，所以f x 的图像关于直线x =π4对称，所以选项B 正确；对于C ：因为f x =sin 2x ⋅log 2sin 2x +cos 2x ⋅log 2cos 2x ，设t =sin 2x ，所以函数转化为g t =t ⋅log 2t +1-t ⋅log 21-t ,t ∈0,1 ,g t =log 2t -log 21-t ，由g t >0得，12<t <1.g t <0得0<t <12.所以g t 在0,12 上单调递减，在12,1 上单调递增，故g (t )min =g 12=-1，即f (x )min =-1，故选项C 正确；对于D ：因为g t 在0,12 上单调递减，在12,1 上单调递增，由t =sin 2x ，令0<sin 2x <12得0<sin x <22，又f x 的定义域为2k π,π2+2k π ,k ∈Z ，解得2k π<x <π4+2k π,k ∈Z ，因为t =sin 2x 在2k π,π4+2k π 上单调递增，所以f x 的单调递减区间为2k π,π4+2k π ,k ∈Z ，同理函数的递增区间为π4+2k π,π2+2k π ,k ∈Z ，所以选项D 正确.故选：BCD .11已知数列a n 的前n 项和为S n ，且2S n S n +1+S n +1=3，a 1=α0<α<1 ，则()A.当0<α<13-14时，a 2>a 1B.a 3>a 2C.数列S 2n -1 单调递增，S 2n 单调递减D.当α=34时，恒有nk =1S k -1 <54【答案】ACD【解析】由题意可得：S n +1=32S n +1，a 1=α，由S n +1=32S n +1可知：S n +1=1⇔S n =1，但S 1=α∈0,1 ，可知对任意的n ∈N *，都有S n ≠1，对于选项A ：若0<α<13-14，则a 2-a 1=S 2-2a 1=32α+1-2α=3-2α-4α22α+1=4α+1+13 13-14-α2α+1>0，即a 2>a 1，故A 正确；对于选项B ：a 3-a 2=S 3-2S 2+S 1=6α+32α+7-62α+1+α=α-1 4α2+32α+39 2α+1 2α+7<0，即a 3<a 2，故B 错误.对于选项C ：因为S n +1-1=-2S n -1 2S n +1，S n +1+32=3S n +32 2S n +1，则S n +1-1S n +1+32=-23⋅S n -1S n +32，且S 1-1S 1+32=α-1α+32<0，可知S n -1S n+32是等比数列，则S n -1S n +32=α-1α+32⋅-23n -1，设A =α-1α+32<0，t =232n -2，可得S 2n =3-3At 3+2At =3253+2At -1 ，S 2n -1=1+32At 1-At =521-At-32，因为At =A 232n -2，可知A 23 2n -2 为递增数列，所以数列S 2n -1 单调递增，S 2n 单调递减，故C 正确；对于选项D ：因为S n +1=32S n +1，S n +1-34=32S n +1-34=33-2S n 42S n +1，由S 1=α=34，可得S 2-34>0，即S 2>34，则S 2≤65，即34<S 2≤65；由34<S 2≤65，可得S 3-34>0，即S 3>34，则S 3<65，即34<S 3<65；以此类推，可得对任意的n ∈N *，都有S n ≥S 1=α=34，又因为S n +1-1S n -1=22S n +1，则S n +1-1 ≤22α+1S n -1 =45S n -1 ，所以∑nk =1S k -1 ≤541-45 n <54，故D 正确.故选：ACD .三、填空题：本题共3小题，每小题5分，共15分.12在(1+ax )n (其中n ∈N *,a ≠0)的展开式中，x 的系数为-10，各项系数之和为-1，则n =.【答案】5【解析】由题意得(1+ax )n 的展开式中x 的系数为aC 1n =-10，即an =-10，令x =1，得各项系数之和为(1+a )n =-1，则n 为奇数，且1+a =-1，即得a =-2,n =5，故答案为：513已知椭圆C :x 2a 2+y 2b2=1a >b >0 的左、右焦点分别F 1，F 2，椭圆的长轴长为22，短轴长为2，P 为直线x =2b 上的任意一点，则∠F 1PF 2的最大值为.【答案】π6【解析】由题意有a =2，b =1，c =1，设直线x =2与x 轴的交点为Q ，设PQ =t ，有tan ∠PF 1Q =PQ F 1Q=t3，tan ∠PF 2Q =PQ F 2Q=t ，可得tan ∠F 1PF 2=tan ∠PF 2Q -∠PF 1Q =t -t31+t23=2t t 2+3=2t +3t ≤2t 23t =33，当且仅当t =3时取等号，可得∠F 1PF 2的最大值为π6.故答案为：π614已知四棱锥P -ABCD 的底面为矩形，AB =23，BC =4，侧面P AB 为正三角形且垂直于底面ABCD ，M 为四棱锥P -ABCD 内切球表面上一点，则点M 到直线CD 距离的最小值为.【答案】10-1【解析】如图，设四棱锥的内切球的半径为r ，取AB 的中点为H ，CD 的中点为N ，连接PH ，PN ，HN ，球O为四棱锥P-ABCD的内切球，底面ABCD为矩形，侧面P AB为正三角形且垂直于底面ABCD，则平面PHN截四棱锥P-ABCD的内切球O所得的截面为大圆，此圆为△PHN的内切圆，半径为r，与HN，PH分别相切于点E，F，平面P AB⊥平面ABCD，交线为AB，PH⊂平面P AB，△P AB为正三角形，有PH⊥AB，∴PH⊥平面ABCD，HN⊂平面ABCD，∴PH⊥HN，AB=23，BC=4，则有PH=3，HN=4，PN=5，则△PHN中，S△PHN=12×3×4=12r3+4+5，解得r=1.所以，四棱锥P-ABCD内切球半径为1，连接ON.∵PH⊥平面ABCD，CD⊂平面ABCD，∴CD⊥PH，又CD⊥HN，PH,HN⊂平面PHN，PH∩HN=H，∴CD⊥平面PHN，∵ON⊂平面PHN，可得ON⊥CD，所以内切球表面上一点M到直线CD的距离的最小值即为线段ON的长减去球的半径，又ON=OE2+EN2=10.所以四棱锥P-ABCD内切球表面上的一点M到直线CD的距离的最小值为10-1.故答案为：10-12024届高三二轮复习“8+3+3”小题强化训练(4)一、选择题：本题共8小题，每小题5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已知双曲线的标准方程为x 2k -4+y 2k -5=1，则该双曲线的焦距是()A.1B.3C.2D.4【答案】C【解析】由双曲线方程可知a 2=k -4,b 2=5-k ，所以c 2=k -4+5-k =1，c =1，2c =2．故选：C2在等比数列a n 中，a 1+a x =82，a 3a x -2=81，前x 项和S x =121，则此数列的项数x 等于()A.4B.5C.6D.7【答案】B【解析】由已知条件可得a 1+a x =82a 3a x -2=a 1a x =81，解得a 1=1a x =81 或a 1=81a x =1 .设等比数列a n 的公比为q .①当a 1=1，a x =81时，由S x =a 1-a x q 1-q =1-81q1-q=121，解得q =3，∵a x =a 1q x -1=3x -1=81，解得x =5；②当a 1=81，a x =1时，由S x =a 1-a x q 1-q =81-q 1-q =121，解得q =13，∵a x =a 1q x -1=81×13x -1=35-x =1，解得x =5.综上所述，x =5.故选：B .3对任意实数a ，b ，c ，在下列命题中，真命题是()A.“ac 2>bc 2”是“a >b ”的必要条件B.“ac 2=bc 2”是“a =b ”的必要条件C.“ac 2=bc 2”是“a =b ”的充分条件D.“ac 2≥bc 2”是“a ≥b ”的充分条件【答案】B【解析】对于A ，若c =0，则由a >b ⇏ac 2>bc 2，∴“ac 2>bc 2”不是“a >b ”的必要条件，A 错．对于B ，a =b ⇒ac 2=bc 2，∴“ac 2=bc 2”是“a =b ”的必要条件，B 对，对于C ，若c =0，则由ac 2=bc 2，推不出a =b ，“ac 2=bc 2”不是“a =b ”的充分条件对于D ，当c =0时，ac 2=bc 2，即ac 2≥bc 2成立，此时不一定有a ≥b 成立，故“ac 2≥bc 2”不是“a ≥b ”的充分条件，D 错误，故选：B ．4已知m 、n 是两条不同直线，α、β、γ是三个不同平面，则下列命题中正确的是()A.若m ∥α，n ∥α，则m ∥nB.若α⊥β，β⊥γ，则α∥βC.若m ∥α，m ∥β，则α∥βD.若m ⊥α，n ⊥α，则m ∥n【答案】D【解析】A选项：令平面ABCD为平面α，A1B1为直线m，B1C1为直线n，有：m∥α，n∥α，但m∩n=B1，A错误；B选项：令平面ABCD为平面β，令平面B1BCC1为平面α，令平面A1ABB1为平面γ，有：α⊥β，β⊥γ，而α⊥β，B错误；C选项：令平面ABCD为平面α，令平面A1ABB1为平面β，C1D1为直线m，有：m∥α，m∥β，则α∥β，而α⊥β，C错误；D选项：垂直与同一平面的两直线一定平行，D正确.故选：D5将甲、乙等8名同学分配到3个体育场馆进行冬奥会志愿服务，每个场馆不能少于2人，则不同的安排方法有()A.2720B.3160C.3000D.2940【答案】D【解析】共有两种分配方式，一种是4:2:2，一种是3:3:2，故不同的安排方法有C48C24C222!+C38C35C222!A33=2940.故选：D6若抛物线y2=4x与椭圆E:x2a2+y2a2-1=1的交点在x轴上的射影恰好是E的焦点，则E的离心率为()A.2-12 B.3-12 C.2-1 D.3-1【答案】C【解析】不妨设椭圆与抛物线在第一象限的交点为A，椭圆E右焦点为F，则根据题意得AF⊥x轴，c2=a2-a2-1=1，则c=1，则F1,0，当x=1时，y2=4×1，则y A=2，则A1,2，代入椭圆方程得12a2+22a2-1=1，结合a2-1>0，不妨令a>0；解得a=2+1，则其离心率e=ca=12+1=2-1，故选：C.7已知等边△ABC 的边长为3，P 为△ABC 所在平面内的动点，且|P A |=1，则PB ⋅PC 的取值范围是()A.