双基限时练8

合集下载

2014-2015学年高中数学 双基限时练8 新人教B版必修4

2014-2015学年高中数学 双基限时练8 新人教B版必修4

双基限时练(八)基 础 强 化1.已知cos ⎝ ⎛⎭⎪⎫α+π2=63,则sin α的值为( ) A.33 B .-33 C.63D .-63解析 cos ⎝ ⎛⎭⎪⎫α+π2=-sin α,∴sin α=-63. 答案 D2.已知sin ⎝ ⎛⎭⎪⎫α-π3=13,则cos ⎝ ⎛⎭⎪⎫α+π6的值为( ) A .-13B.13C.233D .-233解析 ∵⎝ ⎛⎭⎪⎫α+π6-⎝ ⎛⎭⎪⎫α-π3=π2. ∴cos ⎝ ⎛⎭⎪⎫α+π6=cos ⎣⎢⎡⎦⎥⎤⎝ ⎛⎭⎪⎫α-π3+π2=-sin ⎝ ⎛⎭⎪⎫α-π3=-13.答案 A3.已知cos α=23,α是第四象限角,则cos ⎝ ⎛⎭⎪⎫α-π2的值为( )A.23 B.53 C .-23D .-53解析 ∵α是第四象限角,∴sin α=-53. cos ⎝ ⎛⎭⎪⎫α-π2=cos ⎝ ⎛⎭⎪⎫π2-α=sin α=-53. 答案 D4.若f (cos x )=cos2x ,则f (sin150°)的值为( )A.12 B .-12C.32D .-32解析 f (sin150°)=f (sin30°)=f (cos60°)=cos120° =-cos60°=-12.答案 B5.已知tan θ=2,则sin ⎝ ⎛⎭⎪⎫π2+θ-cos π-θsin ⎝ ⎛⎭⎪⎫π2-θ-sin π-θ=( )A .2B .-2C .0 D.23解析 原式=cos θ+cos θcos θ-sin θ=21-tan θ=21-2=-2.答案 B6.已知sin ⎝ ⎛⎭⎪⎫α+π12=13,则cos ⎝ ⎛⎭⎪⎫α+7π12的值为( ) A.13 B .-13C .-223D.223解析 ∵⎝ ⎛⎭⎪⎫α+7π12-⎝ ⎛⎭⎪⎫α+π12=π2. ∴cos ⎝ ⎛⎭⎪⎫α+7π12=cos ⎣⎢⎡⎦⎥⎤⎝ ⎛⎭⎪⎫α+π12+π2=-sin ⎝ ⎛⎭⎪⎫α+π12=-13.答案 B7.若cos(π+α)=-13,则sin ⎝ ⎛⎭⎪⎫3π2-α=________.解析 cos(π+α)=-cos α,∴cos α=13.sin ⎝⎛⎭⎪⎫3π2-α=-cos α,∴sin ⎝ ⎛⎭⎪⎫3π2-α=-13.答案 -138.sin 21°+sin 22°+sin 23°+…+sin 288°+sin 289°+sin 290°=________. 解析 设A =sin 21°+sin 22°+…sin 289°+sin 290°, 则A =cos 289°+cos 288°+…+cos 21°+sin 290° =cos 21°+cos 22°+…cos 289°+sin 290°.∴2A =(sin 21°+cos 21°)+(sin 22°+cos 22°)+…+(sin 289°+cos 289°)+2. ∴2A =89+2=91. ∴A =45.5,∴sin 21°+sin 22°+…+sin 289°+sin 290°=45.5. 答案 45.5能 力 提 升9.cos α2=-1-cos 2⎝⎛⎭⎪⎫π-α2,则α2是第________象限角(设α是第二象限角).解析 由cos α2=-1-cos 2⎝⎛⎭⎪⎫π-α2,得cos α2=-sin 2⎝ ⎛⎭⎪⎫π-α2, 即cos α2=-⎪⎪⎪⎪⎪⎪sin ⎝ ⎛⎭⎪⎫π2-α2=-⎪⎪⎪⎪⎪⎪cos α2,∴cos α2<0,即α2为第二、三象限角.∵α为第二象限角, ∴α2为第一、三象限角. ∴α2为第三象限角. 答案 三10.已知sin(5π-θ)+sin ⎝⎛⎭⎪⎫5π2-θ=72,求sin 3⎝ ⎛⎭⎪⎫π2+θ-cos 3⎝ ⎛⎭⎪⎫3π2-θ的值.解析 ∵sin(5π-θ)+sin ⎝ ⎛⎭⎪⎫5π2-θ=72,∴sin θ+cos θ=72. ∴sin 3⎝⎛⎭⎪⎫π2+θ-cos 3⎝ ⎛⎭⎪⎫3π2-θ=cos 3θ+sin 3θ=(sin θ+cos θ)(sin 2θ-sin θcos θ+cos 2θ)=(sin θ+cos θ)⎣⎢⎡⎦⎥⎤1+1-sin θ+cos θ22=5716.11.若sin θ=33,求cos π-θcos θ⎣⎢⎡⎦⎥⎤sin ⎝ ⎛⎭⎪⎫32π-θ-1+ cos2π-θcos π+θsin ⎝ ⎛⎭⎪⎫π2+θ-sin ⎝ ⎛⎭⎪⎫3π2+θ的值.解析 cos(π-θ)=-cos θ,sin ⎝ ⎛⎭⎪⎫32π-θ=sin ⎣⎢⎡⎦⎥⎤π+⎝ ⎛⎭⎪⎫π2-θ=-sin ⎝ ⎛⎭⎪⎫π2-θ=-cos θ,cos(2π-θ)=cos θ, cos(π+θ)=-cos θ,sin ⎝ ⎛⎭⎪⎫π2+θ=cos θ, sin ⎝⎛⎭⎪⎫3π2+θ=sin ⎣⎢⎡⎦⎥⎤π+⎝ ⎛⎭⎪⎫π2+θ=-sin ⎝ ⎛⎭⎪⎫π2+θ=-cos θ.∴原式=-cos θcos θ-cos θ-1+cos θ-cos θcos θ+cos θ=11+cos θ+11-cos θ=2sin 2θ=2⎝ ⎛⎭⎪⎫332=6. 12.化简:(1)sin ⎝ ⎛⎭⎪⎫32π+αcos ⎝ ⎛⎭⎪⎫π2-αcos 10π+α+sin11π-αcos ⎝ ⎛⎭⎪⎫52π+αsin π+α;(2)cos ⎝⎛⎭⎪⎫3k +13·π+α+cos ⎝ ⎛⎭⎪⎫3k -13·π-α(k ∈Z ).解析 (1)原式=-cos αsin αcos α+sin αcos ⎝ ⎛⎭⎪⎫π2+α-sin α=-sin α+sin α=0.(2)当k =2n ,n ∈Z 时,原式=cos ⎝ ⎛⎭⎪⎫k π+π3+α+cos ⎝ ⎛⎭⎪⎫k π-π3-α=cos ⎝ ⎛⎭⎪⎫2n π+π3+α+cos ⎝ ⎛⎭⎪⎫2n π-π3-α =cos ⎝ ⎛⎭⎪⎫π3+α+cos ⎝ ⎛⎭⎪⎫-π3-α=cos ⎝ ⎛⎭⎪⎫π3+α+cos ⎝ ⎛⎭⎪⎫π3+α =2cos ⎝⎛⎭⎪⎫π3+α;当k =2n +1,n ∈Z 时,原式=cos[(2n +1)π+π3+α]cos ⎣⎢⎡⎦⎥⎤2n +1π-π3-α =cos ⎝ ⎛⎭⎪⎫π+π3+α+cos ⎝ ⎛⎭⎪⎫π-π3-α=-cos ⎝ ⎛⎭⎪⎫π3+α-cos ⎝ ⎛⎭⎪⎫π3+α =-2cos ⎝ ⎛⎭⎪⎫π3+α. 品 味 高 考13.已知sin ⎝ ⎛⎭⎪⎫5π2+α=15,那么cos α=( )A .-25B .-15C.15D.25解析 利用诱导公式化简已知条件即可.sin ⎝ ⎛⎭⎪⎫5π2+α=cos α,故cos α=15,故选C.答案 C。

北师大版高中英语北必修四:双基限时练【8】(含答案)

北师大版高中英语北必修四:双基限时练【8】(含答案)

双基限时练(八)Unit 10CommunicationWorkshop & Culture CornerⅠ.单词拼写1.Electric lights have ________ (取代) candles for many years in our village.答案replaced2.Jim was reading an English ________ (课本) in the classroom.答案textbook3.Thank you again, I am ________ (真正的) grateful for all your help.答案truly4.There isn't enough room on the ________ (磁盘) to download MP3 from the Internet.答案disk5.That woman's babbling (唠叨) is ________ (难以置信的) boring.答案incrediblyⅡ.单句语法填空(不多于3个单词)1.If you just spend your time ________ (weigh) gains and losses, maybe you'll get nothing in the end.答案与解析(in) weighing weigh“称……的重量;考虑;掂量”。

2.I am ________ (true) sorry that things have to end like this.答案与解析truly句意:事情变成这样的结局,我很抱歉。

truly“真正地”,常用于表示说法或感觉。

3.It is very likely that we may get ________ (involve) in the trouble.答案与解析involved主语we,与involve之间有被动关系,故应用表被动的过去分词形式,又由于句意表示“卷入麻烦之中”,故应与介词in搭配。

