选择填空限时训练一教师用1

限时训练(一)Ⅰ.单项填空1.Without the wise leadership of the Party，it hard to achieve the GDP growth target for 2022.A.would have beenB.would beC.wereD.had been答案 B解析考查虚拟语气。

句意为：要是没有党的英明领导，很难实现2022年国内生产总值增长的目标。

此处without短语相当于虚拟条件句，根据题干中的“2022”判断是对将来的虚拟，主句用would＋do的形式，故B项正确。

2.Parents’top priority is to help their kids develop the ability to show in their choices of friends.binationB.admirationC.inspirationD.discrimination答案 D解析考查名词词义辨析。

combination结合；admiration钦佩；inspiration灵感；discrimination 区别，识别。

句意为：父母的首要任务是帮助他们的孩子发展在选择朋友时的识别能力。

根据句意可知D项正确。

3.As Emerson puts it，we travel the world over to find the beautiful，we must carry it with us or we find it not.A.unlessB.untilC.thoughD.because答案 C解析考查连词词义辨析。

句意为：如爱默生所言，我们环游世界是为了寻找美，但是我们必须与美同行才能发现它。

unless除非；until直到；though尽管；because因为。

根据句意可知，这里表示一种让步关系，故选C。

4.I felt like giving up.I probably ，but my Dad whispered，“Come on! You can make it.”A.would haveB.wouldC.shouldD.should have答案 A解析考查过去将来时。

最新2019-2020学年七年级英语上册Unit 5单元限时训练时间：45分钟分值：100分第Ⅰ卷Ⅰ.单项选择(20分)1．—Do you play soccer?—Yes, I play it after school.A．the B．a C．/ D．an2．—your father have a car?—Yes, he ________.A．Does; do B．Do; does C．Does; does D．Does; is3．Peter's father often watches basketball games TV.A．in B．on C．at D．of4．He likes soccer, but he doesn't it on TV.A．look B．play C．watch D．look5．Does Tony a ruler?A．have B．has C．are D．is6．—What do you like better?—Basketball.A．food B．subject C．sport D．music7．Let's stay at home and computer games.A．plays B．to play C．play D．playing8．Mr.Black doesn't like the lecture(演讲)．It is .A．interesting B．boring C．fun D．relaxing9．We need one ping-pong ball and two .A．ping-pongs bats B．ping-pongs bat C．ping-pong bats D．ping-pong bat 10．—Let's play volleyball after class.—.Let's go.A．No, I don't B．That sounds interesting C．OK！It's boring D．Yes, we do Ⅱ.完形填空(20分)Peter and Tom __1__ a good friend.His __2__ name is Bob.They like sports.Peter thinks baseball __3__ interesting.He often plays it __4__ his classmates.He __5__ it.Tom and Bob like play-ing ping-pong.They have a pair of(一副) __6__.They think ping-pong is __7__.They __8__ play soccer because it's difficult.Peter and Bob don't __9__ TV because it's boring.__10__ they watch sports games on TV.( )1.A.is B．are C．have D．has( )st B．first C．family D．friend's( )3.A.is B．are C．has D．have( )4.A.for B．at C．with D．of( )5.A.play B．does C．have D．likes( )6.A.ping-pong bats B．baseball bats C．baseballs bats D．ping-pongs bats ( )7.A.difficult B．Boring C．interesting D．not easy( )8.A.don't B．aren't C．doesn't D．isn't( )9.A.look B．see C．read D．watch( )10.A.And B．For C．But D．WithⅢ.阅读理解(10分)My name is Jim.I'm in Class 2, Grade 7.My parents are English teachers.I have no brothers or sisters.I have an uncle.His name is Bob Green.He has three basketballs, two soccer balls and five ping-pong balls.Look！His ping-pong balls are under the table.He plays ping-pong every day.But I don't play ping-pong.It's boring.I have two soccer balls.And I play soccer with my friends everyday.1．Jim's are English teachers.A．sisters B．brothers C．parents D．uncles2．Jim has no .A．soccer balls B．uncle C．parents D．brothers3．Bob plays every day.A．soccer B．volleyball C．ping-pong D．basketball4．Bob's ping－pong balls are .A．under the bed B．on the bed C．on the table D．under the table5．Which of the following is TRUE(正确的)？A．Jim is in Grade Two. B．Jim's father is Bob.C．Bob has three ping-pong bats. D．Jim plays soccer every day.第Ⅱ卷Ⅳ.任务型阅读(10分)John Smith doesn't play sports.He has two children.His son, Jim, likes tennis and soccer.And he has a sports collection(收藏)．(2)He is in the school tennis club(俱乐部)．(3)He_plays_tennis_with_his_friends_every_day.Cindy, his daughter, likes baseball and tennis.____ She only watches them on TV.1．写出文中喜欢球类运动的人物。

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．。

长郡教育集团初中课程中心2019—2020学年度初三第一次限时检测物理注意事项：1.答题前，请考生先将自己的姓名、准考证号填写清楚，并认真核对条形码上的姓名、准考证号、考室和座位号；2.必须在答题卡上答题，在草稿纸、试题卷上答题无效；3.答题时，请考生注意各大题题号后面的答题提示；4.请勿折叠答题卡，保持字体工整、笔迹清晰、卡面清洁；5.答题卡上不得使用涂改液、涂改胶和贴纸；6.本学科试卷共四大题，考试时量60分钟，满分100分。