-32,92B.-12,112C.[1,4]D.[1,7]【答案】B【解析】如下图构建平面直角坐标系，且A -32,0 ，B 32,0 ，C 0,32，所以P (x ,y )在以A 为圆心，1为半径的圆上，即轨迹方程为x +322+y 2=1，而PB =32-x ,-y ,PC =-x ,32-y ，故PB ⋅PC =x 2-32x +y 2-32y =x -34 2+y -34 2-34，综上，只需求出定点34,34 与圆x +322+y 2=1上点距离平方范围即可，而圆心A 与34,34 的距离d =34+32 2+34 2=32，故定点34,34与圆上点的距离范围为12,52，所以PB ⋅PC ∈-12,112.故选：B 8设a 、b 、c ∈0,1 满足a =sin b ，b =cos c ，c =tan a ，则()A.a +c <2b ，ac <b 2B.a +c <2b ，ac >b 2C.a +c >2b ，ac <b 2D.a +c >2b ，ac >b 2【答案】A【解析】∵a 、b 、c ∈0,1 且a =sin b ，b =cos c ，c =tan a ，则c =tan a =tan sin b ，先比较a +c =sin b +tan sin b 与2b 的大小关系，构造函数f x =sin x +tan sin x -2x ，其中0<x <1，则0<sin x <1，所以，cos1<cos sin x <1，则f x =cos x +cos xcos 2sin x -2=cos x -2 cos 2sin x +cos x cos 2sin x，令g x =cos x -1-12x 2 ，其中x ∈0,1 ，则g x =x -sin x ，令p x =x -sin x ，其中0<x <1，所以，p x =1-cos x >0，所以，函数g x 在0,1 上单调递增，故g x >g 0 =0，所以，函数g x 在0,1 上单调递增，则g x =cos x -1-12x 2 >0，即cos x >1-12x 2，因为x ∈0,1 ，则0<sin x <sin1，所以，cos sin x >1-12sin 2x =1-121-cos 2x =121+cos 2x ，所以，cos 2sin x >141+cos 2x 2，因为cos x -2<0，所以，cos x -2 cos 2sin x +cos x <14cos x -2 1+cos 2x 2+cos x=14cos 5x -2cos 4x +2cos 3x -4cos 2x +5cos x -2 =14cos x -1 3cos 2x +cos x +2 <0，所以，对任意的x ∈0,1 ，f x =cos x -2 cos 2sin x +cos xcos 2sin x <0，故函数f x 在0,1 上单调递减，因为b ∈0,1 ，则f b =sin b +tan sin b -2b <f 0 =0，故a +c <2b ，由基本不等式可得0<2ac ≤a +c <2b (a ≠c ，故取不了等号)，所以，ac <b 2，故选：A .二、选择题：本题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项符合题目要求.全部选对的得6分，部分选对的得部分分，有选错的得0分.9某大学生做社会实践调查，随机抽取6名市民对生活满意度进行评分，得到一组样本数据如下：88、89、90、90、91、92，则下列关于该样本数据的说法中正确的是()A.均值为90B.中位数为90C.方差为2D.第80百分位数为91【答案】ABD【解析】由题意可知，该组数据的均值为x =88+89+90+90+91+926=90，故A 正确；中位数为90+902=90，故B 正确；方差为s 2=1688-90 2+89-90 2+90-90 2×2+91-90 2+92-90 2 =53，故C 错误；因为6×80%=4.8，第80百分位数为91，故D 正确.故选：ABD .10设M ，N ，P 为函数f x =A sin ωx +φ 图象上三点，其中A >0，ω>0，ϕ <π2，已知M ，N 是函数f x 的图象与x 轴相邻的两个交点，P 是图象在M ，N 之间的最高点，若MP 2+2MN ⋅NP=0，△MNP 的面积是3，M 点的坐标是-12,0 ，则()A.A =2B.ω=π2C.φ=π4D.函数f x 在M ，N 间的图象上存在点Q ，使得QM ⋅QN <0【答案】BCD【解析】MP 2+2MN ⋅NP =MP 2-2NM ⋅NP =MP 2-2NM ⋅12NM =T 4 2+A 2 -T 22=A 2-3T 216=0，而S △MNP =AT 4=3，故A =3，T =4=2πω，ω=π2，A 错误、B 正确；-12⋅π2+φ=k π，φ=k π+π4(k ∈Z )，而ϕ <π2，故φ=π4，C 正确；显然，函数f x 的图象有一部分位于以MN 为直径的圆内，当Q 位于以MN 为直径的圆内时，QM⋅QN<0，D 正确，故选:BCD .11设a 为常数，f (0)=12,f (x +y )=f (x )f (a -y )+f (y )f (a -x )，则()．A .f (a )=12B .f (x )=12成立C f (x +y )=2f (x )f (y )D .满足条件的f (x )不止一个【答案】ABC 【解析】f (0)=12,f (x +y )=f (x )f (a -y )+f (y )f (a -x )对A ：对原式令x =y =0，则12=12f a +12f a =f a ，即f a =12，故A 正确；对B ：对原式令y =0，则f x =f x f a +f 0 f a -x =12f x +12f a -x ，故f x =f a -x ，对原式令x =y ，则f 2x =f x f y +f y f x =2f x f y =2f 2x ≥0，故f x 非负；对原式令y =a -x ，则f a =f 2x +f 2a -x =2f 2x =12，解得f x =±12，又f x 非负，故可得f x =12，故B 正确；对C ：由B 分析可得：f x +y =2f x f y ，故C 正确；对D ：由B 分析可得：满足条件的f x 只有一个，故D 错误.故选：ABC .三、填空题：本题共3小题，每小题5分，共15分.12在复平面内，复数z =-12+32i 对应的向量为OA ，复数z +1对应的向量为OB ，那么向量AB 对应的复数是．。

强化技能限时训练(二)语法填空(有提示词类试题)+阅读理解(限时30分钟)Ⅰ.语法填空( A )1. She was put under house arrest two years ago but remained a powerful (symbolic) in last year’s election.2. She wished that he was as easy (please) as her mother,who was always delighted with perfume.3.As time went on,the old man’s health is getting from bad to (bad).4.The (tall) of the two girls is the girl I always talk about with you.5.Only (environment) problems are discussed at the meeting.6.It is a well-known fact that smoking is (harm) to our health.7.As we all know,English is (wide) used in our daily life.8. Mary felt (please),because there were many empty seats in the room.9. What was so (impress)about Jasmine Westland’s victory was that she came first in the marathon bare-footed.10. She turned off her phone and (rejoin) them at the table.答案与剖析:1.symbol 此题考查派生词的用法。