人教A版数学必修一双基限时练8.docx

人教A版数学必修一双基限时练8.docx

高中数学学习材料马鸣风萧萧*整理制作双基限时练(八)1.函数f (x )=|x +1|的图象为( )解析 f (x )=⎩⎪⎨⎪⎧-x -1,x ≤-1,x +1,x >-1,观察图象可得.答案 A2.映射f :A →B ,在f 作用下A 中元素(x ,y )与B 中元素(x -1,3-y )对应,则与B 中元素(0,1)对应的A 中元素是( )A .(-1,2)B .(0,3)C .(1,2)D .(-1,3)解析 由题意知⎩⎪⎨⎪⎧ x -1=0,3-y =1,解得⎩⎪⎨⎪⎧x =1,y =2,所以与B 中元素(0,1)对应的A 中元素是(1,2).答案 C3.已知f (x )=⎩⎪⎨⎪⎧x -5 (x ≥6),f (x +2) (x <6)(x ∈N ),那么f (3)等于( )A .2B .3C .4D .5解析 f (3)=f (3+2)=f (5)=f (5+2)=f (7)=7-5=2. 答案 A4.已知集合A ={x |0≤x ≤2},B ={y |0≤y ≤4},下列对应关系不能构成从集合A 到集合B 的映射的是( )A .y =2xB .y =32x C .y =x 2D .y =4x -1解析 由映射的定义知,在D 中,当x =2时,y =2×4-1=7,而7∉B ,也就是说集合A 中的元素有的在B 中无对应元素.因此,D 不能构成从A 到B 的映射.答案 D5.函数f (x )=⎩⎪⎨⎪⎧2x 2 (0≤x ≤1),2(1<x <2),3(x ≥2)的值域是( )A .RB .[0,+∞)C .[0,3]D .{y |0≤y ≤2,或y =3}解析 作出分段函数的图象易知. 答案 D6.下列各图表示的对应,构成映射的个数是( )A .3B .4C .5D .6解析 从A 到B 的映射有:①,②,③. 答案 A7.设函数f (x )=⎩⎪⎨⎪⎧x 2+2 (x ≤2),2x (x >2),若f (x 0)=8,则x 0=________.解析 ⎩⎪⎨⎪⎧ x 0≤2,x 20+2=8,或⎩⎪⎨⎪⎧x 0>2,2x 0=8.即⎩⎪⎨⎪⎧ x 0≤2,x 0=±6,或⎩⎪⎨⎪⎧x 0>2,x 0=4. ∴x 0=-6,或x 0=4. 答案 -6或48.已知函数f (x )的图象如下图所示,则f (x )的解析式是________. 解析 由图可知,图象是由两条线段组成,当-1≤x <0时,设f (x )=ax +b ,将(-1,0),(0,1)代入解析式,则⎩⎪⎨⎪⎧-a +b =0,b =1.得⎩⎪⎨⎪⎧a =1,b =1.∴f (x )=x +1;f (x )=-x . 当0≤x ≤1时,设f (x )=kx ,将(1,-1)代入,则k =-1.答案 f (x )=⎩⎪⎨⎪⎧x +1, -1≤x <0,-x , 0≤x ≤19.若定义运算a ⊙b =⎩⎪⎨⎪⎧b ,a ≥b ,a ,a <b ,则函数f (x )=x ⊙(2-x )的值域是________.解析 由题意得f (x )=⎩⎪⎨⎪⎧x ,x <1,2-x ,x ≥1.画函数f (x )的图象,得值域是(-∞,1].答案 (-∞,1]10.设函数f (x )=⎩⎪⎨⎪⎧x +2,x ≤-1,x 2,-1<x <2,2x ,x ≥2,(1)在直角坐标系中画出f (x )的图象; (2)若f (t )=3,求t 的值; (3)求f (x )在(-2,1)上的值域. 解 (1)函数f (x )的图象如下图.(2)当x ≤-1时,f (x )=x +2,∴f (t )=t +2=3. ∴t =1,不符合题意舍去;当-1<x <2时,f (x )=x 2,∴f (t )=t 2=3.∴t =3或t =-3,t =-3不符合题意舍去. 故t =3;当x ≥2时,f (x )=2x ,∴f (t )=2t =3. ∴t =32,不符合题意舍去. ∴t 的值为 3.(3)由(1)中图象知x ∈(-2,1)时,值域为[0,1].11.如图所示,在边长为4的正方形ABCD 边上有一点P ,沿着折线BCDA 由B 点(起点)向A 点(终点)移动.设P 点移动的路程为x ,△ABP 的面积为y =f (x ).(1)求△ABP 的面积与P 移动的路程的函数关系式; (2)作出函数的图象,并根据图象求f (x )的最大值.解 (1)函数的定义域为(0,12).当0<x ≤4时,S =f (x )=12×4×x =2x ;当4<x ≤8时,S =f (x )=12×4×4=8;当8<x <12时,S =f (x )=12×4×(12-x )=24-2x .∴函数解析式为f (x )=⎩⎪⎨⎪⎧2x ,x ∈(0,4],8,x ∈(4,8],24-2x ,x ∈(8,12).(2)图象如图所示.从图象可以看出f (x )max =8.12.某市乘出租车计费规定:2公里以内5元,超过2公里不超过8公里的部分按每公里1.6元计费,超过8公里以后按每公里2.4元计费.(1)写出乘车路程x (公里)与收费y (元)之间的函数关系式; (2)若甲、乙两地相距10公里,则乘出租车从甲地到乙地共需要支付车费多少元?解 (1)由题意,得y =⎩⎪⎨⎪⎧5 (0<x ≤2),5+1.6×(x -2) (2<x ≤8),14.6+2.4×(x -8) (x >8),即y =⎩⎪⎨⎪⎧5 (0<x ≤2),1.8+1.6x (2<x ≤8),2.4x -4.6 (x >8).(2)∵甲、乙两地相距10公里,即x =10>8, ∴应付车费y =2.4×10-4.6=19.4(元).即乘出租车从甲地到乙地共需要支付车费19.4元.。

高二数学 双基限时练8

高二数学 双基限时练8

双基限时练(八)1.函数f (x )=x +2cos x 在[0,π2]上的最大值点为( ) A .x =0 B .x =π6 C .x =π3 D .x =π2解析 令f ′(x )=1-2sin x =0,则sin x =12,又x ∈[0,π2],∴x =π6,又f (0)=2,f (π6)=π6+3,f (π2)=π2,∴f (π6)最大,∴最大值点为x =π6. 答案 B2.函数f (x )=x 3-3ax -a 在(0,1)内有最小值,则a 的取值范围是( )A .0≤a <1B .0<a <1C .-1<a <1D .0<a <12解析 ∵f ′(x )=3x 2-3a =3(x 2-a ), 依题意f ′(x )=0在(0,1)内有解. ∴0<a <1. 答案 B3.函数f (x )=x 3-3x (|x |<1)( ) A .有最大值,但无最小值 B .有最大值,也有最小值 C .无最大值,也无最小值 D .无最大值,但有最小值解析 f ′(x )=3x 2-3=3(x +1)(x -1), 当-1<x <1时,f ′(x )<0.∴f (x )在(-1,1)上是减函数,没有最值. 答案 C4.已知函数f (x )=-x 2-2x +3在区间[a,2]上的最大值为154,则a 等于( )A .-32 B.12C .-12D .-12或-32解析 f (x )=-x 2-2x +3=-(x +1)2+4,易知,f (x )的图象是开口向下的抛物线,对称轴x =-1,而f (-1)=4>154,f (2)=-5<154,∴-1<a <2.由f (a )=-(a +1)2+4=154,解得a =-12,或a =-32(舍去).答案 C5.函数y =x e -x ,x ∈[0,4]的最大值为( ) A .0 B.1e C.4e 4D.2e 2解析 ∵y =x e -x ,∴y ′=e -x +x e -x (-x )′=(1-x )e -x . ∵e -x >0,∴当x ∈(0,1)时,y ′>0;当x ∈(1,4)时,y ′<0.故当x =1时,y 有极大值1e .又当x =0时,y =0;当x =4时,y =4e 4.∴最大值为1e .答案 B6.函数f (x )=sin x +cos x 在x ∈⎣⎢⎡⎦⎥⎤-π2,π2时,函数的最大值、最小值分别是________.解析 f ′(x )=cos x -sin x ,x ∈[-π2,π2],令f ′(x )=0,得x =π4,又f (π4)=2,f (-π2)=-1,f (π2)=1,即最大值为2,最小值为-1.答案2,-17.函数f (x )=12x -x 3在区间[-3,3]上的最小值是________. 解析 f ′(x )=12-3x 2=3(4-x 2), 令f ′(x )=0,得x =±2, 而f (-3)=-36+27=-9, f (-2)=-24+8=-16, f (2)=24-8=16, f (3)=36-27=9. ∴最小值是-16. 答案 -168.设f (x ),g (x )是定义在[a ,b ]上的可导函数,且f ′(x )>g ′(x ),令F (x )=f (x )-g (x ),则F (x )在[a ,b ]上的最大值为________.解析 F ′(x )=f ′(x )-g ′(x )>0, ∴F (x )在[a ,b ]上是增函数. ∴最大值为F (b )=f (b )-g (b ). 答案 f (b )-g (b )9.在平面直角坐标系xOy 中,已知P 是函数f (x )=e x (x >0)的图象上的动点,该图象在点P 处的切线l 交y 轴于点M ,过点P 作l 的垂线交y 轴于点N ,设线段MN 的中点的纵坐标为t ,则t 的最大值是________.解析 如图所示,设点P (x 0,e x 0),则 f ′(x 0)=e x 0(x 0>0).∴f (x )=e x (x >0)在点P 处的切线方程为 y -e x 0=e x 0(x -x 0),令x =0,得 M (0,e x 0-x 0e x 0).过点P 与l 垂直的直线方程为 y -e x 0=-1e x 0(x -x 0),令x =0,得N (0,e x 0+x 0e x 0).∴2t =e x 0-x 0e x 0+e x 0+x 0e x 0=2e x 0-x 0e x 0+x 0e -x 0,则(2t )′=2e x 0-e x 0-x 0e x 0+e -x 0-x 0e -x 0=(1-x 0)(e x 0+e -x 0).∵e x 0+e -x 0>0,∴当1-x 0>0时,即0<x 0<1时,(2t )′>0,∴2t 在(0,1)上单调递增;当1-x 0<0,即x 0>1时,(2t )′<0, ∴2t 在(1,+∞)上单调递减. 故当x 0=1时,2t 有最大值e +1e , 即t 的最大值为12(e +1e ).答案 12(e +1e )10.已知a ∈R ,f (x )=(x 2-4)(x -a ). (1)求f ′(x );(2)若f ′(-1)=0,求f (x )在[-2,2]上的最值;(3)若函数f (x )在(-∞,-2]和[2,+∞)上是递增的,求a 的取值范围.解 (1)由原式得 f (x )=x 3-ax 2-4x +4a , ∴f ′(x )=3x 2-2ax -4. (2)由f ′(-1)=0,得a =12, 此时f (x )=(x 2-4)(x -12), f ′(x )=3x 2-x -4.由f ′(x )=0,得x =43,或x =-1. 又f (43)=-5027,f (-1)=92, f (-2)=0,f (2)=0,∴f (x )在[-2,2]上的最大值为92,最小值为-5027.(3)f ′(x )=3x 2-2ax -4的图象是开口向上的抛物线,且过定点(0,-4).由条件得f ′(-2)≥0,f ′(2)≥0,即⎩⎪⎨⎪⎧4a +8≥0,8-4a ≥0,∴-2≤a ≤2. 故a 的取值范围是[-2,2].11.已知a 是实数,函数f (x )=x 2(x -a ).(1)若f ′(1)=3,求a 的值及曲线y =f (x )在点(1,f (1))处的切线方程;(2)求f (x )在区间[0,2]上的最大值. 解 (1)f ′(x )=3x 2-2ax , ∵f ′(1)=3-2a =3,∴a =0. 又当a =0时,f (1)=1,f ′(1)=3,∴曲线y =f (x )在(1,f (1))处的切线方程为3x -y -2=0. (2)令f ′(x )=0,解得x 1=0,x 2=2a3.当2a3≤0,即a ≤0时,f (x )在[0,2]上单调递增,从而f (x )max =f (2)=8-4a .当2a3≥2,即a ≥3时,f (x )在[0,2]上单调递减,从而f (x )max =f (0)=0.当0<2a 3<2,即0<a <3时,f (x )在[0,2a 3]上单调递减,在[2a3,2]上单调递增,从而f (x )max=⎩⎪⎨⎪⎧8-4a ,0<a ≤2,0,2<a <3,)综上所述,f (x )max=⎩⎪⎨⎪⎧8-4a ,a ≤2,0,a >2.)12.已知函数f (x )=x 3-ax 2-3x .(1)若f (x )在区间[1,+∞)上是增函数,求实数a 的取值范围; (2)若x =-13时是f (x )的极值点,求f (x )在[1,a ]上的最大值. 解 (1)∵f ′(x )=3x 2-2ax -3≥0在[1,+∞)上恒成立, ∴⎩⎪⎨⎪⎧--2a 2×3≤1,f ′(1)=3-2a -3≥0,即⎩⎪⎨⎪⎧a ≤3,a ≤0⇒a ≤0, ∴a 的取值范围是(-∞,0]. (2)若x =-13是f (x )的极值点, 则f ′(-13)=3(-13)2+23a -3=0, ∴a =4,∴f (x )=x 3-4x 2-3x . f ′(x )=3x 2-8x -3 =(3x +1)(x -3).令f ′(x )=0得,x 1=-13,或x 2=3. f ′(x ),f (x )随x 变化的情况如下表:新课标第一网系列资料 。

