高二年级第二学期期中练习 2017 4
2017年春高二下学期期中考(含答案)
2017年春高二下学期期中考班级:姓名: 号数: 难度: 成绩:一、选择题:本题共12小题,每小题4分,共48分.在每小题给出的四个选项中,第1~8题只有一项符合题目要求,第9~12题有多项符合题目要求.全部选对的得4分,选对但不全的得2分,有选错的得0分.1、某用电器两端的正弦交变电压的表达式为311sin100u t V π=().关于这个交变电压,下列说法中正确的是( )A .有效值为311VB .有效值为440VC .最大值为311VD .最大值为440V 2、如图所示,理想变压器原线圈的匝数11000n =匝,副线圈的匝数2200n =匝.原线圈两端所加的电压1220U V =时,副线圈两端的电压2U 为( )A .1100VB .44VC .440VD .22V 3、关于下图i t -函数图像,下列说法中正确的是( )A .该函数图像表示的是直流电B .该电流的周期是0.01sC .用电流表测量该电流得到的值为5AD .用电流表测量该电流得到的值为 4、关于磁通量,下列说法中正确的是( )A .过某一平面的磁通量为零,该处磁感应强度不一定为零B .磁通量不仅有大小,而且有方向,所以是矢量C .磁感应强度越大,磁通量越大D .磁通量即使磁感应强度5、面积是20.50m 的导线环,处于磁感应强度为24.010T -⨯的匀强磁场中,环面的法向量与磁场夹角为30︒,穿过导线环的磁通量等于( )A .22.510Wb -⨯B .21.010Wb -⨯C .21.510Wb -⨯D .2410Wb -⨯6、在图甲所示的电路中,理想变压器原线圈两端的正弦交变电压变化规律如图乙所示.已知变压器原、副线圈的匝数比12:10:1n n =,串联在原线圈电路中电流表1A 的示数为1A ,下列说法正确的是( )A .电压表V 的示数为2002VB .变压器的输出功率为20WC .100HzD .电流表2A 的示数为10A 7、对于正弦式交流电,下列说法正确的是( ) A .电流在一周期内方向改变两次,大小随时间变化 B .电流在一周期内方向改变一次,大小随时间变化C .线圈在中性面时穿过线圈的磁通量最大,电动势最大D .线圈在垂直于中性面的位置磁通量为零,电动势为零8、理想变压器在正常工作时,原、副线圈中不一定相同的物理量是( ) A .交变电流的频率 B .原线圈的磁通量变化率和副线圈的磁通量变化率C .原线圈的输入功率和副线圈的输出功率D .原线圈的感应电动势和副线圈的感应电动势9、将4Ω的电阻接到内阻不计的交流电源上,该电源电动势e 随时间t 变化规律如图所示,下列说法中正确的是( )A .电路中交变电流的频率为2.5HzB .2C .电阻消耗的电功率为2WD .10、理想变压器原线圈两端电压不变,当副线圈电路中的电阻减小时,一下说法正确的是( ) A .输出电流增大,输入电流减小 B .输出电流增大,输入电流增大 C .输出电压保持不变 D .输出功率和输入功率都增大 11、下列关于电磁感应现象的说法中,正确的是( )A .磁通量变化率越大感应电动势就越大B .导体相对磁场运动,导体内一定产生感应电流C .感应电动势与匝数无关D .磁通量为零,感应电动势不一定为零 12、如图,当通电直导线MN 中的电流突然增大时(方向未知),则可确定的是( )A .线框中感应电流的方向B .线框各边受磁场力的方向C .线框整体受磁场力的方向D .线框中电流方向、受磁场力的方向皆不可确定二、简答题(共12分,每题6分)1、什么是电磁感应现象?产生感应电流的条件是什么?/B T 2、简述牛顿三大定律的基本内容三、计算题(共40分,每题10分)1、输送4400kW 的电功率,采用110kV 高压输电,输电导线中的电流是多少?如果用110V 电压输电,输电导线中的电流将是多少?若输电线阻值为10Ω,则电功率的损耗分别为多少?你能得出什么结论?2、如图1所示,一个单匝矩形线圈长10.2L m =,宽20.1L m =,匀强磁场垂直线圈平面向里.磁感强度B 随时间t 变化的规律如图2所示.求 1)当1t s =时,穿过矩形线圈中的磁通量; 2)线圈中感应电动势,并画出感应电流方向./t s3、如图所示,MN 、PQ 是两根足够长的光滑平行金属导轨,导轨间距为l ,导轨所在的平面与水平面夹角为 ,M 、P 间接阻值为R 的电阻.匀强磁场的方向与导轨所在平面垂直,磁感应强度大小为B .质量为m 、阻值为r 的金属棒放在两导轨上,在平行于导轨的拉力作用下,以速度v 匀速向上运动.已知金属棒与导轨始终垂直并且保持良好接触,导轨阻值不计,重力加速度为g ,求: 1)金属棒产生的感应电动势E ;2)通过电阻R 的电流I ; 3)拉力F 的大小. 4、如右图所示,质量为m 带电量为q 的粒子在只受电场力的作用下,从A 点静止开始运动,沿直线运动到B 点。