一、单项选择题（每题3分，共45分）1.下列说法中，正确的是（）A.机械能为零的物体，内能一定也为零B. “破镜难重圆”是因为固体分子间存在着斥力C.桂花盛开时花香四溢，是由于分子在不停地做无规则运动D.液体很难被压缩是因为分子间存在引力2.西昌卫星发射中心用长征三号运载火箭于2019年4月20日将第四十四颗北斗导航卫星成功送上太空.下列说法中正确的是（）A.火箭使用液氢燃料，主要是利用了液氢的比热容大B.火箭升空过程中，燃料燃烧释放的内能全部转化为火箭的机械能C.北斗导航卫星加速升空过程中，其动能和重力势能均增大D.发射架周围的“白气”是由大量水蒸气汽化而成的3.关于温度、热量、内能，以下说法正确的是（）A.一个物体的温度升高，一定是外界对物体做功B.温度高的物体，内能一定大C.温度越高的物体含有的热量越多D.物体的内能增大，可能是从外界吸收了热量4.如图所示，关于下列实验中所描述的物理过程的说法正确的是（）A B C DA.试管内的水蒸气推动塞子冲出去时，水蒸气的内能增大，温度升高B.抽去玻璃隔板，两瓶中气体逐渐混合均匀，说明空气比二氧化氮的密度大C.给瓶内打气，空气推动塞子跳起来时，瓶内空气的内能减小，液化成“白雾”D.迅速下压活塞，厚玻璃管中的硝化棉燃烧，说明管内空气被压缩时，它的内能增大，温度升高5.如表列出一些物质的比热容，根据表中数据，下列判断正确的是（）A.不同物质的比热容一定不同B.物质的物态发生变化，比热容不变C.铝和铜升高相同的温度，铝吸收的热量更多D.质量相等的水和煤油吸收相同的热量，煤油的末温可能比水的末温低6.关于热机，下列说法正确的是（）A.因为能量是守恒的，所以热机的效率可以达到100%B.柴油机的汽缸顶部有喷油嘴，吸气冲程时吸进柴油和空气的混合气体C.提高热机的效率，可采用热值高的燃料D.活塞从汽缸的一端运动到另一端的过程叫做一个冲程7.如图是汽油机一个工作循环四个冲程的简图，则这四个冲程正确的排序应为（）甲乙丙丁A.甲乙丙丁B.乙甲丁丙C.乙丁甲丙D.丙乙甲丁8.A、B是两个轻质泡沫小球，C是用毛皮摩擦过的橡胶棒，A、B、C三者之间相互作用时的场景如图所示，由此判断（）A.小球A一定带正电B.小球B一定带正电C.小球B可能不带电D.小球A可能不带电9.根据生活经验，判断下列说法正确的是（）A.教室中三个开关控制六盏灯，一个开关控制两盏灯，因此这六盏灯是两两串联再并联B.马路上的路灯晚上一起亮，早晨一起灭，因此它们是串联的C.高大建筑物上的一串串装饰灯同时亮，同时灭，因此任意一串上的灯是并联的D.在家庭电路中，灯与控制它的开关是串联的，灯与插座是并联的10.下列说法不正确的是（）A.电路中有电源不一定有电流B.规定正电荷定向移动的方向为电流的方向C.绝缘体不容易导电是因为内部没有大量的电荷D.在一个闭合电路中，在电源外部电流从电源正极流出，经过用电器，流向电源负极11.两个验电器，甲带正电，乙不带电，用一带绝缘柄的金属棒分别连接甲、乙验电器的金属球，则（）甲乙A.自由电子由乙转移到甲，电流方向由乙向甲，甲的箔片张角减小B.自由电子由乙转移到甲，电流方向由甲向乙，甲的箔片张角减小C.自由电子由甲转移到乙，电流方向由甲向乙，乙的箔片张角增大D.原子核由甲转移到乙，电流方向由甲向乙，乙的箔片张角增大12.如图所示，图中两灯规格不同，能测出通过灯L2电流的电路是（）A B C D13.如图为某品牌共享电动助力单车，使用前，市民可通过二维码扫码开锁.之后，转动按钮即开关S1闭合，电动机启动，助力车开始运行.在夜间行车时，闭合开关S2行车灯亮起.下列电路设计符合上述要求的是（）A B C D14.如图甲所示的电路中，闭合开关，灯泡发光，电流表A1和A2的指针位置如图乙、丙所示，则通过灯泡L1和L2的电流分别为（）甲乙丙A.1.16A 0.34AB.1.5A 0.34AC.1.5A 1.7AD.0.3A 0.34A15.小阳用三个电流表和两个小灯泡做实验，检验并联电路干路电流是否等于各支路电流之和，其连接的电路如图所示，这个电中有一根导线接错了，这根接错的导线是（ ）A.aB.bC.cD .d二、填空题（每空2分，共20分）16.如图所示，将气球在头发上摩擦后小女孩的头发会飘起来，若气球带负电，这是因为在摩擦的过程中气球 （选填“得到”或“失去”）电子，这种使物体带电的方法叫做 .17.用两只相同的酒精灯分别给两个相同的烧杯中500 g 的甲液体和300 g 的乙液体进行加热，得到的温度随加热时间变化的图象如图所示，酒精燃烧过程中把储存的化学能转化成 能，若甲、乙两种液体中其中一种为水，则另一种液体的比热容是 ()J /kg ⋅℃.[()34.210J /kg c =⨯⋅水℃]18.一台单缸四冲程汽油机，飞轮的转速为2 400 r/min，该汽油机每秒钟对外做功次；如图所示的汽油机，该冲程中燃气的内能（选填“增加”“减小”或“不变”）.19.如图所示，电路中有规格不同的三个灯泡.要使灯泡L2和L3串联，只需闭合的开关是；若电路工作一段时间后，灯泡L2突然熄灭了，灯泡L3仍发光，产生这种现象的原因可能是 .（填写一种即可）20.将一杯热水倒入盛有一些冷水的容器中，冷水的温度升高了10 ℃，这是通过的方式改变了冷水的内能.又向容器内倒入同样一杯热水，冷水的温度又升高了6℃.如果再向容器内倒入同样三杯热水，则冷水温度可再升高℃（不计热量损失）.三、实验题（每空2分，共20分）21.为了探究物体温度升高时吸收热量的多少与哪些因素有关，实验室中准备了以下仪器：两个规格相同的电加热器，两个相同的酒精灯，两个相同的烧杯，两支温度计，手表，铁架台，火柴，适量的水和煤油.（1）为了完成实验，还需要的仪器有；（2）为了准确地控制物体吸收热量的多少，在上述提供的热源中，选比较合理.实验中通过比较来确定吸收热量的多少；（3）某小组实验中记录了实验数据（见下表），分析第1、3次或第2、4次实验数据可得出物体温度升高时吸收热量的多少与有关.22.小余和小乐按照课本要求探究“并联电路中干路电流与各支路电流的关系”，他们设计的电路如图1所示.（1）在连接电路的过程中，开关应该；（2）在实验连线之前，小余发现电流表的指针偏转情况如图2所示，根据你的分析，你认为原因可能是；图1图2图3（3）在实验中，要求电流表A1测量干路电流，请用笔画线将如图3中的电路连接完整；（4）他们连好电路后用电流表分别测出A、B、C处的电流，见下表：他们下一步应该做的是 . A.整理器材，结束实验B.改变电流表的量程或换电流表再测几次C.换用不同规格的小灯泡，再测出几组电流值23.为了探究分子的运动快慢和哪些因素有关，小明进行了如下实验：先取来相同的两只烧杯，在其中放入质量相同的冷水和热水，再各滴入一滴红墨水，然后观察两杯水颜色变化的快慢.甲乙（1）实验中观察到的现象是 现象，得到的结论是：温度越高分子的无规则运动越剧烈. （2）为了保证这个实验的结论真实可靠，做这个实验时，应注意不能 ，避免外界条件的干扰. 四、综合题（24题6分，25题9分，共15分）24.可燃冰是一种新型能源，它是水和天然气在高压低温情况下形成的类冰状结晶物质，主要成分是甲烷，可燃冰的生成有三个基本条件：首先要求低温，可燃冰在0~10℃时生成，超过20℃便会分解，海底温度一般保持在2~4℃左右；其次是高压，可燃冰在0℃时，只需海水产生29个标准大气压即可生成，而以海洋的深度，此高压容易保证；最后是充足的气源，海底的有机物沉淀，其中丰富的碳经过生物转化，可产生充足的气源，其开采是世界难题，据中央电视台2017年5月18日报道，我国可燃冰已试采成功，技术世界领先.1 m 3可燃冰可以释放出175 m 3的天然气，已知()7333.610J /m , 4.210J /kg q c =⨯=⨯⋅天然气水℃.则：（1）求1m 3可燃冰释放出的天然气完全燃烧放出的热量；（2）若将上述的天然气通过燃气灶给水加热，可使41g 10k ⨯的水从20℃升高到75℃，求水吸收的热量； （3）求该燃气灶的热效率.（用百分比表示，小数点后保留一位） 25.近年来，我国汽车工业发展迅猛，各种新型汽车不断投放市场.以下是我国某品牌轿车的测试报告，已知燃料完全燃烧放出热量的公式是Q qV =，式中V 表示燃料的体积.请计算：（g 取10N/kg ）（1）轿车在测试过程中运动的时间； （2）轿车在测试过程中发动机的功率；（3）若轿车在测试过程中发动机的效率为30%，则消耗该型号汽油的热值； （4）说说提高汽油机效率的途径（说出两种即可）.2019—2020学年度初三第一次限时检测物理参考答案一、单项选择题（每题3分，共45分） 二填空题（每空2分，共20分） 16.得到 摩擦起电 17.内 840 18.20增加19.S 2 L 2被短接20.热传递9三、实验题（每空2分，共20分） 21.（1）天平 （2）相同的电加热器 加热时间（3）物质种类22.（1）断开 （2）电流表没调零（3）如图（把L 2的左端与电流表A 1的“3”接线柱连起来）（4）C 23.（1）扩散（2）搅拌或者晃动烧杯四、计算题（24题6分，25题9分，共15分） 24.（1）96.310J ⨯ （2）92.3110J ⨯ （3）36.7% 25，（1）4 000s（2）42.410J ⨯（3）7410J /L ⨯（4）①尽量使燃料充分燃烧 ②尽量减小各种热量损失 ③机件间保持良好的润滑，减小摩擦。