人教B版高中数学必修二双基限时练8.docx

人教B版高中数学必修二双基限时练8.docx

高中数学学习材料马鸣风萧萧*整理制作双基限时练(八)基 础 强 化1.一个铜质的五棱柱的底面积为16 cm 2,高为4 cm ,现将它熔化后铸成一个正方体铜块(不计损耗),那么铸成的铜块的棱长为( )A .2 cm B.43 cm C .4 cmD .8 cm解析 设正方体的棱长为a ,则16×4=a 3, ∴a =4.∴正方体的棱长为4 cm. 答案 C2.一个圆台的轴截面等腰梯形的腰长为a ,下底长为2a ,对角线长为3a ,则这个圆台的体积是( )A.733πa 3 B.7123πa 3 C.783πa 3D.7243πa 3解析 如图,由AD =a ,AB =2a ,BD =3a ,知∠ADB =90°.取DC 中点E ,AB 中点F ,分别过D 点、C 点作DH ⊥AB ,CG ⊥AB ,知DH =32a .∴HB =3a 2-34a 2=32a .∴DE =HF =12a .∴V 圆台=π3⎝ ⎛⎭⎪⎫14a 2+12a 2+a 2·32a =7243πa 3.答案 D3.一空间几何体的三视图如图所示,则该几何体的体积为( )A .2π+2 3B .4π+2 3C .2π+233D .4π+233解析 该空间几何体为一圆柱和一四棱锥组成的,圆柱的底面半径为1,高为2,体积为2π,四棱锥的底面边长为2,高为3,所以体积为13×(2)2×3=233,所以该几何体的体积为2π+233.答案 C4.如图所示,在上、下底面对应边的比为1:2的三棱台中,过上底面一边作一个平行于对棱的平面A 1B 1EF ,这个平面分三棱台成两部分的体积之比为( )A .1:2B .2:3C .3:4D .4:5解析 设棱台上底面面积为S ,由上、下底面边的比为1:2,可知下底面面积为4S .设棱台的高为h ,则V 台=13h (S +S ·4S +4S )=73Sh , ∵棱柱A 1B 1C 1-FEC 的体积为V =S ·h , ∴V 柱V 台-V 柱=Sh 73Sh -Sh=34.答案 C5.如图,是一个几何体的主视图、侧视图、俯视图,且正视图、侧视图都是矩形,则该几何体的体积是( )A .24B .12C .8D .4解析 该几何体是一个长方体挖去了一个三棱柱后剩下的几何体,则其体积为2×3×4-12×(2×3)×4=12.答案 B6.一个几何体的三视图如图所示,则这个几何体的体积等于( )A .4B .6C .8D .12解析 由三视图想象并画出直观图后计算.根据三视图画出直观图如图所示,此几何体是一个四棱锥V -ABCD ,VD ⊥底面ABCD ,底面为一个直角梯形.V V -ABCD =13S ABCD ·VD =13×(4+2)×22×2=4. 答案 A7.已知某几何体的俯视图是如图所示的矩形,主视图是一个底边长为8、高为4的等腰三角形,左视图是一个底边长为6、高为4的等腰三角形,则该几何体的体积为________.解析 该几何体是一个四棱锥,底面是矩形,面积为48,高为4,所以它的体积V =13×48×4=64.答案 648.如图,一个圆柱和一个圆锥的底面直径和它们的高都与一个球的直径相等,这时圆柱、圆锥、球的体积之比为________.解析 设球的半径为R ,∴V 柱=πR 2·2R =2πR 3,V 锥=13πR 2·2R =2πR 33,V 球=43πR 3,∴V 柱:V 锥:V 球=2πR 3:2πR 23:4πR 33=3:1:2.答案 3:1:2能 力 提 升9.在四面体ABCD 中,以A 为顶点的三条棱两两相互垂直,且其长分别为1,6,3,四面体的四个顶点在同一个球面上,则这个球的体积为________.答案 32π310.下列三个图中,左边是一个正方体截去一个角后所得多面体的直观图.右边两个是主视图和左视图.(1)请按照画三视图的要求画出该多面体的俯视图(不要求叙述作图过程);(2)求该多面体的体积(尺寸如图). 解 (1)俯视图如图所示,(2)V =23-13×12×2×2×1=8-23=223.11.正方形ABCD 的边长为1,分别取边BC ,CD 的中点E ,F ,连接AE ,EF ,AF ,以AE ,EF ,F A 为折痕,折叠这个正方形,使B ,C ,D 重合于一点P ,得到一个三棱锥如图所示,求此三棱锥的体积.解 ∵∠B =∠C =∠D =90°, ∴翻折后∠APE =∠EPF =∠APF =90°. ∴Rt △PEF 可以看作是三棱锥的底面, 而AP 可以看作是三棱锥的高.比较发现:AP =1,PE ⊥PF ,PE =PF =12, ∴V A -PEF =13S △PEF ·AP =13×12×12×12×1=124. 12.如图所示正三棱台ABC-A1B1C1中,AB:A1B1=1:2,求三棱锥A1-ABC,B-A1B1C,C-A1B1C1的体积之比.解13.某四棱台的三视图如图所示,则该四棱台的体积是()A .4 B.143 C.163 D .6解析 由四棱台的三视图可知,台体上底面S 1=1×1=1,下底面积S 2=2×2=4,高h =2,代入台体的体积公式V =13(S 1+S 1S 2+S 2)h =13×(1+1×4+4)×2=143.答案 B。

人教A版选修2-3双基限时练8

人教A版选修2-3双基限时练8
D.展开式中第6项的系数最小
解析由二项式系数的性质知,C +C +C +…+C =210=1024.
∴A正确.
又二项式系数最大的项为C ,是展开式的第6项.
∴B正确.
又由通项Tr+1=C a10-r(-b)r=(-1)rC a10-rbr知,第6项的系数-C 最小.
∴D正确.
答案C
5.若n∈N*,( +1)n= an+bn(an,bn∈Z),则bn的值()
答案-8
9.若(x2+ )n展开式的各项系数之和为32,则n=________,其展开式中的常数项为________.(用数字作答)
解析依题意得2n=32,∴n=5,
∵Tr+1=C (x2)5-r·( )r=C x10-5r.
令10-5r=0,得r=2,∴常数项为T3=C =10.
答案510
10.已知 6(k是正整数)的展开式中,常数项小于120,则k=________.
A.一定是奇数B.一定是偶数
C.与n的奇偶性相反D.与n的奇偶性相同
解析取n=1,n=2,验证知A正确.
答案A
6.设m为正整数,(x+y)2m展开式的二项式系数的最大值为a,(x+y)2m+1展开式的二项式系数的最大值为b,若13a=7b,则m=()
A.5B.6
C.7D.8
解析(x+y)2m的展开式共有2m+1项,其中最大的二项式系数为C ,(x+y)2m+1的展开式中共2m+2项,其中二项式系数最大的项为中间两项C =C ,依题意得a=C ,b=C ,由13a=7b,得13C =7C =7(C +C ),∴6C =7C ,即6C =7 ·C ,解得m=6.
答案B
7.如图是一个类似杨辉三角的递推式,则第n行的首尾两个数均为________.

双基限时练8.pptx

双基限时练8.pptx
首 项 为 a1 , 公 差 为 d , 则 根 据 题 意 ,得
a1+4d=10, a1+14d=25.
解得 a1=4,d=32. ∴an=4+32(n-1)=23n+25. ∴a25=23×25+25=40. 11.(1)求等差数列 3,7,11,…的第 4 项与第 10 项. (2)100 是不是等差数列 2,9,16,…的项?如果是,是第几项?如 果不是,说明理由. 解 (1)由 a1=3,d=7-3=4, n=4, 得 a4=3+(4-1)×4=15; n =10 时,得 a10=3+(10-1)×4=39.
学海无 涯
米? 解 设从 2007 年年底开始,n 年后该市每年新建的住房面积为
an 万平方米.
由题意,得{an}是等差数列,首项 a1=400,公差 d=50.
所以 an=a1+(n-1)d=350+50n.