2017年下学期高二年级期中考试题
2017年下学期高二年级期中考试题历史(文)时量:60分钟满分:100分命题人:杨询一、选择题(满分60分,每题2分。
)1.某位先秦思想家曾有如下评论:齐桓公致力连击诸侯,设法合作维持传统封建秩序,是个执行正道而不行诈谋的国君;晋文公虽协助周王定乱,却僭越礼制,想仿行天子礼仪,是个内行诈谋却装作遵循正道的国君。
该思想家应属于A.儒家B.法家C.道家D.墨家2.“桂可食,故伐之;漆可用,故割之。
人皆知有用之用,而莫知无用之用也。
”(引自《庄子·人间世》)下列与此主张同属于一个思想流派的是A.“心外无物,心外无理”B.“不期修古,不法常可”C.“人之性恶,其善者伪也”D.“民之难治,以其上之有为,是以难治”3.有学者认为:“法家之学在先秦诸子中是最为新颖的,但亦有落伍之处。
国家和社会的利益,只是在一定的限度内是一致的,过此以往,便相冲突。
法家不知此义,误以为两者始终是一致的。
所以说法家的思想,也是落伍的。
”这一认识的主旨是A.揭示了法家国家与社会利益并重的实质B.批判了法家的专制思想C.辩证分析了法家在统一国家中的作用D.否定了法家的进步意义4.《春秋繁露卷一·楚庄王第-》日:“由此观之,正朔(帝王新颁历法)、服色之改,受命应天制礼作乐之异,人心之动也,二者离而复合,所为一也。
”该主张的主要目的是A.提出天行有常探究客观规律B.强调纲常名教维护等级秩序C.宣扬天人感应加强君主专制D.主张知行合一重建儒家信仰5.周敦颐曾论及“孔颜乐处”,他说:“颜子‘一箪食,一瓢饮,在陋巷,人不堪其忧而不改其乐’。
夫富贵,人所爱也;颜子不爱不求,而乐乎贫者,独何心哉?天地间有至贵至爱可求而异乎彼者,见其大而忘其小焉尔。
见其大则心泰,心泰则无不足;无不足,则富贵贫贱,处之一也。
”可见理学对“乐”的追求是A.心外无理,心外无乐B.存天理、灭人欲C.格物致知,格苦致乐D.精神超越,身心和谐6.顾炎武指责李贽“无忌惮而敢于叛圣人”,黄宗羲谴责李贽“非名教之所能羁络”,王夫之以“循礼”“秉礼”为“君子之道”,以“尊其尊、卑其卑、位其位”为理想政治局面,并强烈主张严君子、小人之辨。
潍坊市2016-17学年高二下学期4月期中考试试题数学文
2016-2017学年度第二学期普通高中模块监测高二数学 (文) 2017.4本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.试卷共4页,满分150分,考试时间120分钟. 山东省中学联盟提供 注意事项:1.答题前,考生在答题卡上务必将自己的姓名、准考证号涂写清楚.2.第Ⅰ卷,每小题选出答案后,用2B 铅笔把答题卡上对应题目的答案标号涂黑,如需改动,用橡皮擦干净后,再选涂其他答案标号,在试题卷上作答无效.一、选择题:共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.若复数i z 2-3=,则z 的共轭复数z 为A. i 23+-B. i 23--C.23i -+D.32i + 2.用反证法证明命题“设a ,b 为实数,则方程03=++b ax x 至少有一个实根”时,要做的假设是A .方程03=++b ax x 没有实根 B .方程03=++b ax x 至多有一个实根 C .方程03=++b ax x 至多有两个实根 D .方程03=++b ax x 恰好有两个实根 3.已知函数)(x f y =,下列说法错误..的是 A.00()()y f x x f x ∆=+∆-叫函数值的改变量 B.xx f x x f x y ∆-∆+=∆∆)()(00叫该函数在[x x x ∆+00,]上的平均变化率 C.)(x f 在点0x 处的导数记为y ' D.)(x f 在点0x 处的导数记为)(0x f ' 4.以下说法错误..的是 A.