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人教B版高中数学必修二双基限时练20

人教B版高中数学必修二双基限时练20

双基限时练(二十)基础强化1.经过点(3,a ),(-2,0)的直线与直线x -2y +3=0垂直,则a 的值为( )A.52 B.25 C.10 D .-10解析a -03-(-2)=-2,∴a =-10.答案 D2.已知A (2,4)与B (3,3)关于直线l 对称,则直线l 的方程为( )A .x +y =0B .x -y =0C .x +y -6=0D .x -y +1=0解析 k AB =4-32-3=-1,AB 中点⎝ ⎛⎭⎪⎫52,72,∴直线l 的斜率为1,且经过点⎝ ⎛⎭⎪⎫52,72,∴y -72=x -52,即x -y +1=0.答案 D3.已知直线mx +4y -2=0与2x -5y +n =0互相垂直,垂足为(1,p ),则m -n +p 的值为( )A .24B .20C .0D .-4解析 2m -20=0,∴m =10. ∴10+4p -2=0,∴p =-2. ∴2+10+n =0,∴n =-12. ∴m -n +p =20. 答案 B4.△ABC 的顶点是A (3,6),B (2,3),C (-2,4),则AB 边上的高线所在直线方程为( )A .x +3y -10=0B .x +3y +10=0C .3x +y +2=0D .3x -y +2=0 解析 k AB =6-33-2=3,∴k 高=-13.∴高线所在直线:y -4=-13(x +2),即x +3y -10=0.答案 A5.已知点M (0,-1),点N 在直线x -y +1=0上,若直线MN 垂直于直线x +2y -3=0,则N 点的坐标是( )A .(-2,-3)B .(2,1)C .(2,3)D .(-2,-1)解析 k MN =2,∴l MN :y =2x -1.⎩⎪⎨⎪⎧x -y +1=0,y =2x -1, ∴x =2,y =3,∴N (2,3).答案 C6.入射光线在直线l 1:2x -y -3=0上,经过x 轴反射后所在直线为l 2,再经过y 轴反射后所在直线为l 3,则直线l 3的方程为( )A .x -2y +3=0B .2x -y +3=0C .2x +y -3=0D .2x -y +6=0解析 根据光的反射原理,l 1与l 2关于x 轴对称,l 2与l 3关于y 轴对称,∴直线l 1与l 3关于原点对称.∵l 1:2x -y -3=0,∴l 3:2x -y +3=0. 答案 B7.过点(1,3)且与直线x +2y -1=0垂直的直线方程为_________________________________________________________.解析 直线x +2y -1=0的斜率为-12,故所求直线的斜率为2,∴y -3=2(x -1), 即2x -y +1=0. 答案 2x -y +1=08.若直线(m +2)x +3my +1=0与直线(m -2)x +(m +2)y -3=0互相垂直,则m =________.解析 由(m +2)(m -2)+3m (m +2)=0,得(m +2)·(4m -2)=0,∴m =-2或12.答案 -2或12能力提升9.M (-1,0)关于直线x +2y -1=0的对称点M ′的坐标为________.解析 设M ′的坐标为(x 0,y 0),∴⎩⎪⎨⎪⎧y 0x 0+1×⎝ ⎛⎭⎪⎫-12=-1,x 0-12+y 02×2-1=0,∴⎩⎪⎨⎪⎧x 0=-15,y 0=85.∴M ⎝ ⎛⎭⎪⎫-15,85. 答案 M ⎝ ⎛⎭⎪⎫-15,8510.求经过直线l 1:3x +2y -1=0和l 2:5x +2y +1=0的交点,且垂直于直线l 3:3x -5y +6=0的直线l 的方程.解 方法一 先解方程组⎩⎪⎨⎪⎧3x +2y -1=0,5x +2y +1=0,得l 1与l 2的交点(-1,2),再由l 3的斜率35求出l 的斜率为-53,于是由直线的点斜式方程求出l :y -2=-53(x +1),即5x +3y -1=0.方法二 ∵l ⊥l 3,故l 是直线系5x +3y +C =0中的一条,而l过l 1与l 2的交点(-1,2),故5×(-1)+3×2+C =0,由此求出C =-1,故l 的方程为5x +3y -1=0.方法三 ∵l 过l 1与l 2的交点,故l 是直线系3x +2y -1+λ(5x +2y +1)=0中的一条,将其整理,得(3+5λ)x +(2+2λ)y +(-1+λ)=0.其斜率为-3+5λ2+2λ=-53,解得λ=15,代入直线系方程即得l 的方程为5x +3y -1=0.11.已知A (1,0),B (3,2),C (0,4),点D 满足AB ⊥CD ,且AD ∥BC ,试求点D 的坐标.解 设D (x ,y ),则k AB =23-1=1,k BC =4-20-3=-23,k CD =y -4x ,k DA =yx -1.∵AB ⊥CD ,AD ∥BC , ∴k AB ·k CD =-1,k DA =k BC .∴⎩⎪⎨⎪⎧1×y -4x=-1,y x -1=-23.解得⎩⎪⎨⎪⎧x =10,y =-6,即D (10,-6).12.已知直线l :x +2y -2=0,试求: (1)点P (-2,-1)关于直线l 的对称点坐标;(2)直线l 1:y =x -2关于直线l 对称的直线l 2的方程; (3)直线l 关于点A (1,1)对称的直线方程.解 (1)设点P 关于直线l 的对称点为P ′(x 0,y 0),则线段PP ′的中点M 在直线l 上,且PP ′⊥l .∴⎩⎪⎨⎪⎧y 0+1x 0+2×⎝ ⎛⎭⎪⎫-12=-1,x 0-22+2×y 0-12-2=0,解之得⎩⎪⎨⎪⎧x 0=25,y 0=195,即P ′点的坐标为⎝ ⎛⎭⎪⎫25,195.(2)直线l 1:y =x -2关于直线l 对称的直线为l 2,则l 2上任一点P 1(x ,y )关于l 的对称点P 1′(x ′,y ′)一定在直线l 1上,反之也成立.由⎩⎪⎨⎪⎧y -y ′x -x ′×⎝ ⎛⎭⎪⎫-12=-1,x +x ′2+2×y +y ′2-2=0,得⎩⎪⎨⎪⎧x ′=3x -4y +45,y ′=-4x -3y +85.把(x ′,y ′)代入方程y =x -2并整理,得7x -y -14=0,即直线l 2的方程为7x -y -14=0.(3)设直线l 关于点A (1,1)的对称直线为l ′,则直线l 上任一点P 2(x 1,y 1)关于点A 的对称点P 2′(x ,y )一定在直线l ′上,反之也成立.由⎩⎪⎨⎪⎧x +x 12=1,y +y 12=1,得⎩⎪⎨⎪⎧x 1=2-x ,y 1=2-y .将(x 1,y 1)代入直线l 的方程得,x +2y -4=0, 即直线l ′的方程为x +2y -4=0.品味高考13.如图,△ABC 的顶点B (3,4),AB 边上的高CE 所在直线方程为2x +3y -16=0,BC 边上的中线AD 所在直线方程为2x -3y +1=0,求边AC 的长.解 设点A ,C 的坐标分别为A (x 1,y 1),C (x 2,y 2). ∵AB ⊥CE ,k CE =-23,∴k AB =-1k EC =32.∴直线AB 的方程为3x -2y -1=0.由⎩⎪⎨⎪⎧3x 1-2y 1-1=0,2x 1-3y 1+1=0,得A (1,1).∵D 是BC 的中点,∴D ⎝⎛⎭⎪⎫x 2+32,y 2+42. 而点C 在直线CE 上,点D 在直线AD 上, ∴⎩⎪⎨⎪⎧2x 2+3y 2-16=0,2·x 2+32-3·y 2+42+1=0.∴C (5,2).|AC |=(5-1)2+(2-1)2=17.。