350+50n>820,解得
n>
47 5.
由于 n∈N*,则 n≥10.
所以从 2017 年年底开始,该市每年新建住房的面积开始大于 820
万平方米.
12
4
4
答案 A 4.等差数列{an}的前三项依次为 x,2x+1,4x+2,则它的第 5 项
学海无 涯
为() A.5x+5 C.5
B.2x+1 D.4
解析 由等差中项,得 2(2x+1)=x+4x+2
∴x=0,∴a1=0,a2=1,a3=2,a4=3,a5=4.
答案 D
5.若{an}为等差数列,ap=q,aq=p(p≠q),则 ap+q 为( ) A
A.286
B.287
C.288
D.289
答案 C 3.已知等差数列{an}中,a7+a9=16,a4=1,则 a12 的值是( )

中考英语双基练习 8&9

中考英语双基练习 8&9

中考语法双基检测8选择题:1. Please take the seats over there. These seats are for _________ old.A. theB. anC. aD. /2. Alex, a cousin of __________, will call on _________ soon.A. me, usB. me, mineC. her, hersD. hers, her3. The holiday will last _________ the beginning of July to the end of August.A. onB. fromC. atD. for4.Do you know if _________ spectators went to Germany to watch the World Cup?A. thousandB. two thousandsC. thousand ofD. thousands of5.May I keep the book __________ longer? I haven’t finished it.A. a fewB. fewC. a littleD. little6. Yesterday I met _______.A. the friend of my fatherB. a friend of my father’sC. my father’s one friendD. a friend of my father7. John did ________ in the exam, Tom did even ______.A. bad… worstB. bad…worseC. badly…worstD. badly…worse8. Danny is fat because he eats _____ unhealthy food.A. too muchB. too fewC. too manyD. too little9. The mountain climbers succeeded _______ the way and got to the top finally.A. in findB. to findC. in findingD. finding10. ---I hear there will be a concert of Jay Chou._______ will it start? ---In two days.A. How oftenB. How longC. How soonD. How far11 Could you show me another bag? I like _____ the colour _____ the size of this one.A. both, andB. neither, norC. either, orD. not only, but also12. Take your glasses with you, _____ you won’t enjoy the play.A. soB. orC. andD. for13. More laws ____ to fight pollution in the near future.A. will makeB. have madeC. will be madeD. make14. ----It hardly snows in Guangzhou, ________it? ----______, it doesn’t.A. doesn’t , YesB. does, YesC. doesn’t, NoD. does, No15. The comic strip is ________ funny ______ readers are very interested in it.A. too…toB. so…thatC. very…toD. too…that16. I met Emma yesterday. We _____ each other for more than twenty years.A. don’t seeB. didn’t seeC. haven’t seenD. hadn’t seen17. The dish must have gone bad. It smells ________.A. terriblyB. terribleC. wellD. good18. If Mary _____ shopping this afternoon, please ask her to write a shopping list first.A. will goB. goesC. wentD. has gone19. More and more people are _______ computer games.A. interested to playingB. interested in playingC. keen on to playD. like playing20. The boy was very _______ his mother who won the gold medal at the Olympic Games.A. satisfied withB. afraid ofC. proud ofD. surprised at21. Our school doesn’t allow students to take mobile phones. But still _____ students have one.A. littleB. a littleC. fewD. a few22. The heavy snow made the mountain climbers ______ halfway.A. stopB. stoppedC. stoppingD. to stop23. I’ve heard that _______ will happen at the School Art Festival.A. interesting somethingB. something interestingC. interesting anythingD. anything interesting24. I watched the football match last night. Yes. Liverpool didn’t play ______ they did last week.A. well thanB. more thanC. as well asD. as worst as25. You’d better ______ so much fast food. It is harmful to your health.A. not to eatB. eatC. not eatD. don’t eat26. ---I'm sorry I'm late, because the traffic is bad.---______ But tomorrow, please be on time.A. It's a pleasure.B. You're welcome.C. It doesn't matter.D. That's right.27. Can you tell me ______?A. how much is the coatB. how much does the coat costC. this coat is how muchD. how much this coat costs28. —I passed the important test last week!—_________.A. That’s very kind of youB. Have a good timeC. CongratulationsD. What a shame29. Tom missed the bus, so he failed to get to school on time.A. wasn’t able toB. can’tC. missedD. unable30. They prefers beef to pork.A. like…better thanB. likeC. like to eatD. look likeII. 词性转换:1. The Beijing 2008 Olympic Games opening Ceremony was a big ___________. (succeed)2. The ___________ girl from the left is from Australia. (two)3. The film I saw yesterday had a _____________ ending. (drama)4. The old have worse ____________ than the young.(memorize)5. Our new class teacher is a kind and handsome man of average ____________ .(high)6. At Christmas people usually eat ___________ turkey at dinner. (tradition)7. The exercises of Chapter 2 are much ____________ than those of Chapter 5. (easy)8. The president was deeply moved by the brave girl and went to see her __________. (person) III. 句型转换:1. Jack has to walk to school in the morning. (改为否定句)Jack ______ _____ to walk to school in the morning.2. I haven’t worked out the physics problem. (改为被动语态)The physics problem _____ ______ worked out .3. Mr and Mrs Brown have already been to China twice. (对划线部分提问)_______ _______ times have Mr and Mrs Brown been to China?4. Will Bob join the swimming club? Please tell me. (合并为一句)Please tell me _______ Bob ________ join the swimming club.5. Yao Ming is an excellent Chinese basketball player. (改为感叹句)_______ _____ excellent Chinese basketball player Yao Ming is!6. Junior 3 students began to learn chemistry a year ago. (保持句意基本不变)Junior 3 students _____ learned chemistry ______ a year.中考语法综合训练9I. Choose the right answer:1.___ Johnsons adopted (收养) ___ homeless boy out of charity ten years ago.A, /, a B, /, an C, The, a D, The, an2.The little boy often gives ___ to the headmaster on your people’s problems.A, many advice B, some advice C, two advice D, two piece of advice3.There are ______ fish sold at the supermarket.A, a great deal B, a great number of C, a large amount of D, a lots of4.Peter looked ___ when he learned that he hadn’t passed the final exam.A, sadly B, sad C, happily D, happy5.Jack did badly at the school sports meeting. I did even ___.A, worse B, worst C, more bad D, more badly6.Eddie, my best frind ______ Ben.A, is as high as B, works as careful as C, doesn’t sing as beautifully as D, writes more better than 7.Our hobby group has a rule: all the members must speak and write ___ English.A, in B, into C, through D, at8.A: Is there ___ in today’s newspaper?B: Yes. It says that President Hu Jintao has reached Washington.A, anything important B, any important thingC, important anything D, any thing important9.We ___ let out dirty smoke to pollute the air. It’s harmful to our health.A, mustn’t B, needn’t C, may D, can10.I think these foreign visitors ___ Shanghai for more than two weeks.A, have been in B, have come to C, arrived in D, got to11.Since you are badly ill, you’d better ___.A, give up smoking B, to stop smoking C, not to smoke D, not smoking12.The medicine ___ in cool and dry place.A, must kept B, must keep C, must be kept D, must be keeping13.You ___ choose the colour you like. You ___ follow me.A, need, needn’t B, may, needn’t C, must, should D, can’t, can’t14.My friends ___ some foreigners around the ancient church this time yesterday.A, are showing B, were showing C, showed D, had shown15. By the end of last year, they ___ five tunnels under the Huangpu River.A, built B, have built C, had built D, were built16.Drivers in big cities are asked ___ their cars on “World Car Free Day”.A, not use B, don’t use C, not to use D, no use17.The students of Class One got up early this morning ___ be late for the picnic.A, so as to B, in order that C, so as to not D, in order not to18.My wish ___ a successful basketball player like Yao Ming.A, to become B, is to become C, is become D, becomes19.The video cameras are so nice. I really don’t know ___.A, how to choose B, what to choose C, to choose which D, which to choose20.We don’t do morning exercise if it ___ tomorrow.A, rains B, will rain C, rained D, rain21.They didn’t start the work ___ their teacher came b ack.A, until B, while C, as soon as D, if22. The students from a school in Dalian get on a bus and ask people to help keep the city _______.A) cleanly B)clean C) clear D) clearly23. Now that you have two tickets, you may go with a friend of ______.A) you B) your C) yours D) yourself24. Grandmother ______ wash clothes by hand but now she ______ doing it with the machine.A) used to; used to B) is used to; is used toC) is used to; used to D) used to; is used to25.She didn’t go to see the dolphin show (海豚表演) with us because she ____ it already.A, sees B, saw C, has seen D, had seen26.In old China, many palaces ____ in Beijing, Xi’an and Nanjing.A, built B, have built C, were built D, were building27. --- What did your son say in the letter?--- He told me that he ____ the Disney World the next day.A, will visit B, has visited C, is going to visit D, would visit28. I’d rather use my old mobile phone ________ this new one.A) instead B) instead of C) take place D) take the place of29. --- ______ yourself to some tea after the hard work.--- Thank you very much.A) Enjoy B) Drink C) Cook D) Help30. --- What good weather! Let’s go boating on the l ake.--- __________.A) That’s all right B) Thank you very much C) It doesn’t matter D) That’s a good ideaII. Complete the sentences with the given words in their proper forms1.Jane’s cousin has made several films. He is a wonderful ______________. (act)2.Mr Jackson has received an ______________ to his friend’s birthday party. (invite)3.Although grandma is over 80, she still has a good ______________. (memorize)4.Our holidays in Thailand were really ______________ and unforgettable. (enjoy)5.I’m sure that my uncle will ______________ in finding out a best solution of it. (success)6.Lily is __________ enough to get the prize in the 2006 Shanghai Junior High School English Contest. (luck)7.The new manager is much ___________ than any of them. (young)8.They all looked ________ at the teacher when he told them the good news. ( happy)III. Rewrite the following sentences as required1..A ten-year-old boy hit Mr. Wh ite’s window with a slingshot. (改为一般疑问句)________ a ten-year-old boy _________ Mr. White’s window with a slingshot?2..Cars can run at a speed of 80 kilometres per hour on China’s first cross-sea bridge. (对划线部分提问)________ ________ can cars run on China’s first cross-sea bridge?3. Would you like hamburgers at McDonald’s? Mr. Li asked his son.(改为宾语从句)He asked his son ________ ________ would like hamburgers at McDonald’s.4. They held the APEC 2005 from Nov. 12-19 in Busan, South Korea. (改为被动语态)The APEC 2005 ________ ________ from Nov. 12-19 in Busan, South Korea.5.Jack hardly knows anything about the car accident. (改为反意疑问句)Jack hardly knows anything about the car accident, __________ ___________?6.Don’t throw waste paper on the ground, please. (保持原句意思)You ___________ ___________ to throw waste paper on the ground.。

高中语文散文金黄的稻束地之子半棵树边界望乡(略读)双基限时练《中国古代诗歌散文欣赏》

高中语文散文金黄的稻束地之子半棵树边界望乡(略读)双基限时练《中国古代诗歌散文欣赏》

双基限时练(八) 金黄的稻束地之子半棵树边界望乡(略读)一、基础巩固1.下列加点字的读音,全部正确的一项是( )A.皱.纹(zhòu)肩荷.(hè)襁.褓(qiǎnɡ) 卓.著(zhuō)B.惊蛰.(zhé) 登载.(zài)咋.舌(zé) 粗犷.(ɡuǎnɡ)C.滋.润(zī) 愠.色(yùn)宽宥.(yòu) 质.量(zhì)D.鹧.鸪(zhè) 款识.(shí)油渍.(zì) 帝祚.(zuò)解析A.卓—zhuó。

B.载—zǎi。

D.识—zhì。

答案 C2.依次填入下列各句中横线处的词,最恰当的一组是( )①4月21日,前国际奥委会主席萨马兰奇先生因病去世,国家体育总局副局长刘鹏率领全局工作人员向萨翁________。