高考二轮复习限时训练（五）（时间：60分钟）班级 姓名 得分一、填空题：（12×5分=60分）1、命题“2,210x R x x ∃∈-+≤”的否定形式为 ．2、已知U={1，2，3，4，5，6}，集合A={2，3}，集合B={3，5}，则A ∩(U B) = ．3、0tan(1125)-的值是 ．4、如图所示的流程图输出的n 值是 ．5、若复数z 满足(2)5i z -= (i 是虛数单位)，则z= ．6、函数[]sin()(0,3y x x ππ=+∈)的单调减区间是 ．7、方程x x 28lg -=的根)1,(+∈k k x ，k ∈Z ，则k = ．8、已知向量(1,2),(2,3)a b == ，若()()a b a b λ+⊥-，则λ= ．9、设奇函数()f x 满足：对x R ∀∈有(1)()0f x f x ++=，则(5)f = ． 10、某城市一年中12个月的平均气温与月份的关系可近似地用三角函数)]6(6cos[-+=x A a y π(x =1，2，3，…，12）来表示，已知6月份的月平均气温最高，为28℃，12月份的月平均气温最低，为18℃，则10月份的平均气温值为 ℃．11、在等比数列{}n a 中，若22a =，632a =，则4a = ．12、在ABC ∆中,角A,B,C 所对的边分别是,,a b c ，若22b c+2a =,且a b=则∠C= ．二．解答题（每题15分，共30分）13．设向量(c o s,s i n )m θθ=，sin ,cos )n θθ= ，),23(ππθ--∈，若1m n ∙=，求：（1）)4sin(πθ+的值； （2）)127cos(πθ+的值．14.如图，四边形ABCD为矩形，AD⊥平面ABE，AE＝EB＝BC＝2，F为CE上的点，且BF⊥平面ACE．（1）求证：AE⊥BE；使得MN∥平面DAE.南师大附校09高考二轮复习限时训练（五）一、填空题1、∀x∈R，x 2－2x+l ＞0 2、{2} 3、1 4、5 5、1 6、[,]6ππ7、3 8、53- 9、0 10、20.5 11、8 12、1050二、解答题13、解：（1）依题意，cos sin )sin cos )m n θθθθ∙=+cos )θθ=+4sin()4πθ=+又1m n ∙= ∴41)4sin(=+πθ…………… ……7分（2）由于),23(ππθ--∈，则)43,45(4πππθ--∈+结合41)4s i n(=+πθ，可得415)4cos(-=+πθ 则7cos()12θπ+11cos[()]43θππ=++11(4242=-⨯-⨯8=-15分14、（1）证明： ABE AD 平面⊥，BC AD //∴ABE BC 平面⊥，则BC AE ⊥…… ……… 3分又 ACE BF 平面⊥，则BF AE ⊥ ∴BCE AE 平面⊥ 又BCE BE 平面⊂ ∴BE AE ⊥ ………………………… 6分（2）在三角形ABE 中过M 点作MG ∥AE 交BE 于G 点,在三角形BEC 中过G 点作GN ∥BC交EC 于N 点,连MN,则由比例关系易得CN ＝CE 31…… …8分MG ∥AE MG ⊄平面ADE, AE ⊂平面ADE, ∴MG ∥平面ADE ……………………10分同理, GN ∥平面ADE∴平面MGN ∥平面ADE ……………12分又MN ⊂平面MGN ∴MN ∥平面ADE∴N 点为线段CE 上靠近C 点的一个三等分点 …………………………………………15分。

一. 词汇语法巩固练习1.We offered to give Sharon a ride home, but she ______, saying that she felt likewalking.A. understoodB. acceptedC. compromisedD. declined2.Nowadays the ______ for travelling is shifted from shopping to food and scenery.A. priorityB. potentialC. proportionD. pension3.His vivid descriptions of country life quickly became popular, which establishedhis ______ as one of America’s greatest writers.A. trustB. contactC. reputationD. theory4.People believe writing poems provides a ______ through which they can expresstheir feelings.A. shelterB. sourceC. channelD. background5.Try to understand what’s actually happening instead of acting on the ______you’ve made.A. assignmentB. associationC. acquisitionD. assumption6.We’d better ______ some sports because sports can give us a perfect build andprotect us from getting illness easily.A. take upB. make upC. pick upD. build up7.It is important to pay your electricity bill on time, as late payments may affectyour ______.A. conditionB. incomeC. creditD. status8.This actor often has the first two tricks planned before performing, and then goesfor ______.A. whicheverB. wheneverC. whereverD. whatever9.You’d better write down the phone number of that restaurant for future ______.A. purposeB. referenceC. progressD. memory10.There will still be lots of challenges if we are to ______ garbage in a short time.A. clarifyB. justifyC. satisfyD. classify11.New technologies have made ____________ possible to turn out new productsfaster and at a lower cost.12.The research group produced two reports based on the survey, but ____________contained any useful suggestions.13.Most colleges now offer first-year students a course specially ____________(design) to help them succeed academically and personally.14.The dancer’s incredible performance had the audience on its feet ____________(clap) for 10 minutes at the end of the show. show provides women a stage ____________ they show their natural talent,rather than their looks.16.The number of smokers, ____________ is reported, has dropped by 17 percent injust one year.17.We hadn’t met for 20 years, but I recognized her ____________ (direct) I sawher.18.The meaning of the word “nice” changed a few times ____________ it finallycame to include the sense “pleasant.”19.You may have heard the old saying, “The more, the merrier. It’s usually true, butnot for travel. When it comes to travel, I say, “The more, the ____________ (messy).” I’m not alone in my ____________ (prefer) for solo(独自的) travel.According to the Daily Mail, there has been ____________ 143% increase in “solo travel” internet searches over the past three years, ____________ makes it one of the fastest growing parts of the travel industry. To me, it’s ____________ (obvious) more convenient to plan and just pack my bag and go straight to my dream place.20.More than 500 people ____________ (fly) in space since Yuri Gagarin paved theway in 1961. 20 people have travelled to the moon, 12 of ____________ have walked on the moon. So far, space stations ____________ (build), and astronauts have learned to live and work in space for many months ____________ even years.二. 七选五Many people think that positive thinking is mostly about keeping one’s head in the sand and ignoring daily problems, trying to look optimistic. In reality it has more to do with the way an individual talks to himself. Self-talk is a constant stream of thoughts of a person, who is often unaware and uncertain of some events, phenomena, people, or even the person himself.