学生作业7月20

学生作业7月20

高2数学复习限时训练(20)1、设P 和Q 是两个集合,定义集合{,}P Q x x P x Q -=∈∉且,如果}1log {2<=x x P ,{21}Q x x =-<,那么P Q -= .2、 一水池有2个进水口, 1个出水口,一个口的进、出水的速度如图甲、乙所示.某天0点到6点,该水池的蓄水量如图丙所示.给出以下3个论断:进水量 出水量 蓄水量甲 乙 丙(1)0点到3点只进水不出水;(2)3点到4点不进水只出水; (3)4点到6点不进水不出水.则一定不正确...的论断是 (把你认为是符合题意的论断序号都填上) . 3、已知函数()y f x =的图象与函数22()log (2)g x x x =++的图象关于直线2x =对称,则(3)f = .4、 若不等式142x x a +--≥0在x ∈[1,2]上恒成立,则a 的取值范围为 .5、 在公差为正数的等差数列}{n a 中,n S a a a a ,0,011101110<<+且是其前n 项和,则使n S 取最小值的n 是 。

6、 数列{a n }的前n 项和S n =n 2+2n -1,则a 1+a 3+a 5+…+a 25= .7、 在ABC ∆中,若=+=C B C B A tan tan ,cos cos 2sin 则 。

8、 设函数0)(),()(3=+-=x f b bx x x f 若方程为常数的根都在区间[-2,2]内,且函数)(x f 在区间(0,1)上单调递增,则b 的取值范围是 。

9、某观测站C 在城A 的南20˚西的方向上,由A 城出发有一条公路,走向是南40˚东,在C 处测得距C 为31千米的公路上B 处,有一人正沿公路向A 城走去,走了20千米后,到达D 处,此时C 、D 间距离为21千米,问这人还需走多少千米到达A 城?。

2013届高考化学二轮总复习限时训练 第20课时 物质的检验、分离与提纯

2013届高考化学二轮总复习限时训练 第20课时 物质的检验、分离与提纯

第20课时物质的检验、分离与提纯1.(2012·江苏)下列有关实验装置进行的相应实验,能达到实验目的的是( )图1图2图3图4A.用图1所示装置除去Cl2中含有的少量HClB.用图2所示装置蒸干NH4Cl饱和溶液制备NH4Cl晶体C.用图3所示装置制取少量纯净的CO2气体D.用图4所示装置分离CCl4萃取碘水后已分层的有机层和水层2.(2012·北京)下列实验中,所选装置不合理的是( )①②③④⑤A.分离Na2CO3溶液和CH3COOC2H5,选④B.用CCl4提取碘水中的碘,选③C.用FeCl2溶液吸收Cl2选⑤D.粗盐提纯,选①和②3.(2011·浙江)下列说法不正确的是( )A.变色硅胶干燥剂含有CoCl2,干燥剂呈蓝色时,表示不具有吸水干燥功能B.“硝基苯制备”实验中,将温度计插入水浴,但水银球不能与烧杯底部和烧杯壁接触C.“中和滴定”实验中,容量瓶和锥形瓶用蒸馏水洗净后即可使用,滴定管和移液管用蒸馏水洗净后,须经干燥或润洗后方可使用D.除去干燥CO2中混有的少量SO2,可将混合气体依次通过盛有酸性KMnO4溶液、浓硫酸的洗气瓶4.下列物质的分离、提纯和鉴别的方法中(必要时可加热),可行的是( )A.以AgNO3溶液和稀盐酸检验溶液中是否含有Cl-B.用升华法分离碘和氯化铵的混合物C.以粗铜为阴极,精铜为阳极,CuSO4溶液为电解液,精炼铜D.用NaOH溶液、蒸馏水和红色石蕊试纸检验溶液中是否含有NH+45.下列操作中错误的是( )A.除去乙酸乙酯中的少量乙酸:加入乙醇和浓硫酸,使乙酸全部转化为乙酸乙酯B.除去苯中的少量苯酚:加入NaOH溶液,振荡、静置分层后,除去水层C.除去CO2中的少量SO2:通过盛有饱和NaHCO3溶液的洗气瓶D.提取溶解在水中的少量碘:加入CCl4,振荡、静置分层后,取出有机层再分离6.下列除去杂质的方法正确的是( )A.除去N2中的少量O2:通过灼热的CuO粉末,收集气体B.除去CO2中的少量HCl:通入Na2CO3溶液,收集气体C.除去FeCl2溶液中的少量FeCl3:加入足量铁屑,充分反应后,过滤D.除去KCl溶液中的少量MgCl2:加入适量NaOH溶液,过滤7.某固体混合物可能含有NH4Cl、KOH、AgNO3、AlCl3中的若干种,加适量水充分搅拌,得无色澄清溶液,取溶液进行如下实验:①蘸取少量原溶液滴在pH试纸上,试纸显蓝色,与标准比色卡比较,pH约为11;②取1 mL 原溶液加入3滴乙醛,水浴加热,试管内壁出现光亮的“银镜”。