②小说、戏剧、影视,当今各种文艺门类似乎都特别青睐孔子,这充分________了传统文化价值的回归和人们对历史的尊重与信仰。

③新近竣工的上海世博会中国馆,如一枚硕大的朱红色篆字中国印,沉静地矗立在黄浦江畔,________出中国元素、中国气派、中国特色。

A.志哀昭示凸显B.致哀昭示凸现C.志哀展示凸现D.致哀展示凸显解析志哀:用某种方式表示哀悼。

致哀:对死者表示哀悼。

昭示:明白地表示或宣布。

展示:清楚地摆出来;明显地表现出来。

凸显:清楚地显露。

凸现:清楚地显现。

答案 D3.下列各句中,加点的成语使用恰当的一句是( )A.求学期间,他春风得意,事事顺心,没料到踏入社会后,几桩生意下来,就被骗得血本无归,于是他总是感叹遇人不淑....,命途多舛。

B.湛蓝色的密歇根湖,一望无垠,浩如烟海....。

据说它的面积有两个半台湾那么大。

C.春天来了,草长莺飞....的呼伦贝尔大草原更具迷人的魅力。

每年吸引来自国内和世界各地的游客和摄影家超过500万人次。

2019年高考物理双基突破:专题08-恒定电流中基本概念(精练)

2019年高考物理双基突破:专题08-恒定电流中基本概念(精练)

1.如图所示为一磁流体发电机示意图, A 、B 是平行正对的金属板,等离子体(电离的气体,由自由电子和阳离子构成,整体呈电中性)从左侧进入,在t 时间内有n 个自由电子落在B 板上,则关于R 中的电流大小及方向判断正确的是A .I =ne t ,从上向下B .I =2net ,从上向下C .I =net ,从下向上D .I =2net,从下向上【答案】A2.一个用半导体材料制成的电阻器D ,其电流I 随它两端电压U 变化的关系图象如图甲所示,若将它与两个标准电阻R 1、R 2并联后接在电压恒为U 的电源两端,3个用电器消耗的电功率均为P .现将它们连接成如图乙所示的电路,接在该电源的两端,设电阻器D 和电阻R 1、R 2消耗的电功率分别为P D 、P 1、P 2,它们之间的关系为A .P 1=4P DB .P D =P4C .PD =P 2 D .P 1<4P 2【答案】D【解析】由于电阻器D 与两个标准电阻R 1、R 2并联后接在电压恒为U 的电源两端时,三者功率相同,则此时三者电阻相同。

当三者按照题图乙所示电路连接时,电阻器D 两端的电压小于U ,由题图甲图象可知,电阻器D 的电阻增大,则有R D >R 1=R 2,而R D 与R 2并联,电压相等,根据P =U 2R,P D <P 2,C 错误;由欧姆定律可知,电流I D <I 2,又I 1=I 2+I D ,根据P =I 2R ,P 1>4P D ,P 1<4P 2,A 错误,D 正确;由于电阻器D 与电阻R 2的并联电阻R <R 1,所以D 两端的电压小于U 2,且D 阻值变大,则P D <P4,B 错误。

6.(多选)半径为R 的橡胶圆环均匀带正电,总电荷量为Q ,现使圆环绕中心轴线以角速度ω匀速转动,则由环产生的等效电流应有A .若ω不变而使电荷量Q 变为原来的2倍,则电流也将变为原来的2倍B .若电荷量不变而使ω变为原来的2倍,则电流也将变为原来的2倍C .若使ω、Q 不变,将橡胶环拉伸,使环半径增大,电流将变大D .若使ω、Q 不变,将橡胶环拉伸,使环半径增大,电流将变小 【答案】AB7.电池甲和乙的电动势分别为E 1和E 2,内电阻分别为r 1和r 2,已知E 1<E 2,若用甲、乙电池分别向某个电阻R 供电,则在这个电阻上所消耗的电功率相同。

2021年高中英语 双基限时练8 新人教版必修1

2021年高中英语 双基限时练8 新人教版必修1

Ⅰ.单词拼写1.She ________(煮)the eggs soft.答案boiled2.What's the weather ________(预报)today?答案forecast3.A ________(包裹) of clothes were put into the car.答案parcel4.He got a position with an ________(保险) pany.答案insuranceⅡ.用所给动词的适当形式填空1.________(boil) some water for us, please.答案Boil2.Water ________(boil) at 100 degrees centigrade.答案boils3.The pot is ________(boil) away on the fire.答案boiling4.She ________(boil) with anger at his word.答案boiled5.Hurricane has been ________(forecast) for tomorrow afternoon.答案forecast/forecasted6.Spring is ing. It ________(get) warmer and warmer.答案is getting7.My aunt ________(e) to see us. She'll be here soon.答案is ing8.You are always ________(change) your mind.答案changing9.According to the timetable, the train for Athens ________ (leave) at seven o'clock in the evening.答案leaves10.I'd hold on to that house for the time being; house prices ________ (rise) sharply at the moment.答案is rising11.— When does the train start?— It ________ (start) in 15 minutes according to the timetable.答案starts12.— Are you still busy?— Yes, I ________ (finish) my work and it won't take long.答案am finishing13.— When will they leave?— They ________ (leave) very soon.答案are leaving14.— What would you do if it ________ (rain) tomorrow?— We have to carry it on, since we've got everything ready.答案rains15.— You've left the light on.— Oh, I ________ (go) and turn it off.答案will goⅢ.语法填空Woman: Good morning, sir. Can I help you?Man: I wonder __1__ you could give me some information about active holidays.Woman: Active holidays, sir? Can you tell me exactly __2__ you mean, please?Man: Well, you see, __3__ I go on holidays, I like to get plenty of exercise. I don't like __4__ (sit) around and doing nothing. What I mean is that I'm the sort of man __5__ enjoys swimming, water skiing, thosesorts of things.Woman: Yes, sir, very __6__ (interest). Um ... active holidays, let me see. Oh, yes, what about diving, sir? We can offer you two weeks off the coast of Cornwall: one week diving and one week fishing with the local fishermen.Man: Fishing? Is there any chance of __7__ (get) in a bit of sailing?Woman: I'm sorry we don't do many sailing holidays, sir. Mostly they __8__ (organize) by the sailing school. But rowing, yes. Are you interested in rowing, sir?Man: Well, I __9__ (do) a lot of rowing when I was at university.Woman: Why don't you look through these brochures, sir, and see if there's anything __10__ interests you?Man: OK. Thanks.答案 1.if 2.what 3.when 4.sitting 5.who 6.interesting 7.getting 8.are organized 9.did 10.thatⅣ.阅读理解AI have a rule for travel: Never carry a map. I prefer to ask fordirections. Foreign visitors are often puzzled in Japan because most streets there don't have names. In Japan, people use landmarks in their directions instead of street names. For example, the Japanese will say to travelers, “Go straight down to the corner. Turn left at the big hotel and go past a fruit market. The post office is across from the bus stop.”In the countryside of the American Midwest, usually there are not many landmarks. There are no mountains, so the land is very flat (平坦的). In many places there are no towns or buildings within miles. Instead of landmarks, people will tell you directions and distance. In Kansas or Iowa, for example, people will say, “Go north two miles. Turn east, and then go another mile.”People in Los Angeles, California, have no idea of distance on the map: They measure distance by means of time, not miles. “How f ar away is the post office?” you ask. “Oh,” they answer, “It's about five minutes from here.” You say, “Yes, but how many miles away is it?” They don't know.People in Greece sometimes do not even try to give directions because visitors seldom understand the Greek language. Instead of giving you the direction, a Greek will often say, “Follow me.” Then he'll lead youthrough the streets of the city to the post office.Sometimes a person doesn't know the answer to your question. What happens in this situatio n? A New Yorker might say, “Sorry, I have no idea.” But in Yucatan, Mexico, no one answers, “I don't know.”People in Yucatan think that “I don't know” is impolite. They usually give an answer, often a wrong one. A visitor can get very, very lost in Yucatan!One thing will help you everywhere —in Japan, in the United States, in Greece, in Mexico, or in any other place. You might not understand a person's words, but maybe you can understand his body language. He or she will usually turn and then point in the correct direction. Go in that direction, and you may find the post office!1.The passage mainly tells us that ________.A. never carry a map for travelB. there are not many landmarks in the American MidwestC. there are different ways to give directions in different parts of the worldD. New Yorkers often say, “I have no idea,” but people in Yucatan, Mexico, never say this答案与解析 C 文章列举了日本、美国中西部、洛杉矶、希腊、纽约等地问路的回答方式,可知C项最符合文意。

人教A版选修一双基限时练8.docx

人教A版选修一双基限时练8.docx

双基限时练(八)1.设C={复数}、A={实数}、B={纯虚数},全集U=C,那么下列结论正确的是( )A.A∪B=C B.∁U A=BC.A∩∁U B=∅D.B∪∁U B=C答案 D2.已知复数z=a+b i(a,b∈R),则z∈R的充要条件是( ) A.a+b=a-b i B.a+b i=-a+b iC.ab=0 D.a=b=0答案 A3.若(x2-x)+(x-1)i是纯虚数,则实数x的值为( )A.1或0 B.1C.0 D.以上都不对答案 C4.如果(x+y)i=x-1,那么实数x,y的值为( )A.x=1,y=-1 B.x=0,y=-1C.x=1,y=0 D.x=0,y=0答案 A5.(3-1)i的实部是( )A. 3 B.1C.-1 D.0答案 D6.若x,y∈R,且z=x+y i是虚数,则有( )A.x=0,y∈R B.x≠0,y∈RC.x∈R,y=0 D.x∈R,y≠0答案 D7.已知复数z =m 2-3m +(m 2-5m +6)i(m ∈R ),若z <0,则m =________.解析 ∵m ∈R ,且z <0,∴z ∈R ,∴⎩⎪⎨⎪⎧ m 2-5m +6=0,m 2-3m <0,解得m =2.答案 28.复数4-3a -a 2i 与复数a 2+4a i 相等,则实数a 的值为__________.解析 由4-3a -a 2i =a 2+4a i ,得⎩⎪⎨⎪⎧ 4-3a =a 2,-a 2=4a ,解得a =-4.答案 -49.已知实数a ,x ,y 满足a 2+2a +2xy +(a +x -y )i =0,则点(x ,y )的轨迹方程是________.解析 由复数相等的充要条件知,⎩⎪⎨⎪⎧ a 2+2a +2xy =0,a +x -y =0,消去a ,得x 2+y 2-2x +2y =0,即(x -1)+(y +1)2=2.答案 (x -1)2+(y +1)2=210.写出下列复数的实部与虚部,并指出哪些是实数,哪些是虚数,哪些是纯虚数.-3,0,2+3i ,i ,-4i ,sin π2+isin π4.解 -3,0,2+3i ,i ,-4i ,sin π2+isin π4的实部分别为-3,0,2,0,0,1;虚部分别是0,0,3,1,-4,22. -3,0是实数;2+3i ,i ,-4i ,sin π2+isin π4是虚数;其中i ,-4i 是纯虚数.11.若log 2(m 2-3m -3)+ilog 2(m -2)为纯虚数,求实数m 的值.解 由题意得⎩⎪⎨⎪⎧ log 2(m 2-3m -3)=0,log 2(m -2)≠0,∴⎩⎪⎨⎪⎧ m 2-3m -3=1,m -2>0,m -2≠1.解得m =4.12.已知关于x 的方程x 2+(1-2i)x +(3m -i)=0有实根,求实数m 的值.解 设x =a 为方程的一个实数根.则有a 2+(1-2i)a +(3m -i)=0即(a 2+a +3m )-(2a +1)i =0∵a ,m ∈R ,由复数相等的充要条件有⎩⎪⎨⎪⎧ a 2+a +3m =0,2a +1=0,解得⎩⎪⎨⎪⎧ m =112,a =-12.故实数m 的值为112.。