推理一般分为合情推理和演绎推理B.归纳是从特殊到一般的过程,它属于合情推理C.在数学中,证明命题的正确性既能用演绎推理又能用合情推理D.演绎推理经常使用的是由大前提、小前提得到结论的三段论推理 5.某产品的广告费用x (万元)与销售额y (万元)的统计数据如下表:根据上表可得回归直线方程a x yˆ7ˆ+=,若广告费用为10万元,则预计销售额为 A.73万元B.73.5万元C.74万元D.74.5万元6.某品牌电动汽车的耗电量y 与速度x 之间满足的关系式为:),0(402393123>--=x x x x y 为使耗电量最小,则速度为A .30B .40C .50D .607.以下式子正确的个数是:①21)'1(x x =②x x sin )'(cos -= ③2ln 2)'2(xx = ④ 10ln 1)'(lg x x -= A .1个 B.2个 C .3个 D .4个8.已知函数x x x f +=ln )(,则曲线)(x f 在点))1(1(f P ,处的切线与两坐标轴围成的三角形的面积为A .14 B .12C .1D .2 9.甲、乙、丙三位同学被问到是否去过济南、潍坊、青岛三个城市时,甲说:我去过的城市比乙多,但没去过潍坊;乙说:我没去过青岛;丙说:我们三人去过同一城市;由此可判断乙去过的城市为A.济南B.青岛C.济南和潍坊D.济南和青岛 10.函数)(x f 的定义域为R ,导函数)('x f 的图象如右图所示,则函数)(x fA .无极大值点,有四个极小值点B .有1个极大值点,两个极小值点C .有两个极大值点,两个极小值点D .有四个极大值点,无极小值点11.古希腊著名的毕达哥拉斯学派把1、3、6、10 、15、… 这样的数称为“三角形数”,而把1、4、9、16 、25、… 这样的数称为“正方形数”. 从下图中可以发现,任何一个大于1 的“正方形数”都可以看作两个相邻“三角形数”之和.下列等式中,符合这一规律的是4=1+3 9=3+6 16=6+10A. 13316+=B. 16925+=C. 261036+=D. 282149+==-'-'+-+'∈++=)2017(2017)2016()2016()()(),,(1sin )(123f f f f x f x f R b a bx x a x f )(则的导函数,为、已知函数A .2017B .2016C .2D .0第Ⅱ卷 (非选择题 共90分)二、填空题(本大题共4小题,每小题5分,共20分.请把正确答案填在题中横线上.) 13.已知a 是函数x x x f 12)(3-=的极大值点,则=a .14.已知圆的方程是222r y x =+,则经过圆上一点),(00y x M 的切线方程为200r y y x x =+.类比上述性质,可以得到经过椭圆12222=+by a x 上一点P (x 0,y 0)的切线方程为__________.15.欧拉公式cos sin xie x i x =+(i 为虚数单位)是由瑞士著名数学家欧拉发明的,它将指数函数的定义域扩大到复数,建立了三角函数和指数函数的关系,它在复变函数论里占有非常重要的地位,被誉为“数学中的天桥”,根据欧拉公式可知,3i e 表示的复数在复平面中位于 象限.16.对于函数有如下结论:x x x f ln )(= ①该函数为偶函数; ②若()0=2f x ',则0x e =; ③其单调递增区间是),1[+∞e; ④值域是),1[+∞e; ⑤该函数的图像与直线ey 1-=有且只有一个公共点.(本题中e 是自然对数的底数)其中正确的是_____________.(请把正确结论的序号填在横线上)三、解答题(共6小题,满分70分)解答应写出文字说明,证明过程或演算步骤. 17.(本小题满分10分)21()ln 8f x x x =+已知,(Ⅰ)求曲线()f x 在1x =处的切线方程;(Ⅱ)设P 为曲线()f x 上的点,求曲线C 在点P 处切线的斜率的最小值及倾斜角α的取值范围.18.(本小题满分12分)为了解少年儿童的肥胖是否与常喝碳酸饮料有关,现对100名六年级学生进行了问卷调查得到如图联表.且平均每天喝500ml 以上为常喝,体重超过50kg 为肥胖. 已知在全部100人中随机抽取1人,抽到肥胖的学生的概率为0.8. (1)求肥胖学生的人数并将上面的列联表补充完整; (2)是否有95%的把握认为肥胖与常喝碳酸饮料有关?说明你的理由.