中考语文总复习限时训练(二十)名著选段阅读

中考语文总复习限时训练(二十)名著选段阅读

限时训练(二十) 名著选段阅读(时间:20分钟)班级: 姓名:一、阅读下面的名著选段,回答问题。

回到车厂,他懊睡了两天。

决不想上曹宅去了,连个信儿也不必送,曹先生救不了祥子的命。

睡了两天他把车拉出去,心中完全是块空白,不再想什么,不再希望什么,只为肚子才出来受罪,肚子饱了就去睡,还用想什么呢,还用希望什么呢?看着一条瘦得出了棱的狗在白薯挑子旁边等着吃点皮和须子,他明白了他自己就跟这条狗一样,一天的动作只为捡些白薯皮和须子吃。

将就着活下去是一切,什么也无须乎想了。

人把自己从野兽中提拔出,可是到现在人还把自己的同类驱逐到野兽里去。

祥子还在那文化之域,可是变成了走兽。

一点也不是他自己的过错。

他停止住思想,所以就是杀了人,他也不负什么责任。

他不再有希望,就那么迷迷忽忽的往下坠,坠入那无底的深坑。

他吃,他喝,他嫖,他赌,他懒,他狡猾,因为他没了心,他的心被人家摘了去。

他只剩下那个高大的肉架子,等着溃烂,预备着到乱死岗子去。

1.这篇文章的作者是。

2.祥子为何懊睡?3.请从孙侦探、车夫老马、虎妞三个人物中选取其一,谈谈他(她)在祥子从“人”到“走兽”这个过程中产生了怎样的影响。

二、阅读下文,回答问题。

我们婚后的头两年,罗切斯特先生的眼睛一直是瞎的。

也许正因为这样,我们才如此亲近——才结合得如此紧密。

因为那时我就是他的眼睛,正如现在依然还是他的右手一样。

说实在的,我(像他常叫我的那样)就是他的眼珠。

他通过我看大自然,看书。

我也从来不知厌倦地替他仔细察看田野、树木、城镇、河流、云彩、阳光——描摹我们面前的景色。

周围的天气——还用声音向他的耳朵传达了光线已无法向他的眼睛传达的印象。

我永不厌倦地念书给他听,领他到他想去的地方,替他做他想做的事。

在这种效劳中,我尽管感到有点儿悲哀,但却获得一种最为充分、最为强烈的乐趣——因为他要求我为他做事时,并没有感到痛苦羞惭,也没有感到沮丧屈辱。

他是如此真心地爱我,因为他知道,在接受我的照料时,根本用不着勉强。

数学数列限时训练单选20题

数学数列限时训练单选20题

数学数列单项选择题:共20小题,每小题5份,共100分1.数列{a n}为等差数列,且a2+a7+a12=6,则{a n}的前13项的和为()A.52B.C.26D.2.记S n为正项等比数列{a n}的前n项和.若a1=1,4a3=a5,则S10=()A.512B.511C.1023D.10243.数列{a n}是公差为2的等差数列,S n为其前n项和,且a1,a4,a13成等比数列,则S4=()A.8B.12C.16D.244.已知等比数列{a n}满足a1=3,a1+a3+a5=21,则a3+a5+a7=()A.21B.42C.63D.845.记S n为等差数列{a n}的前n项和.若a4+a5=24,S6=48,则{a n}的公差为()A.1B.2C.4D.86.已知等差数列{a n}中,a2=7,a4=15,则前10项的和S10=()A.100B.210C.380D.4007.设{a n}是公差为正数的等差数列,若a1+a2+a3=15,a1a2a3=80,则a11+a12+a13=()A.120B.105C.90D.758.已知数列{a n}满足3a n+1+a n=0,a2=﹣,则{a n}的前10项和等于()A.﹣6(1﹣3﹣10)B.C.3(1﹣3﹣10)D.3(1+3﹣10)9.已知等差数列{a n}前9项的和为27,a10=8,则a100=()A.100B.99C.98D.9710.记S n为等差数列{a n}的前n项和.若3S3=S2+S4,a1=2,则a5=()A.﹣12B.﹣10C.10D.1211.已知等比数列{a n}满足a1=,a3a5=4(a4﹣1),则a2=()A.2B.1C.D.12.等差数列{a n}的前m项和为30,前2m项和为100,则它的前3m项和为()A.130B.170C.210D.26013.等差数列{a n}的首项为1,公差不为0.若a2,a3,a6成等比数列,则{a n}前6项的和为()A.﹣24B.﹣3C.3D.814.已知数列{a n}的前n项和为S n,a1=1,S n=2a n+1,则当n>1时,S n=()A.()n﹣1B.2n﹣1C.()n﹣1D.(﹣1)15.已知等差数列{a n}满足a2+a4=4,a3+a5=10,则它的前10项的和S10=()A.138B.135C.95D.2316.已知各项均为正数的等比数列{a n},a1a2a3=5,a7a8a9=10,则a4a5a6=()A.B.7C.6D.17.等差数列{a n}的公差为2,若a2,a4,a8成等比数列,则{a n}的前n项和S n=()A.n(n+1)B.n(n﹣1)C.D.18.已知S n是等差数列{a n}的前n项和,若a1+a3+a5=3,则S5=()A.5B.7C.9D.1119.等比数列{a n}的前n项和为S n,已知S3=a2+10a1,a5=9,则a1=()A.B.C.D.20.设等差数列{a n}的前n项和为S n,若S m﹣1=﹣2,S m=0,S m+1=3,则m=()A.3B.4C.5D.6数学数列单选答案1.数列{a n}为等差数列,且a2+a7+a12=6,则{a n}的前13项的和为()A.52B.C.26D.【分析】由等差数列的性质可求a7,然后代入到求和公式S==13a7可求.【解答】解:由等差数列的性质可知,a2+a7+a12=3a7=6,故a7=2,则{a n}的前13项的和S===13a7=26.故选:C.