填写板块限时模拟训练01语法填空+应用文写作+读后续写时间：45分钟满分：55分Ⅰ.语法填空（共10小题，每小题1.5分，满分15分）（2022·邵阳一模）Scientists say soft robot jellyfish that can swim through openings ____36____（narrow）than their bodies could be used to observe coral reefs. In future the “jellybots” could be sent into delicate environments, such as coral reefs, without risking damage. Their creators believe they could act ____37____ “guardians of the oceans”.Dr Erik Engeberg, one of the robot’s ____38____（inventor）from Florida Atlantic University in the US, said, “____39____（study）and observing delicate environment has always been hard for researchers. Soft robots can help with this.”The design of the jelly bot is based on ____40____ shape of the moon jellyfish. The team used a system ____41____（drive）by two impeller pumps（叶轮系）to allow the robot to swim.Dr Engeberg said, “Soft robots based on fish and other sea animals ____42____（win）popularity in the research community in the last few years. Jellyfish, are excellentchoices,____43____ that they are very efficient swimmers. Their ____44____（perform）is due to the shape of their bodies,_____45_____ can produce a combination of vortex（漩涡）and jet propulsion（推进）.”Ⅰ.写作（共两节，满分40分）第一节应用文写作（满分15分）（2022·江苏八校联考）假设你是李华，最近你要参加主题为“My Dream University”的英语演讲比赛。