人教版高中语文必修三:《林黛玉进贾府》双基限时练及答案

人教版高中语文必修三:《林黛玉进贾府》双基限时练及答案

人教版高中语文必修三:《林黛玉进贾府》双基限时练及答案D.我带了外甥女过去,倒也便宜..——此中的“便宜”是方便的意思,而“便利此月内”中的“便利”也含有方便的意思。

解析C项中的“宝”应是印章,这些字都是皇帝印玺上的字。

答案 C4.下列各句中加点的词语使用恰当的一项是( )A.新中式婚房回归古典东方之美,一对新人说:“我们不追求家具的雕梁画栋....,无需过多装饰,只求简单,古色古香。

”B.肃穆的气氛,庄重的举动。

在场的所有人都低下了头,敛声屏气....,用一分钟的默哀,向逝世的音乐之王迈克尔·杰克逊送上最真诚的缅怀,他是全球以个人名义捐助善款最多的人。

C.南京——一个靡丽而怀旧的城市。

如果说它有过繁华,那么秦淮河边的洪武路会告诉你多少纨袴膏粱....的一掷千金、纸醉金迷,多少士大夫的理想,随着末世国都一点点丧尽。

D.众嬷嬷引着,便往东转弯,穿过一个东西的穿堂,向南大厅之后,仪门内大院落,上面五间大正房,鳞次栉比....。

解析A项,雕梁画栋:指房屋华丽的彩绘装饰,常用来形容建筑物富丽堂皇。

句中用于家具,使用对象错误。

B项,敛声屏气:指不说话,暂抑呼吸。

形容小心害怕的样子。

不合语境,语境表达的意思是安静、沉默。

C项,纨袴膏粱:指富贵人家的子弟。

适合语境。

D项,鳞次栉比:像鱼鳞和梳子的齿一样,一个挨着一个地排列着,多用来形容房屋等密集。

不合语境。

答案 C5.人物的语言最能体现人物的心理和性格,下面是对有关王熙凤的语言描写的赏析,选出赏析不当的一项( )A.“我来迟了,不曾迎接远客”——这是对王熙凤的出场描写,她的出场“未见其人,先闻其声”,这既表现了她性格的泼辣,也说明了她在贾府中的地位。

B.“天下真有这样标致的人物……竟不像老祖宗的外孙女儿,竟是个嫡亲的孙女”——这句话内涵丰富,可谓一箭三雕,既夸奖了黛玉,又恭维了贾母,还奉承了贾氏三姊妹,足见其圆滑。

C.“在这里不要想家,想要什么吃的、什么玩的,只管告诉我;丫头老婆们不好了,也只管告诉我”——这是王熙凤对黛玉说的话,这番话一方面是为了在贾母面前表现她对黛玉的关心,另一方面也炫耀了她在贾府的地位和权势,暗示黛玉不要小看她。

北师大版高中数学必修五双基限时练8

北师大版高中数学必修五双基限时练8

双基限时练(八)一、选择题1.在等比数列{a n }中,a 3=12,a 7=2,则a 5等于( )A .-1B .1C .±1D .以上均不对解析 设等比数列的公比为q ,∵a 5=a 3q 2, ∴a 3与a 5同号,故a 5=a 3a 7=1. 答案 B2.若数列{a n }是公差为2的等差数列,则数列{2a n }是( ) A .公比为4的等比数列 B .公比为2的等比数列 C .公比为12的等比数列D .公比为14的等比数列解析 2a n +12a n =2(a n +1-a n )=22=4.答案 A3.若数列{a n }为等比数列,则下列四个命题:①数列{a 2n }也是等比数列;②数列{a 2n }也是等比数列;③数列⎩⎨⎧⎭⎬⎫1a n 也是等比数列;④数列{lg|a n |}也是等比数列.其中正确的个数是( )A .1B .2C .3D .4解析 ④不正确,如a n =1,lg|a n |=0,而0,0,0,…不是等比数列.答案 C4.设等差数列{a n }的公差d 不为0,a 1=9d .若a k 是a 1与a 2k 的等比中项,则k =( )A .2B .4C .6D .8解析 由a 1=9d ,a k =9d +(k -1)d =(k +8)d ,a 2k =(2k +8)d ,由(k +8)2=9·(2k +8),得k =4.答案 B5.已知各项均为正数的等比数列{a n }中,a 1a 2a 3=5,a 7a 8a 9=10,则a 4a 5a 6=( )A .5 2B .7C .6D .4 2解析 由a 1a 2a 3=5,a 7a 8a 9=10, 得(a 1a 2a 3)·(a 7a 8a 9)=50.即a 65=50,a n >0.∴a 35=52,即a 4a 5a 6=5 2.答案 A6.已知等比数列{a n }中,各项都是正数,且a 1,12a 3,2a 2成等差数列,则a 9+a 10a 7+a 8的值等于( ) A .1+ 2 B .1- 2 C .3+2 2D .3-2 2解析 ∵{a n }为等比数列,又a 1,12a 3,2a 2成等差数列,∴a 1+2a 2=a 3,∴1+2q =q 2,得q =1+2或q =1-2(舍).又a 9+a 10a 7+a 8=(a 7+a 8)q 2a 7+a 8=(1+2)2=3+2 2. 答案 C 二、填空题7.在各项都是正数的等比数列{a n }中,lg(a 3a 8a 13)=6,则a 1a 15=________.解析 由lg(a 3a 8a 13)=6, 得a 3a 8a 13=106.又a 3a 8a 13=a 38,∴a 8=102.由等比中项知a 1a 15=a 28=(102)2=104.答案 1048.在等比数列{a n }中,公比q 为整数,且a 3+a 8=124,a 5·a 6=-512,则a 5=________.解析 ∵a 5a 6=a 3a 8=-512,又a 3+a 8=124,得⎩⎪⎨⎪⎧a 3=128,a 8=-4,或⎩⎪⎨⎪⎧a 3=-4,a 8=128,又q 为整数,∴⎩⎪⎨⎪⎧a 3=-4,a 8=128.又a 8=a 3q 5=128,得q =-2,∴a 5=a 3q 2=-16. 答案 -169.数列{a n }是等比数列,a n >0,已知a 2a 4+2a 3a 5+a 4a 6=125,则a 3+a 5=________.解析 由a 2a 4+2a 3a 5+a 4a 6=125,得a 23+2a 3a 5+a 25=125,a 3+a 5=±55,又a n >0,故a 3+a 5=5 5.答案 5 5 三、解答题10.在等比数列{a n }中,a 1=98,a n =12,q =23,求n .解 ∵a n =a 1·q n -1,得12=98×⎝ ⎛⎭⎪⎫23n -1,⎝ ⎛⎭⎪⎫23n -1=49,∴n -1=2,得n =3.11.在等比数列{a n }中,各项均为正值,且a 6a 10+a 3a 5=54,a 4a 8=5,求a 4+a 8的值.解 ∵{a n }为等比数列,∴a 3a 5=a 24,a 6a 10=a 28, ∴a 6a 10+a 3a 5=a 24+a 28=54,又a 4a 8=5,∴(a 4+a 8)2=a 24+2a 4a 8+a 28=54+10=64,又a n >0,∴a 4+a 8=64=8.12.已知三个数成等比数列,它们的积为27,它们的平方和为91,求这三个数.解 设这三个数分别为aq,a ,aq ,由题意,得⎩⎪⎨⎪⎧aq·a ·aq =27,a 2q 2+a 2q 2+a 2=91,①②由①得a =3,代入②,得q =±3或q =±13.∴当q =3时,这三个数分别是1,3,9; 当q =-3时,这三个数分别是-1,3,-9; 当q =13时,这三个数分别是9,3,1;当q =-13时,这三个数分别是-9,3,-1.思维探究13.在等比数列{a n }中,a 1=1,公比为q (q ≠0),且b n =a n +1-a n .(1)判断数列{b n }是否为等比数列?说明理由; (2)求数列{b n }的通项公式.解 (1)∵等比数列{a n }中,a 1=1,公比为q ,∴a n =a 1q n -1=q n -1(q ≠0),若q =1,则a n =1,b n =a n +1-a n =0,∴{b n }是各项均为0的常数列,不是等比数列.若q ≠1,由于b n +1b n =a n +2-a n +1a n +1-a n =q n +1-q n q n -q n -1=q n (q -1)q n -1(q -1)=q ,∴{b n }是首项为b 1=a 2-a 1=q -1,公比为q 的等比数列.(2)由(1)可知,当q=1时,b n=0;当q≠1时,b n=b1q n-1=(q-1)·q n-1,∴b n=(q-1)q n-1(n∈N+).。

双基练习题答案

双基练习题答案

双基练习题答案一、选择题1. 以下哪个选项是双基教育的核心内容?A. 体育B. 音乐C. 语文D. 数学答案:D2. 双基教育强调的“双基”指的是什么?A. 基础体能和基础技能B. 基础知识和基本技能C. 基础理论和基础实践D. 基础文化和基本素养答案:B3. 双基教育的实施目的是什么?A. 培养学生的应试能力B. 培养学生的创新能力C. 培养学生的基础知识和基本技能D. 培养学生的道德品质答案:C4. 双基教育在教学中通常采用哪种教学方法?A. 启发式教学B. 讲授式教学C. 互动式教学D. 以上都是答案:D5. 以下哪项不是双基教育的实施原则?A. 面向全体学生B. 注重学生个性发展C. 只注重知识传授D. 因材施教答案:C二、填空题6. 双基教育强调学生应该掌握的________和________。