【点评】本题主要考查了等差数列的性质及求和公式的简单应用,属于基础试题.2.记S n为正项等比数列{a n}的前n项和.若a1=1,4a3=a5,则S10=()A.512B.511C.1023D.1024【分析】结合已知及等比数列的性质可求公比q,然后结合等比数列的求和公式即可求.【解答】解:由4a3=a5可得q2=4,∵q>0,所以q=2,由等比数列的求和公式可得,S10==1023.故选:C.【点评】本题主要考查了等比数列的求和公式及性质的简单应用,属于基础试题.3.数列{a n}是公差为2的等差数列,S n为其前n项和,且a1,a4,a13成等比数列,则S4=()A.8B.12C.16D.24【分析】运用等差数列的通项公式和等比数列的中项性质,解方程可得首项,再由等差数列的求和公式,计算可得所求值.【解答】解:数列{a n}是公差d为2的等差数列,S n为其前n项和,且a1,a4,a13成等可得a42=a1a13,即(a1+6)2=a1(a1+24),解得a1=3,则S4=4a1+6d=4×3+6×2=24.故选:D.【点评】本题考查等差数列的通项公式和求和公式的运用,等比数列的中项性质,考查方程思想和运算能力,属于基础题.4.已知等比数列{a n}满足a1=3,a1+a3+a5=21,则a3+a5+a7=()A.21B.42C.63D.84【分析】由已知,a1=3,a1+a3+a5=21,利用等比数列的通项公式可求q,然后在代入等比数列通项公式即可求.【解答】解:∵a1=3,a1+a3+a5=21,∴,∴q4+q2+1=7,∴q4+q2﹣6=0,∴q2=2,∴a3+a5+a7==3×(2+4+8)=42.故选:B.【点评】本题主要考查了等比数列通项公式的应用,属于基础试题.5.记S n为等差数列{a n}的前n项和.若a4+a5=24,S6=48,则{a n}的公差为()A.1B.2C.4D.8【分析】利用等差数列通项公式及前n项和公式列出方程组,求出首项和公差,由此能求出{a n}的公差.【解答】解:∵S n为等差数列{a n}的前n项和,a4+a5=24,S6=48,∴,解得a1=﹣2,d=4,∴{a n}的公差为4.【点评】本题考查等差数列公式的求法及应用,是基础题,解题时要认真审题,注意等差数列的性质的合理运用.6.已知等差数列{a n}中,a2=7,a4=15,则前10项的和S10=()A.100B.210C.380D.400【分析】由第二项和第四项的值可以求出首项和公差,写出等差数列前n项和公式,代入n=10得出结果.【解答】解:d=,a1=3,∴S10==210,故选:B.【点评】若已知等差数列的两项,则等差数列的所有量都可以求出,只要简单数字运算时不出错,问题可解.7.设{a n}是公差为正数的等差数列,若a1+a2+a3=15,a1a2a3=80,则a11+a12+a13=()A.120B.105C.90D.75【分析】先由等差数列的性质求得a2,再由a1a2a3=80求得d即可.【解答】解:{a n}是公差为正数的等差数列,∵a1+a2+a3=15,a1a2a3=80,∴a2=5,∴a1a3=(5﹣d)(5+d)=16,∴d=3,a12=a2+10d=35∴a11+a12+a13=105故选:B.【点评】本题主要考查等差数列的运算.8.已知数列{a n}满足3a n+1+a n=0,a2=﹣,则{a n}的前10项和等于()A.﹣6(1﹣3﹣10)B.C.3(1﹣3﹣10)D.3(1+3﹣10)【分析】由已知可知,数列{a n}是以﹣为公比的等比数列,结合已知可求a1,然后代入等比数列的求和公式可求【解答】解:∵3a n+1+a n=0∴∴数列{a n}是以﹣为公比的等比数列∵∴a1=4由等比数列的求和公式可得,S10==3(1﹣3﹣10)故选:C.【点评】本题主要考查了等比数列的通项公式及求和公式的简单应用,属于基础试题9.已知等差数列{a n}前9项的和为27,a10=8,则a100=()A.100B.99C.98D.97【分析】根据已知可得a5=3,进而求出公差,可得答案.【解答】解:∵等差数列{a n}前9项的和为27,S9===9a5.∴9a5=27,a5=3,又∵a10=8,∴d=1,∴a100=a5+95d=98,故选:C.【点评】本题考查的知识点是数列的性质,熟练掌握等差数列的性质,是解答的关键.10.记S n为等差数列{a n}的前n项和.若3S3=S2+S4,a1=2,则a5=()A.﹣12B.﹣10C.10D.12【分析】利用等差数列的通项公式和前n项和公式列出方程,能求出a5的值.【解答】解:∵S n为等差数列{a n}的前n项和,3S3=S2+S4,a1=2,∴=a1+a1+d+4a1+d,把a1=2,代入得d=﹣3∴a5=2+4×(﹣3)=﹣10.故选:B.【点评】本题考查等差数列的第五项的求法,考查等差数列的性质等基础知识,考查运算求解能力,考查函数与方程思想,是基础题.11.已知等比数列{a n}满足a1=,a3a5=4(a4﹣1),则a2=()A.2B.1C.D.【分析】利用等比数列的通项公式即可得出.【解答】解:设等比数列{a n}的公比为q,∵,a3a5=4(a4﹣1),∴=4,化为q3=8,解得q=2则a2==.故选:C.【点评】本题考查了等比数列的通项公式,属于基础题.12.等差数列{a n}的前m项和为30,前2m项和为100,则它的前3m项和为()A.130B.170C.210D.260【分析】利用等差数列的前n项和公式,结合已知条件列出关于a1,d的方程组,用m 表示出a1、d,进而求出s3m;或利用等差数列的性质,s m,s2m﹣s m,s3m﹣s2m成等差数列进行求解.