____1____Meanwhile, positive thinking can help to stop negative self-talks and start to form a positive view on an issue. People who regularly practise positive thinking tend to solve problems more effectively. They areless exposed to stress caused by external factors. They tend to believe in themselves and in what they do.____2____People who think positively demonstrate increased life spans, lower rates of depression and anxiety, better physical and psychological health, reduced risks of death from heart problems. Positive thinking also contributes to one’s ability to deal with problems and hardships.____3____For example, researchers have found that in the case of a crisis accompanied by strong emotions, such as a natural disaster, positive thinking can provide a sort of buffer against depression and anxiety. Resilient people who think positively tend to treat every problem as a challenge, a chance for improvement of any kind, or as an opportunity for personal growth. Pessimists, on the contrary, tend to perceive problems as a source of additional stress.____4____ In conclusion, positive thinking is a powerful and effective tool for dealing with hard times and improving the quality of one’s life. It doesn’t have anything to do with ignorant optimism when an individual refuses to notice a problem.____5____ Thinking in a positive, self-encouraging way brings about many benefits to one’s physical and mental health.A. It doesn’t cause any severe emotional discomfort, either.B. Negative self-talk damages self-confidence and decreases self-respect.C. It helps one to remain clear-headed and confident in difficult situations.D. Positive thinking has several beneficial effects on the body and the mind.E. As thinking changes, an individual’s behaviour and habits change as well.F. They often offer a real alternative to the common and regular way of thinking.G. They often feel discouraged long before trying to solve the problem, even if small.三. 完形填空Normal people can hardly imagine what kind of pain orphans feel while growing up – being abandoned, disliked, and laughed at. For all these reasons, many of them build a ____1____ room in their hearts. In this room they can behave as they ____2____. Will Hunting, the main character in the US film Good Will Hunting is the perfect example.The story is simple. Will Hunting is a(n) ____3____ who works as a cleaner at the Massachusetts Institute of Technology in the US. Lambeau, a math professor, discovers that Hunting has a special ____4____ for math and wants him to be asuccess. However, Will gets into a fight and is ____5____ after he attacks a police officer. But Lambeau doesn’t ____6____. He sends Hunting to see a wise therapist, Dr Sean Maguire, to see if he can help the ____7____ young man. After a period of stony (冷漠的) silence, Will ____8____ to the therapist and is finally cured.But Will’s treatment takes a long time. In my opinion, the key to its success is ____9____ in Maguire’s words, “It is not your fault.” He ____10____ the words while staring at Will, making it clear to him that he no longer has any place to hide. They are words that all orphans should hear.Orphans have ____11____ because they feel that no one loves them, and in order to avoid harm, they ____12____ people’s kindness. I feel sorry for Will, as his background has led him into ____13____ and self-abasement (自卑). I am also sympathetic toward Maguire, who has a kind heart.It is ____14____ to love our own kids, but much harder to love those of other people. I suggest that we pay more attention to orphans, not out of ____15____, but out of love.1. A. persuasive B. offensive C. defensive D. aggressive2. A. dislike B. please C. protect D. cure3. A. student B. orphan C. expert D. defender4. A. interest B. appetite C. desire D. talent5. A. arrested B. bullied C. cared D. diagnosed6. A. give in B. give off C. give up D. give out7. A. gifted B. frustrated C. exhausted D. troubled8. A. opens up B. closes down C. breaks down D. picks up9. A. conveyed B. contained C. lied D. included10. A. repeats B. ignores C. avoids D. refuses11. A. confidence B. disadvantages C. obstacles D. barriers12. A. need B. refuse C. lack D. expect13. A. restriction B. resistance C. rejection D. rebellion14. A. normal B. necessary C. urgent D. difficult15. A. policy B. sympathy C. curiosity D. empathy四. 