限时训练刷题练习册一、选择题1. 下列哪个选项是正确的数学表达式？A. 2 + 3 = 5B. 3 × 4 = 12C. 4 ÷ 2 = 6D. 5 - 1 = 3答案：A2. 请从下列选项中选出正确的英语单词拼写。

A. colerB. colorC. collerD. cooler答案：B3. 以下哪个历史事件标志着第二次世界大战的结束？A. 珍珠港事件B. 诺曼底登陆C. 广岛原子弹爆炸D. 柏林墙倒塌答案：C二、填空题1. 地球的自转周期是 ________ 小时。

答案：242. 请填写下列化学元素的符号：铁 ________，氧 ________。

答案：Fe，O3. 根据题目要求，以下句子中缺少的单词是 ________。

- 句子：I am ________ to help you.答案：willing三、简答题1. 请简述牛顿第三定律的内容。

答案：牛顿第三定律指出，对于两个相互作用的物体，它们之间的作用力和反作用力大小相等，方向相反。

2. 请解释什么是光合作用，并简述其过程。

答案：光合作用是植物、藻类和某些细菌利用光能将水和二氧化碳转化为葡萄糖和氧气的过程。

这个过程主要发生在叶绿体中，分为光反应和暗反应两个阶段。

光反应产生ATP和NADPH，暗反应则利用这些能量和还原力将二氧化碳固定并转化为有机物。

四、计算题1. 如果一个圆的半径是5厘米，求这个圆的面积。

答案：圆的面积公式是A = πr²，代入 r = 5 厘米，得到 A = π × 5² = 25π ≈ 78.54 平方厘米。

2. 一个班级有40名学生，其中25名男生和15名女生。

如果随机选择一名学生，求选中男生的概率。

答案：选中男生的概率是男生人数除以总人数，即 P(男生) = 25/40 = 5/8。

结束语本练习册旨在帮助学生通过限时训练提高解题速度和准确率。

希望同学们能够认真完成每一道题目，不断提升自己的学习水平。

金陵中学河西分校八(上)英语限时训练(四)一、选择填空(每小题1分，共15分)1. — What did you and your friends buy in the shop this afternoon?— ________. ________ of us had money for anything.A. Nothing; NoneB. None; No oneC. Nothing; NobodyD. None; Nothing2. — Listen! What's the noise outside?— It________ be some children playing football. But I'm not quite sure.A. mayB. mustC. shouldD. would3. — Could you please ________ here? There are many kids playing here.— Sorry, I didn't know that.A. to smokeB. not to smokeC. smokeD. not smoke4. My parents agreed ________ me ________ at home alone when they were away.A. to let; to stayB. letting; stayC. to let; stayD. letting; to stay5. The boy is so smart that he can ________ many difficult maths problems correctly.A. put outB. cut outC. work outD. give out6. —How is Susan now?— I hear a famous company her with a job, but she refused it.A. preventedB. offeredC. providedD. introduced7. — Not all people coming here are real bird lovers.— Yeah. Some try to make the place a better area ________ others just bring trouble.A. soB. becauseC. whileD. moreover8. — Is Jack good at singing?— Yes, we can often hear him ________ in the classroom.A. singB. singsC. to singD. singing9. Spring Bud Project is an organization that raises money ________ poor young girls return to school.A. helpingB. to helpC. invitingD. to inviteA. on; atB. on; inC. at; onD. at; in11. It snowed heavily last night. Grandpa made a snowman ________ the children in the morning.A. surpriseB. surprisedC. to surpriseD. surprising12. Our teacher often tells us ________ when we have trouble ________ a problem.A. to give up; solvingB. not to give up; working outC. to give up; to work outD. not to give up; to solve13. As is known to all, about seventy-five ________ of the earth ________ covered with water.A. per cent; areB. per cent; isC. per cents; areD. per cents; is14. He spoke loudly ________ everyone in the room ________ him clearly.A. to make; to hearB. to make; hearC. making; to hearD. making; hear15. ________ animals will be in danger and there will be space for them if we don't take any action.A. More and more; fewer and fewerB. Fewer and fewer; more and moreC. More and more; less and lessD. Less and less; more and more二、完形填空Many people live with animals. Some have dogs and cats. Some have fish. Dian Fossey was different from them. She had gorillas. The gorillas didn't live in her house. She lived in ___16___. What a special woman she was!Dian Fossey was born in San Francisco, California, in 1932. She loved ___17___. She went to college to become a veterinarian(兽医). One day, she took a trip to Africa. ___18___ she was there, she became interested in gorillas. Dian loved watching the gorillas. She decided to move there.Dian Fossey was known as the first person to have19contact(接触) with gorillas. She sat with the gorillas. She played with them. They never ___20___ her.Dian became interested in a baby gorilla. She was able to ___21___ him. She named him Digit. Dian was able to watch him grow. Digit liked Dian, too. A few years later, Digit was killed by hunters. Dian was very ___22___ about Digit's death. She decided to do something to ___23___ the gorillas.She started a ___24___ that tried to protect the gorillas. National Geographic wrote a story about Dian and her group. Many people ___25___ money to her to help save the gorillas. She said she would do more things to save gorillas.16. A. hers B. his C. theirs D. ours17. A. bird B. animals C. forests D. plants18. A. Although B. Because C.While D. Unless19. A. friendly B. strict C. serious D. careful20. A. win B. beat C. protect D. hurt21. A. stay away from B. get close to C. look out for D. catch up with22. A. pleased B. sad C. bored D. proud23. A. protect B. kill C. lose D. miss24. A. team B. group C. club D. center25. A.stole B. wrote C. sent D. received三、阅读理解(每小题1分，共5分)(A)Wetlands are any land that is flooded with shallow water all or most of the time. They are a natural water holding system.There are many types of wetlands. Among them, bogs, marshes and swamps are the three n types. The different types of wetlands have different kinds of plants. Only mosses (苔藓) and a few other kinds of plants can grow in bogs. Grassy plants like cattails(香蒲) and reeds (芦苇) are the most common plants in a marsh. A swamp is a forest whose ground is underwater all or most of the time. Unlike bogs or marshes, a swamp is full of trees and bushes. It's not a good idea to go exploring a swamp without a guide. There are hidden pools of water, thick mud, and sometimes big crocodiles looking for their next meal.Hungry crocodiles aren't the only animals that make the wetlands their home. Otters, turtles, frogs, snakes and many other animals do too. The water is home to many kinds of fish and crabs. Birds, including ducks, geese and cranes, use wetlands seasonally during their long migrations.Wetlands are important because they provide habitats for plants and animals. A wetland system can also protect shorelines, make polluted waters clean, prevent floods, and restore underground water supplies.According to WWF, more than half of the world's wetlands have disappeared since the beginning of the 20th century. If this continues, countless plant and animal species will surely die- out. Without wetlands, cities have to spend more money treating water. Don't feel helpless. Try to do your part to protect them right now. Here are some ideas.26. According to the passage, which of the following is TRUE?A. Wetlands are a man-made water holding system.B. There are three types of wetlands in the world.C. People cannot meet with big crocodiles in wetlands.D. Wetlands can protect shorelines and prevent floods.27. What is the writer probably going to talk about after Paragraph 5?A.What are the different types of wetlands?B. What will happen if wetlands disappear?C. What actions can people take to protect wetlands?D. What other kinds of plants and animals live in wetlands?(B)Some of the greatest problems we face today are the destruction(破坏) of our environment. Brown clouds, polluted water, endangered wild animals... these problems seem so huge.So my family do what we can. We take cloth bags to stores instead of using plastic bags. We walk where we don't have to drive ...But does it do any good? When I am the only one in line at the market with cloth bags, am I doing any good? Does my walking to stores make any real difference to the world?I recently learned something about flamingos(火烈鸟) which like to get together in groups of a thousand or more. Every year, when the time comes for migration, a few of them first take off from the lake. But none of the others seem to notice, so the small group returns. However, the next day they try again. This time a few more fly along with them, but most of them still pay no attention, so they return again. They try for several times. Every time a few more birds join in but, since the thousands of others still take no notice, the great migration plan is once more stopped.Then one day something changes. The same small group of birds once again starts flying and a small number more join in just as before, then more. Finally, they all take flight and the migration really begins. What a spectacular(壮观的) sight it must be -thousands of flamingos taking off into the sky at once!A few can make a difference. Even if you're the one to take the first step, and continue trying, others will someday take notice and together we will solve even our greatest problems.28. Why does the small group of flamingos return to the lake?A. Because they are very tired.B. Because they can't find food in other places.C. Because they don't know where to go.D. Because the others don't fly along with them.29. What’s the writer's purpose of writing this article?A. To show that the writer loves to see the migration of flamingos.B. To introduce how flamingos migrate to readers.C. To tell readers to continue trying and it can make a difference.D. To show that we face lots of serious problems today.30. Where would you most probably read this passage?A. A guidebook.B. A movie poster.C. A personal blog (博客).D. A science magazine.四、词汇(每小题2分，共30分)(A)根据汉语提示、音标或首字母提示写单词，使句意完整31. They finally___________ (卖) their house at a good price.32. I invited Tom to dinner and he happily___________(接受) my invitation.33. The man can't sleep well at night and it leads to___________ / ˈsɪəriəs/ problems to his health.34. Don't take p___________ him. He is pretending(假装) to be poor.35. Don't be too strict with yourself. After all, none of us is___________(完美的).36. Our_____(政府) is taking some action to improve people's living conditions.37. My teacher asked me to___________(描述) this picture with my own words.38. The students come from different places. Do you know all their ___________(地址)?(B)用所给词的适当形式填空39. We don't need to hurry because there is still a little time ___________(leave).40. My grandfather likes sleeping with the windows___________(close) in summer.41. The film is very interesting and I lost myself in it from___________(begin) to end.42. He treats everyone with ___________(kind), so he has lots of friends.43. --Is John in his bedroom now? --Yes. He may___________(make) his bed.44. The doctor did everything he could___________(save) the child who was hurt in the accident.45. A nurse took her arm and___________(lead) her to a chair.五、句型转换，每空一词。