答案:基础知识,基本技能7. 双基教育认为,教育应该________学生的全面发展。

答案:促进8. 在双基教育中,教师应该根据学生的________进行教学。

答案:实际情况9. 双基教育倡导的是一种________的教学模式。

答案:全面发展10. 双基教育要求学生在学习过程中,不仅要掌握知识,还要培养________。

答案:基本技能三、简答题11. 简述双基教育的重要性。

答案:双基教育的重要性在于它为学生提供了扎实的基础知识和基本技能,使学生能够在未来的学习和工作中具备必要的能力。

同时,它也有助于培养学生的创新思维和解决问题的能力。

12. 描述一下双基教育在课堂教学中的实施策略。

答案:在课堂教学中实施双基教育,教师应该采用多样化的教学方法,激发学生的学习兴趣,注重学生基础知识的掌握和基本技能的培养。

同时,教师还应该关注学生的个体差异,实施因材施教,确保每个学生都能在学习过程中得到发展。

四、论述题13. 论述双基教育与学生终身发展的关系。

答案:双基教育与学生终身发展密切相关。

首先,双基教育为学生提供了坚实的知识基础和技能基础,这为他们未来的学习和工作奠定了基础。

人教A版高中数学选修一双基限时练8

人教A版高中数学选修一双基限时练8

双基限时练(八)1.椭圆x 24+y 23=1的右焦点到直线y =3x 的距离是( )A.12B.32C.1D. 3解析 椭圆的右焦点(1,0)到直线y =3x 的距离为d =|3-0|3+1=32. 答案 B2.若椭圆a 2x 2-a 2y 2=1的一个焦点是(-2,0),则a 为( )A.1-54B.1+52C.12D.22解析 由a 2x 2-a 2y 2=1,得x 21a 2+y2-2a=1,∴a <0,∵焦点(-2,0),∴1a 2+2a=4,即4a 2-2a -1=0,解得a =1-54,或a =1+54(舍去).答案 A3.设P 是椭圆x 29+y 25=1上一点,M ,N 分别是两圆(x +2)2+y 2=1和(x -2)2+y 2=1上的点,则|PM |+|PN |的最小值和最大值分别为( )A .4,8B .6,8C .8,12D .2,6解析 设椭圆的左、右焦点分别为F 1、F 2,两圆的半径为R ,由题意可知|PM |+|PN |的最大值为|PF 1|+|PF 2|+2R ,最小值为|PF 1|+|PF 2|-2R ,又因为|PF 1|+|PF 2|=2a =6,R =1,所以|PM |+|PN |的最大值为8,最小值为4.答案 A4.若点O 和点F 分别为椭圆x 24+y 23=1的中心和左焦点,点P 为椭圆上的任意一点,则OP →·FP →的最大值为( )A .2B .3C .6D .8解析 由题意,F (-1,0),设点P (x 0,y 0), 则有x 204+y 203=1,解得y 20=3(1-x 204),∵FP →=(x 0+1,y 0),OP →=(x 0,y 0),∴OP →·FP →=x 0(x 0+1)+y 20=x 0(x 0+1)+3(1-x 24)=x 204+x 0+3.此二次函数对应的抛物线的对称轴为x 0=-2, ∵-2≤x 0≤2,∴当x 0=2时,OP →·FP→取得最大值224+2+3=6,选C. 答案 C5.直线y =x +2与椭圆x 2m +y 23=1有两个公共点,则m 的取值范围是( )A .m >1B .m >1且m ≠3C .m >3D .m >0且m ≠3解析 把y =x +2代入x 2m +y 23=1,并整理得(m +3)x 2+4mx +m =0.Δ=16m 2-4m (m +3)=12m (m -1),由Δ>0,得m <0或m >1. ∵m >0且m ≠3,∴m >1且m ≠3. 答案 B6.经过椭圆x 22+y 2=1的一个焦点,作倾斜角为45°的直线l ,交椭圆于A ,B 两点,设O 为坐标原点,则OA →·OB→等于( ) A .-3B .-13C .-13或-3D .±13解析 设椭圆的一个焦点F (1,0),则直线l :y =x -1,代入椭圆方程x 22+y 2=1,并整理得3x 2-4x =0.解得x 1=0,x 2=43,∴y 1=-1,y 2=13.又OA →·OB →=(x 1,y 1)·(x 2,y 2)=x 1x 2+y 1y 2=-13. 答案 B7.椭圆对称轴在坐标轴上,短轴的一个端点与两个焦点构成一个正三角形,焦点到椭圆上的点的最短距离是3,则这个椭圆方程为________.解析 由题意可知⎩⎪⎨⎪⎧a =2c ,a -c =3,∴⎩⎪⎨⎪⎧a =23,c =3,b 2=a 2-c 2=12-3=9.∴椭圆方程为x 212+y 29=1,或y 212+x 29=1.答案x 212+y 29=1,或y 212+x 29=1 8.设P 为椭圆x 24+y 2=1上任意一点,O 为坐标原点,F 为椭圆的左焦点,点M 满足OM →=12(OP →+OF →),则|OM →|+|MF →|=________.解析如图所示,F 0为椭圆的右焦点,连接PF 0, 由OM →=12(OP →+OF →),可知M 为PF 的中点, 则|OM →|=12|F 0P →|, ∴|OM →|+|MF →|=12|F 0P →|+12|PF →|=12(|F 0P →|+|PF →|)=a =2. 答案 29.已知椭圆x 2a 2+y 2b2=1(a >b >0),以坐标原点O 为圆心,短半轴长为半径作圆O ,过椭圆的长轴的一端点P 作圆O 的两条切线,切点为A ,B ,若四边形PAOB 为正方形,则该椭圆的离心率为________.解析 如图,∵四边形OAPB 是正方形,且PA ,PB 为圆O 的切线, ∴△OAP 是等腰直角三角形, 故b =c ,a =2c ,∴e =22. 答案 2210.已知椭圆C :x 2a 2+y 2b 2=1(a >b >0)的离心率e =12,且椭圆经过点N (2,-3).(1)求椭圆C 的方程;(2)求椭圆以M (-1,2)为中点的弦所在直线的方程. 解 (1)由椭圆经过点N (2,-3), 得22a 2+(-3)2b2=1,又e =c a =12,解得a 2=16,b 2=12.∴椭圆C 的方程为x216+y212=1.(2)显然M 在椭圆内,设A (x 1,y 1),B (x 2,y 2)是以M 为中点的弦的两个端点,则x 2116+y 2112=1,x 2216+y 2212=1. 相减得(x 2-x 1)(x 2+x 1)16+(y 2-y 1)(y 2+y 1)12=0.整理得k AB =-12·(x 1+x 2)16·(y 1+y 2)=38,则所求直线的方程为y -2=38(x +1),即3x -8y +19=0.11.已知动点P 到定点F (2,0)的距离与点P 到定直线l :x =22的距离之比为22.(1)求动点P 的轨迹C 的方程;(2)设M ,N 是直线l 上的两个点,点E 与点F 关于原点O 对称,若EM →·FN →=0,求|MN |的最小值.解 (1)设P (x ,y ),依题意,有(x -2)2+y 2|x -22|=22,整理,得x 24+y 22=1, ∴动点P 的轨迹C 的方程为x 24+y 22=1.(2)∵点E 与点F 关于原点O 对称, ∴点E 的坐标为(-2,0).∵M ,N 是直线l 上的两个点,∴可设M (22,y 1),N (22,y 2)(不妨设y 1>y 2). ∵EM →·FN →=0,∴(32,y 1)·(2,y 2)=0, 6+y 1y 2=0,即y 2=-6y 1.由于y 1>y 2,∴y 1>0,y 2<0. ∴|MN |=y 1-y 2=y 1+6y 1≥2y 1·6y 1=2 6.当且仅当y 1=6,y 2=-6时,等号成立. 故|MN |的最小值为2 6.12.如图椭圆E 经过点A (2,3),对称轴为坐标轴,焦点F 1,F 2在x 轴上,离心率e =12.(1)求椭圆E 的方程;(2)求∠F 1AF 2的角平分线所在直线的方程.解 (1)设椭圆E 的方程为x 2a 2+y 2b2=1.(a >b >0)由e =12,得c a =12,b 2=a 2-c 2=3c 2,∴x 24c 2+y23c2=1.将A (2,3)代入,有1c 2+3c2=1,解得c =2,∴椭圆E 的方程为x 216+y 212=1.(2)由(1)知F 1(-2,0),F 2(2,0), ∴直线AF 1的方程为y =34(x +2),即3x -4y +6=0.直线AF 2的方程为x =2.由椭圆E 的图形知,∠F 1AF 2的角平分线所在直线的斜率为正数. 设P (x ,y )为∠F 1AF 2的角平分线所在直线上任一点,则有|3x -4y +6|5=|x -2|, 若3x -4y +6=5x -10,得x +2y -8=0,其斜率为负,不合题意,舍去. 于是3x -4y +6=-5x +10,即2x -y -1=0. ∴∠F 1AF 2的角平分线所在直线的方程为 2x -y -1=0.。

双基能力训练习题及答案

双基能力训练习题及答案

双基能力训练习题及答案1.[ ]class same want bag2.[ ]goodness broken empty get3.[ ]wrong worry only not4.[ ]put mum ruler student5.[ ]say play Sunday today6.[ ]flower window know how7.[ ]China school catch chair8.[ ]nine know think orange1.[ ]A.yellow B.pencilC.fourteen D.sweater2.[ ]A.bedroom B.middleC.very D.Chinese3.[ ]A.basket B.excuseC.begin D.today4.[ ]A.apple B.dutyC.orange D.thirteen5.[ ]A.worry B.aboutC.other D.broken1.“I can see some books there. ”“What o things can you see? ”2.Here are your clothes. Please put them a.3.Don't w. I can help you.4.“Is seventy and twenty niy?”“Yes. That's r.”5.What's w with your kite? 1.[ ]A.sister B.brother C.friend D.father2.[ ]A.hat B.rulerC.eraser D.pencil3.[ ]A.bus B.carC.coat D.jeep4.[ ]A.apple B.bananaC.cake D.orange5.[ ]A.English B.Chinese C.China D.American 1.good(名词)2.empty(反义词)3.right(反义词)4.do not(缩略式)5.he(宾格)6.she(宾格)7.them(主格)8.knife(复数)1.Very much thank you.2.I think not so.3.Don't to worry.4.Put on it, please.5.My ruler is broken. What about you?1.Here are two books. One is yours, is mine. [ ]A.other B.othersC.the other D.the others2.The look nice. Put , please.[ ]A.sock, away them B.socks, them away C.sock, it away D.socks, away them3.—My hat is the tree. What can I do? —Don't worry. I can get it.[ ]A.on, / B.on, toC.in, / D.in, to4.— , Jim?—I can't get down.[ ]A.What's wrong B.Where are you C.What do you do D.How are you 5.—Is the window broken?—I don't think .[ ]A.this B.thatC.it D.so6.Don't thank . Thank Lily. [ ]A.you B.goodC.goodness D./7.Oh ! My desk is broken. [ ]A.dear B.goodC.fine D.right8.Please give the .[ ]A.he, knifes B.he, knives C.him, knifes D.him, knives 9.What's wrong your bike? [ ]A.in B.onC.of D.with10.—Can you mend my car?—.[ ]A.Yes, I don't think so B.Thank youC.No, I think so D.OK1.Sit down. (改为否认句)2.My can write now. (改为一般疑问句)3.He' s a twin. (改为复数句)4.Please put away your shoes. (用人称代词替代画线局部改写此句)5.Bob's pencil is broken. (就画线局部提问)6.She's Tom's sister.(就画线局部提问)7.The boy behind Rose is Jim. (就画线局部提问)8.My shoes are under the bed. (就画线局部提问)—Whose bags are these? Are they 1 ?—Yes, they're mine. I want 2 take them 3 my bedroom. Could you 4 me, please?— 5 .—Can you 6 the red one?—Oh, it's heavy.—What 7 the blue 8 ?—That's OK.— 9 .—You're 10 .Ⅰ.1.D 2.C 3.C 4.D 5.B 6.B 7.B 8.BⅡ.1.C 2.D 3.A 4.D 5.BⅢ.1.other 2.away 3.worry 4.right 5.WrongⅣ.1.C 2.A 3.C 4.C 5.CⅤ.1.goodness 2.full 3.wrong 4.don't 5.him 6.her 7.they 8.KnivesⅥ.1.Thank you very much.2.I don't think so.3.Don't worry.4.Put it on, please5.My ruler is broken. What about yours?Ⅶ.1.C 2.B 3.C 4.A 5.D 6.C 7.A 8.D 9.D 10.D Ⅷ.1.Don't sit down.2.Can Mary write now?3.They're twins.4.Please put them away.5.What's wrong with Bob's pencil?6.Whose sister is she?7.Which boy is Jim?8.Where are your shoes?Ⅸ.1.yours 2.to 3.to 4.help 5.OK(Certainly)6.carry 7.about 8.one 9.Thanks 10.wele。