语篇填空While driving alone through the countryside, Linda saw an old woman by theside of the road, reaching out her hand. It ____1____ (get) dark and raining. “I can’t leave her out in this weather,” Linda said to herself, so she stopped the car.“Shall I offer you ____2____ lift?” Linda asked. The old woman nodded and climbed into the car. After a while Linda asked, “Have you waited for long?” The old woman shook her head. ____3____ (strange) enough, the old woman didn’t say a single word all the way. Her only ____4____ (respond) was always a nod of the head or something else like that.Then Linda saw the lady’s hands, ____5____ were very large and covered with thick hair. She realized ____6____ the lady was a man! After ____7____ (stop) the car, Linda said, “Can’t see that mirror. Would you mind cleaning it ____8____ me?” The lady nodded and opened the door. As soon as the “lady” was out of the car, Linda drove off quickly.When Linda arrived home, she found that the old lady ____9____ (leave) a handbag on the backseat. She opened it and let out a deep breath. Inside it ____10____ two sharp knives.一. 词汇语法巩固练习1-5 DACCD 6-10 ACDBD11-18 it; neither; designed; clapping; where; as; directly; before19 messier; preference; a; which; obviously20 have flown; whom; have been built; or二. 七选五BDCGA三. 完形填空CBBDA CDABA BBDAB四. 语篇填空1-5was getting; a; Strangely; response; which6-10that; stopping; for; had left; were。

限时训练（三十九）一、选择题：本大题共12小题，每小题5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{}220A x x x =--<，{}B x x a =<，若A B ≠∅，则a 的取值范围是（ ）.（A ）(),2-∞ （B ）()2,-+∞ （C ）()1,-+∞ （D ）(]1,2-（2）若12i z =-，则4i1zz=-（ ）.（A ）1 （B ）1- （C ）i （D ）i - （3）设等差数列{}n a 的前n 项和为n S .若13m S -=-，0m S =，14m S +=，则m =（ ）.（A ）4 （B ）5 （C ）6 （D ）7（4）甲、乙两人相约晚7时到8时之间在某地会面，先到者等候另一人20min ，过时离去，则两人能会面的概率是（ ）. （A ）13 （B ）14 （C ）59 （D ）23（5）已知O 为坐标原点，F 是椭圆()2222:10x y C a b a b+=>>的左焦点，,A B 分别是C 的左、右顶点.P 为C 上一点，且PF x ⊥轴.过点A 的直线l 与线段PF 交于点M ，与y 轴交于点E .若直线BM 经过OE 的三等分点（靠近O 点），则C 的离心率为（ ）. （A ）13 （B ）12 （C ）23 （D ）34（6）某四面体的三视图如图所示，该四面体的六条棱中，长度最长的是（ ）.（A ） （B ） （C ） （D ）（7）设函数()2ax bf x x c+=+的图像如图所示，则,,a b c 的大小关系是（ ）. （A ）a b c >> （B ）a c b >> （C ）b a c >> （D ）c a b >>（8）函数()sin()f x A x ωϕ=+()0,0,A ωϕ>><π的部分图像如图所示，若12,,36x x ππ⎛⎫∈- ⎪⎝⎭，12x x ≠，且()()12f x f x =，则()12f x x +=（ ）.（A ）1 （B ）12（C）2 （D正（主）视图俯视图（9）秦九韶是我国南宋时期的数学家，普州（现四川安岳县）人，他在所著的《数书九章》中提出的多项式求值的秦九韶算法，至今仍是比较先进的算法.如图所示的程序框图给出了利用秦九韶算法求多项式值的一个实例，若输入,n x 的分别为5，4.则输出v 的等于（ ）.（A ）569 （B ）2275 （C ）2276 （D ）2272（10）已知点M 是抛物线24yx =上的一点，F 为抛物线的焦点，A 在圆()()22:311C x y -+-=上，则MA MF+的最小值为（ ）.（A ）2 （B ）3 （C ）4 （D ）5 （11）已知{}n a 是等差数列，数列{}n b 满足()*12n n n n b a a a n ++=∈N ，设n S 为{}n b 的前n 项和，若95308a a =>，则当n S 取得最大值时，n =（ ）. （A ）9 （B ）10 （C ）11 （D ）12（12）已知e 为自然对数的底数，若对任意的1,1e x ⎡⎤∈⎢⎥⎣⎦，总存在唯一的[]1,1y ∈-，使得2ln 1e y x x a y -++=成立，则实数a 的取值范围是（）.（A ）1,e e ⎡⎤⎢⎥⎣⎦ （B ）2,e e ⎛⎤ ⎥⎝⎦ （C ）2,e ⎛⎫+∞ ⎪⎝⎭ （D ）21,e e e⎛⎫+ ⎪⎝⎭二、填空题：本题共4小题，每小题5分.（13）在等边ABC △中，AB 在BC 方向上的投影为1-，且2AD DC =，则BD AC ⋅= ＿＿＿.（14）若,,,A B C D 四人站在一排照相，,A B 不相邻的排法总数为k ，则二项式()1kx -的展开式中2x 项的系数为＿＿＿.（15）过平面区域202020x y y x y -+⎧⎪+⎨⎪++⎩内一点P 作圆22:1O x y +=的两条切线，切点分别为,A B ，记APB α∠=，当α最小时，此时点P 的坐标为＿＿＿.（16）设各项均为正数的数列{}n a 的前n 项和n S 满足212nn a S +⎛⎫= ⎪⎝⎭，又cos n n b S n =⋅π，则数列{}n b 的前n 项和nT=＿＿＿.限时训练（三十九）答案部分一、选择题二、填空题 13.2314.66 15. ()4,2-- 16. ()()112n n n +-⋅解析部分（1）解析 依题意，{}12A x x =-<<，{}B x x a =<，如图所示，若AB ≠∅，则1a >-.故选C.（2）解析 解法一：首先计算分母1zz -，由12i z =-得其共轭复数12i z =+，根据复数的运算法则（或者根据共轭复数的性质）知()()1112i 12i 4zz -=-+-=-，所以4ii 1z z=--.故选D.解法二：根据共轭复数的性质，2zz z =为实数，从而1zz -也是实数，所以4i1zz-是纯虚数，故排除选项A 和B.又1zz >，则10z z -<.故选D.评注 本题考查考生最熟悉的知识点之一——复数及其运算.考生只要知道复数共轭的概念，掌握复数的四则运算法则就能得出正确的答案.同时，本题还可以利用复数及其共轭复数的性质，通过简单的逻辑推理得出正确选项. （3）解析 解法一：由题设得13mm m a S S -=-=，114m m m a S S ++=-=，故等差数列{}n a 的公差11m m d a a +=-=，所以由3m a =，0m S =，得()1113102a m m m ma +-=⎧⎪⎨-+=⎪⎩，解得13a =-，7m =.故选D.a-1解法二：等差数列{}n a 的前n 项和()112n n n S na d -=+，故112n S n a d n -=+，所以n S n ⎧⎫⎨⎬⎩⎭成等差数列，于是11211m m m S S S m m m -++=-+，即34011m m -+=-+，解得7m =.故选D. （4）解析 依题意，则设甲到达的时刻为x ，乙到达的时刻为y ，若两人能会面，则20x y -，如图所示，所以，两人能会面的概率为221402521609P ⨯⨯=-=.故选C.（5）解析 解法一：如图所示，记OE 得三等分点Q （靠近点O ）的坐标为()0,m ，则()0,3E m ，从而直线AE 的方程为：13x ya m+=-，直线BQ 的方程为：1x ya m +=.由题意，可设直线AE 与直线BM 的交点M的坐标为()0,c y -，所以013y c a m-+=-，1y c a m-+=， 可得131c c a a ⎛⎫+=- ⎪⎝⎭，即()131e e +=-，得12e =.故选B. 解法二：如图所示，记OE 得三等分点为Q （靠近点O ）.由PF x ⊥轴，知Rt Rt BOQ BFM △∽△，于是OQ OBFM BF =，所以 OQ aFM a c=+①类似地，有Rt Rt AFM AOE △∽△，于是OE aMF a c=- ② 由①②得13OQ a c OE a c -==+，即1113e e -=+，得12e =.