2024年春学期八下英语第一次限时训练试卷一、单项选择（每小题1分，满分15分）（）1．Mike Smith has been in China for fifteen years and he _________ a lot of Chinese friends.A．makes B．made C．has made D．will make（）2．________ exercise every day makes us _________.A．Doing；health B．Doing；healthy C．Do；health D．Do；healthy（）3．Of the three foreigners, one is from London, ________ two are from New York.A．other B．the other C．another D．some other（）4．They ________ in 2008 and they have had a happy family ________ 2008.A．were married; since B．got married; in C．married; from D．got married; since（）5．When the woman had free time, she _________ shopping on Sundays.A．used to going B．get used to going C．was used to going D．was used to go( ) 6. I have ______read the book. Have you read the book _______?A. already, alreadyB. yet, yetC. yet, alreadyD. already, yet( ) 7. Both his parents look sad. Maybe they _________what's happened to him .A. knewB.have knownC. must knowD.will know( ) 8. Please tell me more __________your interview ____________him.A. about, toB. to, withC. about, withD. to, and( )9.It has been two months_______my family moved to Lianyungang.A.forB.sinceC.becauseD.so( )10.---What a terrible experience! ---__________, lucky us, we are safe now.A.AnywayB.OtherwiseC.For exampleD.As a result（）11．What a surprise to see you here!When you back？A．do；get B．did；get C．have；got D．will；get（）12．-I like the dress，but I'm _________ I haven't got enough money．﹣Don't worry．I don't mind ________ you some if you like．A．afraid；lending B．glad；lendingC．afraid；to lend D．glad；to lend（）13．﹣Could you help me with my Geography project？﹣_______.A．No problem. B．You' re welcome.C．My pleasure. D．Don't mention it.（）14．—I went boating on Baijia Lake last Saturday.—_____ Why didn't you tell me earlier？A．You did？B．I hope so. C．Have a good time. D．It's kind of you.（）15．Which of the following is an Opinion？A．There are many museums in Nanjing.B．Qinhuai River runs through our city Nanjing.C．Nanjing has held many important events over the past years.D．Life in Nanjing is getting better in some ways.二、完形填空（每小题1.5分，满分15分）Tina was a seventeen-year-old girl who always wore a bright smile. She had a disease（疾病）and had to use a walker most of the time. Maybe it was because she looked 1 and people didn’t know how to come near to her. Tina usually broke the ice with people she met with a big “Hi”.In one class, I gave the students an assignment (作业) to recite (背诵) a poem. I only made the assignment worth a very small part of their total grade since I knew most of my 2 would not do it anyway. In the class, one by oneeach student 3 to correctly recite the poem. Finally, angry and half joking, I said that the next student who 4 recite the poem had to do three push-ups (俯卧撑). To my 5 , Tina was next. She used her walker to move to the front of the class. 6 she recited, she made a mistake. Before I could say a word, she 7 her walker and started doing push-ups. I wanted to say, “Tina, I was just joking!” But she stood up, continued the poem and she finished the rest perfectly.When she finished, a student asked, “Tina, why did you do that? It’s not an 8 assignment!”“Because I want to be like you guys! To be normal (正常的),” Tina said.Silence fell on the whole room when another student cried out, “Tina, we’re not normal! We are teenagers! We get in 9 all the time.”“I know,” Tina said as a big smile spread across her face. The rest of the students laughed, too.Tina got only a few 10 that day, but she got the love and respect of her classmates. To her, that was worth a lot more than a grade.( ) 1. A. different B. strong C. short D. young( ) 2. A. friends B. students C. teachers D. parents( ) 3. A. started B. planned C. continued D. failed( ) 4. A. shouldn’t B. couldn’t C. needn’t D. wouldn’t( ) 5. A. surprise B. surprised C. excitement D. excited( ) 6. A. Because B. When C. After D. Though( ) 7. A. raised B. changed C. threw D. checked( ) 8. A. active B. useful C. important D. exciting( ) 9. A. attention（注意）B. thought C. touch D. trouble( ) 10. A. points B. chances （机会） C. mistakes D. successes(成功)三、阅读理解（每小题2分，满分30分）Mr.and Mrs.White are my neighbors.Their house is small but they have a large garden.And they sped manyhappy hours working on it every day.We all agree the garden is great.While we spend our dollars in the market，the old couple can pick vegetables in their own garden.Last month，Mr.and Mrs.White invited most of the neighbors to their garden for an all﹣day food festival"，They asked us to bring gloves and arrive early in the morning.We didn't know what would take place.By 9：00 a.m，there were nine of us in the garden picking tomatoes，beans and other vegetables.At lunch，there were lots of laughs.We shared a lot of stories.After dinner，we played games.Before we left，Mr.and Mrs.White gave each of us a shopping bag filled with vegetables.What a great gift!Of course I loved the green vegetables，but I loved the happy time being with my neighbors more.Mr.and Mrs.White were so nice that they helped us learn from each other better.Now Mr.and Mrs.White have a blog about gardening.They teach us to grow vegetables online.And I can't wait to see my tomato plants bear fruit.（）4.Why did Mr.and Mrs.White invite their neighbors to their garden？ A．To go to the birthday party. B．To help pick vegetables.C．To help them learn from each other. D．To learn about their stories.（）5.What did the writer probably think of the food festival？ A．It was too short. B．It was wonderful.C．It was a waste of time. D．It was terrible.（）6.What can we learn from the passage about Mr.and Mrs.White？ A．They have a big house with a garden. B．They have less than eight neighbors.C．They want to open a vegetable store. D．