【解答】解:解法1:设等差数列{a n}的首项为a1,公差为d,由题意得方程组,a1解得d=,a1=,∴s3m=3ma1+d=3m+=210.故选C.解法2:∵设{a n}为等差数列,∴s m,s2m﹣s m,s3m﹣s2m成等差数列,即30,70,s3m﹣100成等差数列,∴30+s3m﹣100=70×2,解得s3m=210.故选C.a1【点评】解法1为基本量法,思路简单,但计算复杂;解法2使用了等差数列的一个重要性质,即等差数列的前n项和为s n,则s n,s2n﹣s n,s3n﹣s2n,…成等差数列.13.等差数列{a n}的首项为1,公差不为0.若a2,a3,a6成等比数列,则{a n}前6项的和为()A.﹣24B.﹣3C.3D.8【分析】利用等差数列通项公式、等比数列性质列出方程,求出公差,由此能求出{a n}前6项的和.【解答】解:∵等差数列{a n}的首项为1,公差不为0.a2,a3,a6成等比数列,∴,∴(a1+2d)2=(a1+d)(a1+5d),且a1=1,d≠0,解得d=﹣2,∴{a n}前6项的和为==﹣24.故选:A.【点评】本题考查等差数列前n项和的求法,是基础题,解题时要认真审题,注意等差数列、等比数列的性质的合理运用.14.已知数列{a n}的前n项和为S n,a1=1,S n=2a n+1,则当n>1时,S n=()A.()n﹣1B.2n﹣1C.()n﹣1D.(﹣1)【分析】利用递推关系与等比数列的通项公式即可得出.【解答】解:∵S n=2a n+1,得S n=2(S n+1﹣S n),即3S n=2S n+1,由a1=1,所以S n≠0.则=.∴数列{S n}为以1为首项,公比为的等比数列∴S n=.故选:A.【点评】本题考查了递推关系与等比数列的通项公式,考查了推理能力与计算能力,属于中档题.15.已知等差数列{a n}满足a2+a4=4,a3+a5=10,则它的前10项的和S10=()A.138B.135C.95D.23【分析】本题考查的知识点是等差数列的性质,及等差数列前n项和,根据a2+a4=4,a3+a5=10我们构造关于基本量(首项及公差)的方程组,解方程组求出基本量(首项及公差),进而代入前n项和公式,即可求解.【解答】解:∵(a3+a5)﹣(a2+a4)=2d=6,∴d=3,a1=﹣4,∴S10=10a1+=95.故选:C.【点评】在求一个数列的通项公式或前n项和时,如果可以证明这个数列为等差数列,或等比数列,则可以求出其基本项(首项与公差或公比)进而根据等差或等比数列的通项公式,写出该数列的通项公式,如果未知这个数列的类型,则可以判断它是否与某个等差或等比数列有关,间接求其通项公式.16.已知各项均为正数的等比数列{a n},a1a2a3=5,a7a8a9=10,则a4a5a6=()A.B.7C.6D.【分析】由数列{a n}是等比数列,则有a1a2a3=5⇒a23=5;a7a8a9=10⇒a83=10.【解答】解:a1a2a3=5⇒a23=5;a7a8a9=10⇒a83=10,a52=a2a8,∴,∴,故选:A.【点评】本小题主要考查等比数列的性质、指数幂的运算、根式与指数式的互化等知识,着重考查了转化与化归的数学思想.17.等差数列{a n}的公差为2,若a2,a4,a8成等比数列,则{a n}的前n项和S n=()A.n(n+1)B.n(n﹣1)C.D.【分析】由题意可得a42=(a4﹣4)(a4+8),解得a4可得a1,代入求和公式可得.【解答】解:由题意可得a42=a2•a8,即a42=(a4﹣4)(a4+8),解得a4=8,∴a1=a4﹣3×2=2,∴S n=na1+d,=2n+×2=n(n+1),故选:A.【点评】本题考查等差数列的性质和求和公式,属基础题.18.已知S n是等差数列{a n}的前n项和,若a1+a3+a5=3,则S5=()A.5B.7C.9D.11【分析】由等差数列{a n}的性质,a1+a3+a5=3=3a3,解得a3.再利用等差数列的前n项和公式即可得出.【解答】解:由等差数列{a n}的性质,a1+a3+a5=3=3a3,解得a3=1.则S5==5a3=5.故选:A.【点评】本题考查了等差数列的通项公式及其性质、前n项和公式,考查了推理能力与计算能力,属于中档题.19.等比数列{a n}的前n项和为S n,已知S3=a2+10a1,a5=9,则a1=()A.B.C.D.【分析】设等比数列{a n}的公比为q,利用已知和等比数列的通项公式即可得到,解出即可.【解答】解:设等比数列{a n}的公比为q,∵S3=a2+10a1,a5=9,∴,解得.∴.故选:C.【点评】熟练掌握等比数列的通项公式是解题的关键.20.设等差数列{a n}的前n项和为S n,若S m﹣1=﹣2,S m=0,S m+1=3,则m=()A.3B.4C.5D.6【分析】由a n与S n的关系可求得a m+1与a m,进而得到公差d,由前n项和公式及S m=0可求得a1,再由通项公式及a m=2可得m值.【解答】解:a m=S m﹣S m﹣1=2,a m+1=S m+1﹣S m=3,所以公差d=a m+1﹣a m=1,S m==0,m﹣1>0,m>1,因此m不能为0,得a1=﹣2,所以a m=﹣2+(m﹣1)•1=2,解得m=5,另解:等差数列{a n}的前n项和为S n,即有数列{}成等差数列,则,,成等差数列,可得2•=+,即有0=+,解得m=5.又一解:由等差数列的求和公式可得(m﹣1)(a1+a m﹣1)=﹣2,m(a1+a m)=0,(m+1)(a1+a m+1)=3,可得a1=﹣a m,﹣2a m+a m+1+a m+1=+=0,解得m=5.故选:C.【点评】本题考查等差数列的通项公式、前n项和公式及通项a n与S n的关系,考查学生的计算能力.。