八年级物理上册课时双基目标训练

八年级物理上册课时双基目标训练

横道中学八年物理阶段性双基目标测试题二年班姓名时间50分钟满分70分成绩:一、选择题(每题2分共12分)1、用天平称出一个塑料瓶的质量,然后将其剪碎再放到天平上称,比较这个物体在形状变化前后的质量()A.变大B.变小C.不变D.无法确定2、用托盘天平测出一枚邮票的质量,下列做法中最合理的是()A.把一枚邮票放在托盘天平上称量B.先测出一枚邮票和一螺母的总质量,再减去螺母的质量C.先测出200枚邮票总质量,再除以200D.先测出200枚邮票的总质量,再加一枚邮票测出201枚邮票的总质量,求其差就是一枚邮票的质量3、在测量铁块的密度时,用了下面几个步骤,则测量步骤合理顺序是()①计算铁块的体积;②记录铁块放入量筒后水面上升到的刻度;③观察量筒中水的体积;④用天平测出铁块的质量;⑤调节好天平;⑥计算铁块的密度.A.⑤④②③①⑥B.④⑤③②①⑥C.③②①⑤④⑥D.⑤④③②①⑥4、托盘天平横梁上都有标尺和游码,向右移动游码的作用是()A.相当于向左调节平衡螺母B.代替指针用来指示平衡C.相当于在左盘中加小砝码D.相当于在右盘中加小砝码5、老花镜掉在地上,镜片裂成两块,但仍留在镜框内,则用这个眼镜看书上的字时()A.一个字会变成两个字B.看不到字C.仍能看到一个完整的字,只是字上有一条缝D.仍能看到一个完整的字6、关于照相机的使用,下列说法正确的是()A.拍摄远景时,应将镜头向前伸B.拍摄近景时,应将镜头向前伸C.晴天拍摄景物时,应开大光圈D.阴天拍摄景物时,应缩短曝光时间二、填空题(每空1分共18分)放大后正立的像.三、计算题(每题5分共10分)15、如图所示,一容积为3×10-4m3的瓶内盛有0.2kg的水,一只口渴的乌鸦每次将一块质量为0.01kg的小石子投入到瓶中,当乌鸦投入了25块相同的小石子后,水面升到瓶口,求:(1)瓶内石块的总体积;(2)石块的密度.16、一个实心铅球的质量是4kg,经测量它的体积是0.57×10-3m3,这个铅球是用纯铅制成的吗?(已知ρ3kg/m3)铅=11.3×10四、简答题(每题2分共6分)17、飞机为什么用铝合金制造而不用钢铁制造?18、实验室有一架托盘天平和砝码,横梁调节螺母已丢失,其他部件都完好准确,将它放在水平桌面上,游码移到零刻度线上时,指针总偏向中央标尺左端,现要利用这架天平准确测出一小金属块的质量,请你简述出一种测量方法.(可选用生活中常见的材料)19、望远镜物镜的直径比我们眼睛的瞳孔大得多,这是为什么?五、作图与实验题(20题2分21题10分22题4分23题5分24题3分共24分)20、如图所示,OO′是凸透镜的主光轴,A′B′是蜡烛AB通过凸透镜成的像,试在图中画出光线AC经凸透镜后的折射光线,并用作图的方法找出右侧焦点的位置.21、小华在观察凸透镜成像的条件的实验中:(1)把蜡烛、凸透镜、光屏从左向右依次放在同一上,点燃蜡烛并调整烛焰、凸透镜、光屏的高度,使它们的中心大致在同一.(2)确定像的位置时,沿凸透镜主光轴移动光屏,直到光屏上出现明亮、的烛焰像;(3)从蜡烛到凸透镜距离等于2倍焦距处开始,向焦点方向移动蜡烛,在这过程中,像到透镜的距离(增大/减小/不变),像(变大/变小/不变);(4)实验得到的结论是:凸透镜成放大实像的条件是:;凸透镜成缩小实像的条件是:;凸透镜放大虚像的条件是:.(以上三空均填字母符号)(5)小华看到装满清水的透明玻璃酒杯很像一个凸透镜(如图),就想探究它是否具有凸透镜的性质.请你帮他想出一个验证的方法.。

【最新】人教版英语选修六双基限时练【8】Unit 3 Part Ⅱ(含答案)

【最新】人教版英语选修六双基限时练【8】Unit 3 Part Ⅱ(含答案)

最新教学资料·人教版英语双基限时练(八)Unit 3Part Ⅱ基础夯实Ⅰ.完成句子1.I found ________ ________ ________ ________ ________ ________ ________ ________ in a week. (finish)我发现一周内完成我们的工作很难。

答案it difficult for us to finish our work2.________ ________ ________ ________ they should start early. (suggest)建议他们早点动身。

答案It was suggested that3.________ ________ ________ crying over spilt milk. (use)覆水难收。

答案It's no use4.It is easy ________ ________ ________ ________ ________ ________. (work)做出那道数学题很容易。

答案to work out the maths problem5.________ ________ ________ ________ ________ to the hospital. (it)去医院走路需要两个小时。

答案It is a two-hour walk6.The manager thinks ________ ________ ________ ________ ________ ________ ________ ________ this week. (hold)经理认为本周再召开一次会议很有必要。

答案it very necessary to hold the meeting again7.________ ________ ________ ________ ________ since we met last time. (it)自从我们上次见面以来已过了两周。

  1. 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
  2. 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
  3. 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。

双基限时练(八)
1.“a⊥α,则a垂直于平面α内的任一直线”是()
A.全称命题B.特称命题
C.不是命题D.真命题
答案 A
2.命题“所有能被2整除的整数都是偶数”的否定是()
A.所有不能被2整除的整数都是偶数
B.所有能被2整除的整数都不是偶数
C.存在一个不能被2整除的整数是偶数
D.存在一个能被2整除的整数不是偶数
解析全称命题的否定是特称命题,因此A,B不正确.全称命题“∀x,P(x)”的否定是特称命题“∃x,綈p(x)”,因此D正确.答案 D
3.已知命题p:∀x∈R,sin x≤1,则()
A.綈p:∃x∈R,sin x≥1
B.綈p:∀x∈R,sin x≥1
C.綈p:∃x∈R,sin x>1
D.綈p:∀x∈R,sin x>1
答案 C
4.下列语句中,判断正确的个数是()
①全称命题“∀n∈Z,2n+1是奇数”是真命题
②特称命题“∃x∈R,x2是无理数”是真命题
③命题“∀n∈Z,2n+1是奇数”的否定是“∃n∈Z,2n+1不是奇数”
④命题“∃x∈R,x2是无理数”的否定是“∀x∈R,x2是有理数”
A.1 B.2
C.3 D.4
答案 D
5.已知命题p:∀x∈R,cos x≤1,则()
A.綈p:∃x∈R,cos x≥1
B.綈p:∀x∈R,cos x≥1
C.綈p:∃x∈R,cos x>1
D.綈p:∀x∈R,cos x>1
答案 C
6.命题“对任意的x∈R,x3-x2+1≤0”的否定是()
A.不存在x∈R,x3-x2+1≤0
B.存在x∈R,x3-x2+1≤0
C.存在x∈R,x3-x2+1>0
D.对任意的x∈R,x3-x2+1>0
答案 C
7.命题“函数都有最大值”的否定是________.
答案有的函数没有最大值
8.命题“至少有一个正数满足方程x2+2(a-1)x+2a+6=0”的否定是________.
答案∀x∈R+,方程x2+2(a-1)x+2a+6=0不成立
9.若命题p:∀x∈R,
1
x-2
<0,则綈p:____________.
解析 綈p :∃x 0∈R ,使1x 0-2
>0或x 0-2=0. 最容易出现的错误答案是:存在x 0∈R ,使1x 0-2
≥0. 答案 ∃x 0∈R ,使1x 0-2
>0或x 0-2=0 10.命题“存在x ∈R ,使2x 2-3ax +9<0”为假命题,求实数a
的取值范围.
解析 ∵“存在x ∈R ,使∴它的否定:“对任意的∴只要Δ=9a 2-4×2×9≤故实数a 的取值范围是[-11.设集合A =断其真假.
(1)p :∀n ∈A ,n <12;
(2)q :∃n ∈{奇数},使n ∈A .
解 (1)綈p :∃n ∈A ,使n ≥12.
∵当n =12时,綈p 成立,
∴綈p 是真命题.
(2)綈q :∀n ∈{奇数},n ∉A .綈q 是假命题.
12.若p (x ):sin x +cos x >m ,q (x ):x 2+mx +1>0,如果∀x ∈R ,p (x )为假命题,q (x )为真命题,求实数m 的取值范围.
解 由于sin x +cos x =2sin(x +π4)∈[-2,2],又∀x ∈R ,p (x )
为假命题,即对任意x ∈R ,sin x +cos x >m 不成立,所以m > 2.
又对任意x ∈R ,q (x )为真命题,即对任意x ∈R ,x 2+mx +1>0
恒成立,所以Δ=m2-4<0,
即-2<m<2.
故∀x∈R,p(x)为假命题,q(x)为真命题,应有2<m<2.
新课标第一网系列资料。

相关文档
最新文档