They teach people gardening online.COur life has become easier and more colorful because of mobile phones and the Internet.However，a number of worrying problems have come along with them.Many car accidents are caused by mobile phones.Some drivers use mobile phones to get on the Internet while driving.Looking at their mobile phones and not at the road，these drivers have become a danger to us all.Many of the dangers of the Internet are well known.Children who visit websites without their parents' permission sometimes talk to elder people they don't know and put themselves in great danger.People with blogs sometimes post information of people they don't know on their websites without permission.Some people even use the Internet to get personal information that we do not want them to have.Mobile phone and the Internet safety tips：•Use a"hands﹣free" phone if you need to use a mobile phone when you drive.•In difficult driving situations such as in rain or snow，do not use your phone at all.•Do not make very important phone calls while driving.They take your attention（注意力）off the road.•Tell your children not to talk with anyone they do not know on the Internet.•In an embarrassing situation（尴尬场合）be careful with people around you.•Don't give out any personal information on the Internet.（）8.Which of these dangers of mobile phones or the Internet is NOT mentioned in the passage？ A．You can have a car accident.B．Someone can post your information on the Internet.C．Someone can get your personal information.D．Someone you do not know can call you.（）9.A "hands﹣free" mobile phone is probably a mobile phone that .A．is free to use. B．can be used without your hands.C．can be used with only one hand. D．can be used only in ears.（）10.What advice is given in the passage to stop someone from taking an embarrassing picture of you？ A．Be careful with the people around you.B．Don't give out your personal information.C．Use a "hands﹣free"mobile phone.D．Don't talk with the people you don't know.（）11.The underlined word "permission"in the article may mean .A．陪伴B．监督C．允许D．反对DChinese tourists flock(涌入) to Morocco(摩洛哥) during Spring Festival In the Chinese restaurant of Chefchaouen, one of the top tourist attractions(旅游景点) in Morocco, an area of less than 100 square meters was full of more than a hundred Chinese tourists, with a lot more waiting outside, during the Chinese Spring Festival."Boss, can I have my Kung-Pao Chicken now?" "Boss, is my beef noodles ready?" Some visitors were not patient enough to wait long for their ordered foods. Restaurant owner said that during the Spring Festival, her business was much better than usual.Another Chinese restaurant owner in Fes, Morocco's central city, had the same experience. The waiters said during the Spring Festival, a large number of Chinese tourists waited in lines for meals every day.In the southern Moroccan town of Marrakech, tourist buses full of Chinese tourists can be found everywhere during the Spring Festival.In Majorelle Garden, a famous tourist attraction, two girls from Ningbo, a city in East China's Zhejiang province, said they chose Morocco as their Spring Festival travel place to taste the country's special food.Morocco's Minister (部长) of Tourism Mohamed Sajid said it hopes to attract(吸引) 500,000 Chinese tourists each year by 2020. Sajid also said Morocco is working to improve hotels, transports and tourist reception. He said direct flights between Morocco and China are being discussed, which is one of the key factors (因素) to attract Chinese tourists.( )12.Where can we probably find this article?A. In a storybook.B. In a sports magazine.C. In a newspaper.D. In a cookbook.( )13. What do the two girls from Ningbo come to Morocco for?A. shoppingB. Moroccan foodC. tourist attractionsD. Chinese restaurants( )14. Which is one of the key factors to attract Chinese visitors?A. hotelsB. transportsC. foodD. direct flights( )15. Which are famous places of interest in Morocco?A. ChefchaouenB. MarrakechC. Majorelle GardenD. All of the above四、基础知识考查（1-10小题每小题1分，11-15小题每题2分，满分20分）A. 根据中文提示写出单词1.It’s our duty to protect our________________ (环境).2.You should____________ (归还)the books to the library as soon as possible.3.Our study____________ (条件)is much better than before.4.I want to go____________ (在国外)for further study.5.They have been good friends since they were at a____________ (小学教育的)school.B. 用所给单词的适当形式填空6.The trip to this Asian country is really a （pleasure）one.7.It's a long time since the readers （gather）at at the gate of the bookshop.8.I can't open the door.Who （take）away my key.9. （use）a website to learn about a new place is also a good way.10.What grate fun it is （see）the elephants take bananas from the visitors.C. 根据中文提示补全句子11.The film Avatar （结束）for a quarter so far.12.We often （梦想环游世界）without a passport in the future.13.Would you please （不要乘坐直达航班）to Beijing？14.The UK is （由…组成）England，Scotland，Wales and Northern Ireland.15.I tried to pull one hand free and finally （设法弄断）the ropes.五、任务型阅读（满分20分）A.阅读回答问题（每小题2分，计10分）About 10 years ago, when I was a student in college, I was working in a shop near my school. One day, I saw an elderly couple come in with a little girl in a wheelchair (轮椅).As I looked at this girl, I saw that she was seated in her chair. I realized she was disabled,without any arms or legs, just a head, neck and body. She was wearing a little white dress with red dots.As the couple pushed her up to me, I was looking down at the cash register. I turned my head toward the girl and gave her a wink. As I took the money from her grandparents, the girl gave me the biggest smile I had ever seen. All of a sudden, her disability disappeared. All I saw was this beautiful girl, whose smile gave me a new sense of life. She took me from being a poor, unhappy college student and brought me into her world of smiles, love and warmth.I'm a successful businessman now. Whenever I am down and think about the troubles of the world,I think about that little girl's smile and the life lesson she taught me. Smile is the strongest power!1. When did the story happen?____________________________________________________2. Why was the girl seated in a wheelchair?____________________________________________________3. What did the little girl's smile give the writer?____________________________________________________4. What is the writer now?_____________________________________________________5. What can you learn from the story?______________________________________________________B）阅读短文，根据首字母提示，在空格里填入适当的单词。