2020届中考语文总复习限时训练 微写作训练

2020届中考语文总复习限时训练 微写作训练

微写作训练(时间:30分钟)班级:姓名:1.根据下面的对话,代班长拟写一则通知。

(只写正文,不超过60字)班长:区科技图书馆正式对外开放了,周六的社会实践活动就去那儿吧。

团支书:好啊。

听说里面增设了八大行星体验展台,以前北京才有,难得一见。

班长:是啊,那里还可以借乐高玩具,凭个人借书卡每次可借10天。

这几天去的人很多。

团支书:那我们就去那里做志愿者吧,帮助工作人员整理图书,打扫卫生,维持秩序。

你看我们什么时候出发?班长:上午8点整,在学校门口集中出发。

团支书:好的。

写通知和点名的事就归你了。

2.校学生会准备出一期“献给母校”的专刊。

主编邀请你以一个初三毕业生的名义为“感恩”栏目写一段话,向在三年的学习生活中帮助和关爱过你的人表示感谢。

请你从班主任、任课教师、校医、保安、图书管理员中任选一个,表达感谢的心意。

要求:(1)内容要求真实具体,语言要准确得体。

(2)字数在150~200之间。

(3)不要出现所在学校的校名或师生姓名。

3.请给下面的新闻材料拟写标题,并简述理由。

××网2016年4月22日讯4月24日是首个中国航天日。

为迎接首个航天日,航天专家在国家博物馆学术报告厅作专题报告,总结了中国载人航天、火箭发射、深空探测等方面的航天成就,强调了中国进入太空的能力目前已经达到航天大国的水平,并指出“十二五”期间,中国每年的航天发射数量稳居世界前三,尤其值得所有中国人自豪的是,近20年中国长征火箭的发射成功率居全球之最。

4.班级拟开展“说教养”主题演讲活动,请参考下列材料,结合自己的认识,写一篇不超过200字的演讲稿。

材料一:教养是指一般文化和品德的修养。

有教养的人必具有良好的外在行为和美好的道德品质。

要完善个人修养,首先要致力于读书求学,提高自身的认知水平。

材料二:待人要谦虚,礼仪要得体,遇困境不气馁,获大奖不骄傲。

(傅雷)良好的教养不仅来自家庭和学校,而且可以得之于自身。

(利哈乔夫)材料三:周总理每次外出视察工作,离开当地时总是与身边服务人员一一握手道谢。

【高一英语】暑假作业限时训练有素20(附解析)

【高一英语】暑假作业限时训练有素20(附解析)

高一英语暑假作业20核心词汇积累核心单词1._____________受害人;牺牲者,牺牲品2._____________违反(法律、协议等),侵犯3._____________暴力,暴力行为4._____________美德,优点5._____________生死攸关的,重大的,生命的,至关重要的6.volunteer n. _____________v. _____________7.wander vi. _____________8.welfare n. _____________9.withdraw vt. _____________10.witness vt. _____________n. _____________重点短语_____________出现;开大(音量)_____________开小(音量);拒绝_____________翻转_____________交出;上交_____________求助于;翻到;转向_____________拿起,占用;开始从事_____________捡起;偶然学会;收听;(用车)接人_____________建立;增强我读我练1.I’m Li Hua, a student from Class 2, Grade 3.I’m writing to apply for the position as a student ____________(志愿者).2.Our city _____________(见证) four different seasons, plenty of sunshine and good rainfall, but in winter you may feel a little cold.3.It is wrong to see _____________ (violent) as the only way to solve problems.4.I was once ashamed to admit my _____________(weak), but now I have no such feeling.5.He _____________(拒绝) my suggestion and turned over the business to his daughter.6.Practising Chinese kung fu can not only ______________(增强体质), but also develop one’s character.精选语法小练阅读下面材料,在空白处填入适当的内容(1 个单词)或括号内单词的正确形式。