故选B.（6）解析 由三视图还原几何体四棱锥D ABC -，如图所示，由主视BE =，图知CD ABC ⊥平面，设AC 的中点为E ，则BE AC ⊥，且2AE CE ==，由左视图得4CD =，BE =DCBA在Rt BCE△中，4BC===，同理4AB=，在Rt BCD△中，BD===在Rt ACD△中，AD===综上，四面体的六条棱中，长度最长的是故选A.（7）解析因为函数()f x的定义域为R，所以0c>，且()00f=，得0b=.又()111afc==+，因此1a c c=+>，所以a c b>>.故选B.（8）解析依题意，1A=，22Tπ=，得2Tωπ=π=，所以2ω=，()sin(2)f x xϕ=+，又函数()f x的图像的对称轴方程为36212xππ-π==，则sin16ϕπ⎛⎫+=⎪⎝⎭，得262kϕππ+=+π，k∈Z，所以23k kϕ2π=π+∈Z，，又ϕ<π，故3ϕ2π=，所以()sin23f x x2π⎛⎫=+⎪⎝⎭.又()()12f x f x=，12,,36x xππ⎛⎫∈- ⎪⎝⎭，则12362212x xππ-++π==-，所以126x xπ+=-，则()12sin2663f x x fπ⎡π2π⎤⎛⎫⎛⎫+=-=⨯-+=⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦.故选C.（9）解析因为输入的5n=，4x=，故1v=，4i=，满足进行循环的条件；1448v=⨯+=，3i=，满足进行循环的条件；84335v=⨯+=，2i=，满足进行循环的条件；3542142v=⨯+=，1i=，满足进行循环的条件；14241569v=⨯+=，0i=，满足进行循环的条件；569402276v=⨯+=，1i=-，不满足进行循环的条件，故输出的v值为2276.故选C.（10）解析依题意，由抛物线定义得MF d=（M到准线:1l x=-），则MF MA+=()()11:1:113MM MA AM d A l x d C l x+=-=--=点到准线点到准线.故选B.评注 求圆锥曲线的最值一般用几何法+定义法或函数法.本题利用抛物线定义类比，将MF转化为点M 到准线:1l x =-的距离是关键，在多次转化过程中，要注意取等号的条件是否同时取得. （11）解析 设等差数列{}n a 的首项为1a ，公差为d ，由9538aa =得()113848a d a d +=+，即 1525a d =-，则5132405a a d d =+=->，得0d <，10a >.故12110a a a >>>>>1213a a >>.因此910110a a a >，1011120a a a <，1112130a a a >，1213140a a a <，且()()()111213101112111210131112111211121111222105da a a a a a a a a a a a a a a a a d a a +=+=+=+=>， 所以当11n =时，n S 取最大值.故选C. （12）解析 设()ln 1f x x x a =-++，当1,1e x ⎡⎤∈⎢⎥⎣⎦，()110f x x '=-，所以()f x 在1,1e ⎡⎤⎢⎥⎣⎦上单调递增，则()()11e f f x f ⎛⎫ ⎪⎝⎭，即()1,e f x a a ⎡⎤∈-⎢⎥⎣⎦.设()2e yg y y =，因为对任意1,1e x ⎡⎤∈⎢⎥⎣⎦，总存在唯一的[]1,1y ∈-，使得2ln 1e yx x a y -++=成立，所以1,e a a ⎡⎤+⎢⎥⎣⎦是()g y 的不含极值点的单调区间的子集，因为()()2e y g y y y '=+，所以当[)1,0y ∈-时，()0g y '<，()g y 在[)1,0-上单调递减，若(]0,1y ∈时，()0g y '>，()2e yg y y =在(]0,1上单调递增.又因为()()11e 1e g g -=<=，所以11,,e e e a a ⎡⎤⎛⎤-⊆ ⎢⎥⎥⎣⎦⎝⎦，解得2e e a <.故选B.（13解析 因为AB 在BC 方向上的投影为1-，且ABC △为等边三角形，所以BC AB ==2AC =.又2AD DC =，所以()2233BD BA AD BA AC BA BC BA =+=+=+-=2133BC BA +，AC BC BA=-，故()2212333BD AC BC BA BC BA BC ⎛⎫⋅=+⋅-=-⎪⎝⎭2211333BC BA BA BC BA ⋅+⋅-=22211333BC BC BA BA -⋅-=2211222332⨯-⨯⨯⨯- 212233⨯=.（14）解析 依题意，,A B 不相邻的排法有2232A A 12=种，则二项式()121x -的展开式中2x 项的系数为212C 66=.（15）解析 如图阴影部分所示，表示二元一次不等式组确定的平面区域，当APB α∠=最小值时，得2APO α∠=也取最小值，此时1sin 2OA OP PO α==，即线段OP 最长，也即当点P 离圆心O 最远时，α最小，此时点P 位于2020x y y -+=⎧⎨+=⎩的交点()4,2--.故答案为()4,2--.（16）解析 在212n n a S +⎛⎫= ⎪⎝⎭中，令1n =可得1a =，还可得()241n n S a =+， ()21141n n S a --=+()*2,n n ∈N ，所以()()()2211411n n n n S S a a ---=+-+，即2211422n n n n n a a a a a --=+--，也即2211220n n n n a a a a -----=，()()()1112n n n n n n a a a a a a ---+-=+.又()*0n a n >∈N ，则()*122,n n a a nn --=∈N ，所以{}n a 是首项为1，公差为2的等差数列，则得()*21n a n n =-∈N ，所以2212n n a S n +⎛⎫== ⎪⎝⎭，()22cos 1n n b n n n =π=-. 当n 为偶数时，有()()()22222212341nT n n ⎡⎤=-++-+++--+=⎣⎦()137212n n n ++++-=； 当n 为奇数时，有()()2211122nn n n n n T T nn --+=-=-=-.故()()112nn n n T +=-⋅.。

俯视图侧视图正视图高考二轮复习限时训练（十）（时间：60分钟）班级 姓名 得分一．填空题（每题5分，12小题共60分） 1．命题p ：“,12≤>∃xx ”的否定是2. 已知2{|2,}P y y x x x R ==+∈，{|Q x y ==，则P Q ⋂=3. 设命题p ：｜4x －3｜≤1；命题2:(21)(1)0q x a x a a -+++≤．若p ⌝是q ⌝的必要不充分条件，则实数a 的取值范围是4.若函数.右图给出的是计算201614121++++的值的一个流程图，其中判断框内应填入的条件是________5.若一个底面为正三角形、侧棱与底面垂直的棱柱的三视图如下图所示，则这个棱柱的体积为__________6.1019113sin)tan()3423πππ----的值是 ．7．函数-=)(x f )0(123≠++m mxx 在)2,0(的极大值为最大值，则m 的取值范围是 ．8.已知[0,]π∈x ．若向量(2cos 1,2cos 22)=++ a x x 和向量(cos ,1)=-b x 垂直，则x 的值为 ．9.若l 为一条直线，αβγ，，为三个互不重合的平面，给出下面三个命题：①αγβγαβ⊥⊥⇒⊥，；②αγβγαβ⊥⇒⊥，∥；③l l αβαβ⊥⇒⊥，∥． 其中正确命题序号是_________10．给出以下五个命题：(1)1)55(,22*=+-∈∀n n N n ．(2)设全集U={1,2,3,4,5,6}，集合{}{}3,4,3,6A B ==，则}6,5,3,2,1{)(=⋃B A C U ． (3)定义在R 上的函数()y f x =在区间(1,2)上存在唯一零点的充要条件是0)2()1( f f(4)已知A B C ∆所在平面内一点P (P 与,,A B C 都不重合)满足PA PB PC BC ++=， 则A C P ∆与B C P ∆的面积之比为2．其中正确命题的序号是___________ 11．已知函数)0(22cos 2sin ≠++=ab x b x a y 的一条对称轴方程为6π=x ，则函数)0(22cos 2sin ≠++=ab x b x a y 的位于对称轴方程6π=x 左边的第一个对称中心___________12．给出下列四个命题: ①若z ∈C,22z z =,则z ∈R; ②若z ∈C,z z =-,则z 是纯虚数;③若z ∈C,2zzi =,则z=0或z=i ; ④若121212,,z z C z z z z ∈+=-则120z z =.其中真命题的个数为 ． 二．解答题（每题15分，共30分）13.已知向量(cos ,sin ),(cos ,sin ),a b a b ααββ==-=（1）求的值)cos(βα-． （2）若202παβπ<<<<-，且αβsin ,135sin 求-=的值．14.已知数列{}n x 的首项13x =，通项()2*,,nn x p np n N p q =+∈为常数，且成等差数列。