40分钟限时练习（1）一．选择题（共8小题，满分24分，每小题3分）1．（3分）﹣8的倒数是（ ）A ．8B ．18C ．−18D ．﹣8 2．（3分）若√x+2x有意义，则实数x 的取值范围为（ ） A ．x ＞﹣2 B ．x ≥﹣2 C ．x ＞﹣2且x ≠0 D ．x ≥﹣2且x ≠03．（3分）2022年11月5日，“长征三号”运载火箭在西昌卫星发射中心点火起飞，随后将“中星19号”卫星准确送入高度为35800千米的预定轨道，发射任务取得圆满成功．该卫星主要为跨太平洋重要航线、东太平洋海域及北美西海岸等覆盖区域提供通信服务．数据35800用科学记数法表示为（ ）A ．0.358×105B ．358×102C ．3.58×104D ．3.58×105 4．（3分）分式方程1x−2=3x 的解为（ ） A ．x ＝3 B ．x ＝2C ．x ＝1D ．无解 5．（3分）已知点（﹣2，3）在反比例函数y =k x 的图象，则下列各点也在该图象上的是（ ）A ．（2，3）B ．（1，﹣6）C ．(6，−12)D ．（0，0）6．（3分）计算2sin 30°的值为（ ）A ．1B ．√3C ．2D ．2√37．（3分）一个多边形的内角和的度数可能是（ ）A ．1700°B ．1800°C ．1900°8．（3分）下列图形中，既是轴对称图形也是中心对称图形的有（ ）A ．4个B ．3个C ．2个D ．1个二．填空题（共8小题，满分32分，每小题4分）9．（4分）因式分解：ma 2﹣2am +m ＝ ．10．（4分）化简：3m 2n9m = ．11．（4分）如图，四边形OABC 是矩形，OC 在x 轴上，OA 在y 轴上，函数y ＝x 的图象与AB 交于点D （3，3），点E 是射线BC 上一点，沿DE 折叠点B 恰好落在函数y ＝x 的图象上，且BE ＝2CE ，则点B 的坐标为 ．12．（4分）已知下列命题：①若a2＝b2，则a＝b；②2022年全年鄂尔多斯市一般公共预算累计完成842.8亿元，用科学记数法表示为8.428×1010元；③二元一次方程2x+y＝6的正整数解有3对；④连接两点之间的线段叫做两点之间的距离．是真命题的是.（只填序号）13．（4分）为了落实“双减”政策，减轻学生作业负担，某学校领导随机调查了九（1）班学生每天在作业上共花费的时间，随机调查了该班10名学生，其统计数据如下表：则这10名学生每天在作业上花费的平均时间是小时．时间（小时）43210人数2421114．（4分）如图，四边形ABCD中，E、F、G、H依次是各边中点，O是四边形内一点，若S四边形AEOH＝3，S四边形BFOE＝4，S四边形CGOF＝5，则S四边形DHOG＝．15．（4分）如图，在▱ABCD中，∠ABC＜90°，⊙O与它的边BA，BC相切，射线BO交边AD于点E．当AB＝6，AD＝8时，DE的长等于．16．（4分）如图，点A是反比例函数y=kx（k≠0，x＜0）图象上一点，过点A作AB⊥y轴于点D，AD＝DB．若点C为x轴上任意一点，且S△ABC＝9，则k的值为．三．解答题（共4小题，满分44分）17．（10分）计算：（1）（√12−√13）×√3+(12)0；（2）（m ﹣1）2﹣m （m ﹣3）．18．（10分）计算．（1）{x −y =12x +5y =9；（2）3x +2≤﹣2（x ﹣2）．19．（12分）为了解市民对全市创卫工作的满意程度，某中学数学兴趣小组在全市甲、乙两个区内进行了调查统计，将调查结果分为不满意，一般，满意，非常满意四类，回收、整理好全部问卷后，得到下列不完整的统计图．请结合图中信息，解决下列问题：（1）求此次调查中接受调查的人数，并补全条形统计图．（2）若本市人口300万人，估算该市对市创卫工作表示满意和非常满意的人数．（3）兴趣小组准备从调查结果为不满意的4位市民中随机选择2位进行回访，已知4位市民中有2位来自甲区，另2位来自乙区，请用列表或用画树状图的方法求出选择的市民均来自同区的概率．20．（12分）如图，在△ABC中，AB＝AC，D是边BC延长线上一点，连接AD．AE∥BD，∠BAC＝∠DAE，连接CE交AD于点F．（1）若∠D＝36°，求∠B的度数；（2）若CA平分∠BCE，求证：△ABD≌△ACE．。