高一语文限时训练20

高一语文限时训练20

高一语文限时训练20一、基础夯实(一)阅读下面的文字,完成1——3题。

高考过后,考生和家长又围绕着志愿填报忙碌起来。

面对那么多学校和专业该如何选择,让一些考生和家长犯难,这也让帮助考生填报志愿的辅导市场——“大数据预测录取概率”、“专家一对一咨询”等填报服务,要价数千到数万。

天价志愿咨询的最大卖点,就是大数据。

据调查发现,很多机构商家宣传大数据来自教育部、考试院,但被相关部门否认。

事实上,全国还没有哪家单位或企业建立起真正的高考志愿“大数据库”。

所谓数据,真实的情况是,高校网站、媒体每年都会公布一些录取信息,还有一些其他公开渠道的数据。

商业机构把这些数据收集起来,进行整合,冒充来自特殊渠道的大数据。

众所周知,高考录取分数线的划定,受到许多不确定因素的影响。

某一区域内的高校招生计划、当地考生的志愿填报情况、高考试卷的难易程度等都有可能让分数线出现波动。

拿往年的“大数据”来套用今年的分数线,无异于。

至于商家倾情推销的“专家”,也未必靠谱,他们的眼里恐怕更多的是利益。

的事情是整顿市场乱象。

从媒体调查的情况来看,关于高考志愿填报辅导市场,不是傻子太多、骗子不够用,( )。

有关部门应及时出手,不能任由高考志愿填报市场野蛮生长。

此外,也应尽力拓展渠道,为考生提供正规、科学的填报志愿辅导服务。

1.依次填入文中划线处的成语,全都恰当的一项是()A.炙手可热五花八门刻舟求剑当务之急B.应运而生五花八门刻舟求剑迫在眉睫C.应运而生琳琅满目固步自封迫在眉睫D.炙手可热琳琅满目固步自封当务之急2.文中画波浪线的句子有语病,下列修改最恰当的一项是()A.调查发现,很多机构商家宣传的大数据来自教育部、考试院,但被相关部门否认。