一. 词汇语法巩固练习1.When there was a fire, I would investigate and determine if it was accidental or________ (purpose) set.2.To be honest, I really miss walking down the streets. . . the smell, the atmosphereand ________ (important), the places where I used to go to!3.The doors of the garden, like the windows, are also carved in different shapes tobring more vigor and elegance to the ________ (surround).4.I can't stop the ________ (pass) of time just as I can't end my memory of you. Iwish the light breeze would carry my best regards to you.5.It was the last thing that I expected, because people think of Kenya as a prettysafe, ________ (order) place where tourists go on holiday.6.Sometimes Mother nature's storms can be violent and ________ (destroy) butother times she can be very beautiful, like when flowers blossom.7.If the sealed jar had light and a steady temperature, there was ________ (theory)no reason why the lichens(地衣；苔藓) couldn't live until the sun dies.8.He snatched the mail form my hand, ________ (open) the mail and pointed to themail inside. “You were after this.”9.For over a decade the KindSpring user community has focused on inner________ (transform), while ________ (collect) changing the world with ________ (generous), gratitude and trust.10.When you read a book, you've probably noticed that a brand new book has arather special smell, which differs from ________ of an older book.11.Today, the forms of books have been changing ________ (drama) since the birthof eBooks. They might be convenient, ________ you can't give your copy to others as a present, and they don't smell as nice.12.Though it is 30 years ________ we last met, I still remember the scene ________we got separated on a rainy day.13.________ the fact that peer pressure is most commonly seen as a very negativeissue, it can ________ (see) positively in some situations.14.High school years are a large part in ________ (shape) the person you are goingto become. A school often provides education for you, and it can also be a place ________ you can develop relationships with many different people, join clubs, and participate in a ________ (vary) of different sports. These years can makeyou become a high school student ________ (fill) with many different emotions, including happiness, anger, sadness, envy, loneliness, stress, and a numerous amount more. This is also a time ________peer pressure is most commonly put upon a vast ________ (major) of teens.15.If you have a job, ________(devote) yourself to it and finally you’ll succeed.16.Now the popularity of pasta has been confirmed in a survey ________ places it asthe world’s favourite food ahead of meat, rice and pizza.17.He spent so many unforgettable days with them ________he was unwilling topart with them.18.The number of smokers, ________is reported, has dropped by 17 percent in justone year.19.Many experts hold the view that teachers’ development is ________ the key to abetter education lies.20.Golf is rapidly becoming more popular. Near some towns and cities new coursesare being built in ________was farmland.二. 七选五I am not sure how many books I have reread, but perhaps it is fewer than the average person. ___1___ The source material, though, is of course not.I used to take the same approach to books as I did to travel: don’t go to the same place twice. Life is too short. ___2___ Then I realized that the fact that life is short might work the other way around, too: if you know you enjoy something, or somewhere, then why not return?Recently I reread Joseph Heller’s Catch-22. I was inspired to do so when reminded of how he’d respond when people rudely asked him why he’d never wri tten anything as good: “Who has?” Catch-22 pretty much saved my life when I first read it. ___3___ I had dropped out of school twice, didn’t leave the house at all and didn’t have a life. It felt as though I hadn’t laughed in such a long time.___4___ It managed to take me out of the dark world, and though its themes are, of course, serious, its cleverness cheered me greatly. I related to its characters who are themselves trapped. I am now planning to reread the sort of books that inspired me in my own writing.I won’t take a break altogether from reading the most recent releases. I love the smell of new books fresh from the printers. ___5___A.Catch-22 had me laughing.B.My favourites are secondhand editions.C.There is discomfort in reading recently-released books.D.At that time I was an extremely depressed 17-year-old.E.For me, the pleasure of rereading is a newly discovered one.F.There is so much to read and so much to see and experience.G.However, I have determined to dip more frequently into the old ones三. 