限时训练(一)Ⅰ.单项填空1.Without the wise leadership of the Party，it hard to achieve the GDP growth target for 2022.A.would have beenB.would beC.wereD.had been答案 B解析考查虚拟语气。

句意为：要是没有党的英明领导，很难实现2022年国内生产总值增长的目标。

此处without短语相当于虚拟条件句，根据题干中的“2022”判断是对将来的虚拟，主句用would＋do的形式，故B项正确。

2.Parents’top priority is to help their kids develop the ability to show in their choices of friends.binationB.admirationC.inspirationD.discrimination答案 D解析考查名词词义辨析。

combination结合；admiration钦佩；inspiration灵感；discrimination 区别，识别。

句意为：父母的首要任务是帮助他们的孩子发展在选择朋友时的识别能力。

根据句意可知D项正确。

3.As Emerson puts it，we travel the world over to find the beautiful，we must carry it with us or we find it not.A.unlessB.untilC.thoughD.because答案 C解析考查连词词义辨析。

句意为：如爱默生所言，我们环游世界是为了寻找美，但是我们必须与美同行才能发现它。

unless除非；until直到；though尽管；because因为。

根据句意可知，这里表示一种让步关系，故选C。

4.I felt like giving up.I probably ，but my Dad whispered，“Come on! You can make it.”A.would haveB.wouldC.shouldD.should have答案 A解析考查过去将来时。