B.据调查,很多机构商家宣传的大数据来自于教育部、考试院,但被相关部门否认。

C.据调查发现,很多机构商家宣传大数据来自教育部、考试院,但被相关部门所否认。

D.据调查发现,很多机构商家宣传大数据来自教育部、考试院,但相关部门予以否认。

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限时训练20Last January I was given a chance to work in another country. At first, I lived a 36 life, but soon I met a group of new friends and they were really nice.Then something 37 happened. One of our friends suffered from kidney stones (肾结石), which had to be 38 surgically as soon as possible. But it was not as 39 as that! Our friend could not 40 the operation, which put him into deep 41 . I didn’t know how to help him because I didn’t have any money either. 42 , I never lost hope. Instead I tried to think of other 43 and told him not to worry because everything would be all right.Before I left the hospital, I talked to the surgeon and asked him to do the surgery, saying I would pay the expenses the next day 44 I had that kind of money. Well, I must have been 45 because the surgeon agreed to do it! But how was I going to get that much money?All I knew was that I needed to help my friend! Although it could be 46 to deal with this kind of situation, I decided to make some efforts. I tried to borrow money but failed. I tried 47 my things but no one wanted them. Then I had the idea of talking to my 48 . I asked him if I could have my next two months’ salary in advance—without even thinking on how I would 49 for two months with no money. He was surprised and asked me the reason. I told him that I had a friend who was really in need. 50 I convinced him too and then he gave my two months’ salary without any 51 !Upon getting the money, I ran to the hosp ital and paid all the bills. My friend didn’t know how to thank me. I just said, “Come on! Give me a 52 !”Friendship may not only be found at the happy moment but sometimes you can find it on the 53 time in your life! I’ve done a good thing by bein g a (an) 54 friend and I think it is well worth the 55 .36. A. private B. colorful C. lonely D. normal37. A. strange B. wrong C. unexpected D. unfair38. A. lifted B. reduced C. examined D. removed39. A. terrible B. painful C. simple D. special40. A. stand B. consider C. perform D. afford41. A. confusion B. anxiety C.embarrassment D.dissatisfaction42. A. Thus B. Otherwise C. However D. Therefore43. A. reasons B. options C. symptoms D. expectations44. A. even if B. as if C. so that D. in case45. A. encouraging B. annoying C. astonishing D. convincing46. A. challenging B. disturbing C. amazing D. scaring47. A. saving B. handling C. donating D. selling48. A. colleague B. relative C. boss D. surgeon49. A. adapt B. survive C. continue D. wander50. A. Luckily B. Slightly C. Constantly D. Occasionally51. A. analysis B. dignity C. hesitation D. remark52. A. favor B. hug C. promise D. reply53. A. best B. longest C. strangest D. toughest54. A. aggressive B. sensitive C. supportive D. attentive55. A. effort B. sympathy C. pressure D. toleranceAThe common cold is the world’s most widespre ad illness, which is plagues(疫病) that flesh receives.The most widespread fallacy(谬误) of all is that colds are caused by cold. They are not. They are caused by viruses passing on from person to person. You catch a cold by coming into contact, directly or indirectly, with someone who already has one. If cold causes colds, it would be reasonable to expect the Eskimos to suffer from them forever. But they do not. And in isolated arctic regions explorers have reported being fre e from colds until coming into contact again with infected people from the outside world by way of packages and mail dropped from airplanes. During the First World War soldiers who spent long periods in the trenches(战壕), cold and wet, showed no increased tendency to catch colds.In the Second World War prisoners at the notorious Auschwitz concentration camp(奥斯维辛集中营), naked and starving, were astonished to find that they seldom had colds. At the Common Cold Research Unit in England, volunteers took part in Experiments in which they gave themselves to the discomforts of being cold and wet for long stretches of time. After taking hot baths, they put on bathing suits, allowed themselves to be with cold water, and then stood about dripping wet in drafty room. Some wore wet socks all day while others exercised in the rain until close to exhaustion. Not one of the volunteers came down with a cold unless a cold virus was actually dropped in his nose.If, then, cold and wet have nothing to do with catching colds, why are they more frequent in the winter?Despite the most pains-taking research, no one has yet found the answer. One explanation offered by scientists is that people tend to stay together indoors more in cold weather than at other times, and this makes it easier for cold viruses to be passed on.No one has yet found a cure for the cold. There are drugs and pa in suppressors(止痛片) such as aspirin, but all they do is relieve the symptoms.1. The writer offered _______ examples to support his argument.A. 4B. 5C. 6D. 32. Which of the following does not agree with the chosen passage?A. The Eskimos do not suffer from colds all the time.B. Colds are not caused by cold.C. People suffer from colds just because they like to stay indoors.D. A person may catch a cold by touching someone who already has one.3. Arctic explorers may catch colds when _______.A. they are working in the isolated arctic regionsB. they are writing reports in terribly cold weatherC. they are free from work in the isolated arctic regionsD. they are coming into touch again with the outside world4. V olunteers taking part in the experiments in the Common Cold Research Unit _______.A. suffered a lotB. never caught coldsC. often caught coldsD. became very strong5. The passage mainly discusses _______.A. the experiments on the common coldB. the fallacy about the common coldC. the reason and the way people catch coldsD. the continued spread of common coldsBAs the summer time approaches, more and more commercials and advertisements can be seen on the mass media encouraging students to join English study tours abroad. The purpose of these tours is to provide students with an opportunity to learn English in a native speaking environment and use English in real situations. Parents always believe that their children will automatically speak good English after attending these study tours. However, for me, as an English major and former participant of many English study tours, I find these summer tours ineffective.Parents who send their children abroad for a summer to study English usually believe that the native speaking environment, including the courses, teachers, and host families, can help their children to learn English better. This might be true if the students really study hard and they really get an all-English environment. But the fact is that, most of the time, the tour part is more important than the study part. Although these tours take place in foreign countries, students are rarely exposed in an all-English environment. You will find that, in the morning classes, Taiwanese students would like to stick together and speak Chinese with each other, even though there are some students from other countries. In the afternoon, the students go on a sightseeing excursion(观光游览) with their companions from Taiwan; they speak Chinese of course. In the evening, when students return to the host families, they would stay in their room and share the day with their roommate, another Taiwanese student. People who speak the same language with you is like a log in the ocean when living in a foreign country where you can not express yourself well, so that you would naturally stick with them. This is especially true with children. Another unrealistic expectation parents have for these tours is that the experience of living abroad can make their children more independent and mature. However, maturity does not come overnight. It takes time and practice. The most important of all is that they need a good mentor. None of these exist in the study tours.66. What’s the author’s attitude towards the English study tours abroad?A. Doubtful.B. Positive.C. Negative.D. Neutral.67. Which of the following is the most important thing for children who study in an English environment according to the writer?A. Host families.B. Hard work.C. Courses.D. Teachers.68. Why are students rarely exposed in an all-English environment?A. The study tour itself makes it less important for students to expose themselves.B. Staying with the students from the same country makes study easy abroad.C. They only want to stay with their English teachers.D. They are not sure about their spoken English.69. Which of the following words is closest in meaning to the underlined word“mentor”?A. companion adult, grown-up, matureB. bossC. doctorD. adviser70. What makes the parents send their children to join English study tours abroad according to the text?A. Parents’ misunderstanding to the ads.B. Parents’ lack of information.C. Parents’ unrealistic expectation.D. Parents’ social experience.CThe London Organizing Committee of the Olympic Games announced the details of the lighting of the Olympic Flame and travel around the UK to be started in May.Organized by the Hellenic Olympic Committee (HOC)(希腊奥委会),the Olympic Flame will be ignited mid-morning on May 10 using the sun's rays at the Temple of Hera in Olympia,in an hour-long ceremony among the historic ruins of the home of the ancient Games.From the Lighting Ceremony in Olympia, the HOC will take the Olympic Flame on an 8-day relay around Greece before arriving in Athens on May 17 for the official handover ceremony. During an early evening hour-long ceremony, the Olympic Flame will be handed over to a London 2012 representative.British Airways, the official airline partner of the London 2012 Games, will carry the Olympic Flame on board flight “BA2012”一a golden aircraft from Athens into the UK on May 18. The Flame will arrive into the UK at Royal Naval Air Station Culdrose on early evening of May 18. On the morning of May 19,the Flame will then travel from RNAS to Land's End ready for the start of the London 2012 Olympic Torch Relay.The Olympic Flame will travel 8,000 miles across the UK to within 10 miles of 95 percent of the population^ 8,000 torchbearers (火炬手)will carry the Olympic Flame for approximately 300 meters each, passing the Olympic Flame from torch to torch until it reaches the Olympic Stadium on the eve of 27 July.56. The underlined word “ignited” in Paragraph 2 is closest in meaning to “____”.A. changedB. replacedC. litD. kept57. The official handover ceremony of the Olympic Flame will last ____.A. an hourB. one eveningC. a weekD. 8 days58. We learn from the text that ____.A. the Olympic Flame relay will take place only in the UKB. the 2012 London Olympic torch relay will last over three monthsC. the Olympic Flame will travel from Greece to the UK by landD. most people in the UK needn't travel far to see the Olympic Flame59. What is the best title for the text?A. Welcome to enjoy the 2012 London Olympic torch relayB. Olympic Flame lighting and UK arrival details announcedC. The hard journey of Olympic Flame from home to LondonD. A brief introduction to the history of Olympic Flame relay36-40 CCDCD 41-45 BCBBD 46-50 ADCBA 51-55CBDCABCDAC CBADC CADB。

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