完形填空For one year, the Dannemiller family gave up buying any unnecessary purchases.In an effort to get back in touch with what they call their family mission, which includes “growing in faith together and serving others to create a world without ___1___”, parents Scott and Gabby Dannemiller decided to ___2___ spending money on things like toys, books, clothing, or anything that wasn’t a necessity or an experience.Overall, the family successfully ___3___ the plan. Though the experiment took place in 2013, the family ___4___ tries to live by the lessons they learned while cutting out unnecessary spending. “By focusing on experiences ___5___ purchases, we grew together in faith as a family, we were able to ___6___ others and we were able to give more of our time and treasure to people who really need it, ” Scott says.“My daughter’s birthday is next month, and she asked if we could go to visit her uncle’s farm and ride a horse, ___7___ just asking for a horse stuffed animal, ” he says. “Now we look at ___8___ and say ‘will that really add value to our life, or is it something we will just need to find space for and take care of?’ ”If you’re trying to teach your children to focus ___9___ on physical stuff, Scott says it’s helpful to tweak(稍微改进) your ___10___ when kids ask for things. “We used to say ‘that's too ___11___’, but that made our kids think OK, we need more money, and when we get more money we can have it, ” he says. “We ___12___ to ‘we don’t need that’, and that helped them understand.”When ___13___ spending, Scott says the most important thing is to focus not on what your family is giving up, but what it is gaining. “It’s not about what you're___14___,” he says. “The question should be, ‘What are we going to replace that with?’ Then, make sure you are adding something to your life that the people in your family ___15___. For us, that was time together.”1. A. need B. faith C. delay D. pity2. A. slow B. begin C. increase D. stop3. A. applied to B. subscribed to C. responded to D. stuck to4. A. even B. ever C. still D. just5. A. instead of B. in contrast to C. in parallel with D. regardless of6. A. treat B. serve C. please D. satisfy7. A. more than B. or else C. or rather D. rather than8. A. purchases B. possessions C. treasures D. earnings9. A. less B. occasionally C. frequently D. more10.A. appearance B. atmosphere C. language D. identity11.A. worthy B. cheap C. worthless D. expensive12.A. admitted B. referred C. shifted D. took13.A. cutting off B. cutting down C. cutting up D. cutting away14.A. lacking B. losing C. obtaining D. finding15.A. own B. preserve C. owe D. value四. 语篇填空AirCar is the latest generation flying car ____1____ transforms from road vehicle into air vehicle in less than 3 minutes. Useful for leisure and self-driving ____2____(journey), the vehicle can go from driving to lying mode with the click of a button. The fifth generation flying car designed by Professor Stefan Klein ____3____(complete) two flight tests at Piestany airport in Slovakia so far.____4____(weigh) 1,100kg, the two-seat AirCar can carry an additional load of 200kg per flight. ____5____(power) by a BMW 1.6L engine, the car-plane has an effective power output of 140HP. The estimated travel range of the vehicle is 1,000km and the flight ____6____(consume) is 18 liters (升) per hour. AirCar can transform from a ground-based vehicle to aircraft with its speed ____7____(reach) up to 200 km/h. But most importantly, the stability and controllability of the AirCar is ____8____(access) to any pilot.“With AirCar you will arrive at your destination ____9____ the trouble ofgetting a ride to airport and passing through commercial security. You can drive your AirCar to the golf course, the office, the shopping centre or your hotel and park ____10____ in a normal parking space,” said Anton Zajac, Klein Vision's co-founder, investor and pilot.一. 词汇语法巩固练习1.purposely2.most importantly/ more importantly3.surroundings4.passage5.orderly6.destructive7.theoretically8.opened9.transformation; collectively; generosity10.that11.dramatically; but12.since; where13.Despite; be seen14.shaping; where; variety; filled; when; majority15.devote16.that/which17.that18.as19.where20.what二. 七选五EFDAG三. 完形填空ADDCA BAAAC DCBBD四. 语篇填空1. that2. journeys3. has completed4. Weighing5. Powered6. consumption7. reaching8. accessible9. without 10. it。