附:参考公式:22112212211212()n n n n n n n n n χ++++-=19.(本小题满分12分)(Ⅰ)已知ABCD 是复平面内的平行四边形,并且A ,B ,C 三点对应的复数分别是i +3,i i ---1,2,求D 点对应的复数.(Ⅱ)已知复数,2121i Z Z Z ==,并且|,|||,22||21z z z z z -=-=求z.20.(本小题满分12分)已知函数xe xf bx +=)(过点),1(e .(Ⅰ)求()y f x =的单调区间; (Ⅱ)当0x >时,求()f x x的最小值; (Ⅲ)()0f x mx m R m -=∈试判断方程(且为常数)的根的个数.注意:选修部分需要考生选做2道题.要求考生从21和22选一个题解答;然后从23和24中选一个题解答.如果多做,则按所做的前2个题计分.作答时请在答题纸上写清题号. 山东省中学联盟提供21.(本小题满分12分)选修4-4:坐标系与参数方程 在直角坐标系xoy 中,圆C 的参数方程1cos (sin x y ϕϕϕ=+⎧⎨=⎩为参数).以原点O 为极点,x 轴的正半轴为极轴,建立极坐标系.(I )求圆C 的普通方程和极坐标方程; (II )射线OM :4πθ=与圆C 的交于O 、P 两点,求P 点的极坐标.22.(本小题满分12分)选修4-5:不等式选讲设函数()3f x x a x =-+,其中0a >.(Ⅰ)当1a =时,求不等式()32f x x ≥+的解集; (Ⅱ)若不等式()0f x ≤的解集为{}|1x x ≤- ,求a 的值.23.(本小题满分12分)选修4-4:坐标系与参数方程在直角坐标系xOy 中,以原点O 为极点,x 轴的正半轴为极轴,建立极坐标系.已知曲线1C :4cos ,3sin ,x t y t =+⎧⎨=-+⎩ (t 为参数), 2C :6cos ,2sin ,x y θθ=⎧⎨=⎩(θ为参数). (Ⅰ)化1C ,2C 的方程为普通方程,并说明它们分别表示什么曲线; (Ⅱ)若1C 上的点P 对应的参数为2t π=-,Q 为2C 上的动点,求线段PQ 的中点M 到直线3:cos sin 8C ρθθ=+. 24.(本题满分12分)选修4—5 不等式选讲已知不等式|2||2|18x x ++-<的解集为A . (Ⅰ)求集合A ;(Ⅱ)若,a b A ∀∈,(0,)x ∈+∞,不等式4a b x m x+<++ 恒成立,求实数m 的取值范围.2016-2017学年第二学期普通高中模块高二文科数学答案2017.4一、选择(60分)1——5 DACCB 6——10 BBAAC 11——12 DC 二、填空(20分) 13、 -2 14、12020=+b yy a x x15、二 16、②③⑤ 三、解答(共70分)+111(1).1,288x f ∞==⋯⋯⋯17.本题满分10分.解:(1)函数的定义域是(0,).当时,故切点坐标是().分又11'4y x x =+ 则曲线()f x 在1x =处的切线的斜率是1151444k =+=⋯⋯⋯分 此时的切线方程是:10x-8y-9=0 ………5分(2)设曲线()f x 的切线的斜率为k,11'21,74tan 1,801042k y x x ππααπα==+≥=⋯⋯⋯≥⋯⋯⋯≤≤∴≤<⋯⋯⋯则分即分分18、山东省中学联盟提供 解:(1………2分………6分(2)将2×2列联表中的数据代入公式计算,得22112212211212()n n n n n n n n n χ++++-==100×(60×10-20×10)270×30×80×20 ………8分 =10021≈4.762.………10分由于4.762>3.841,所以有95%的把握认为肥胖与常喝碳酸饮料有关.………12分19.(1)解:由题意点A B C 、、的坐标分别为3,1,02,11---()(,)(,),…………1分设点D 坐标为(,)x y ,由中点坐标公式知3(1)0,1(1)2,x y +-=++-=-+ ∴2,2x y ==, ………3分 .∴D 点对应的复数为22i +.………4分(2)解:设,bi a z +=因为22|z |=所以2222=+b a ① ………5分 又因为i z 2,2z |,z -z ||z -z |2121===…………………………………………………….…6分 所以|)2(||2|i b a bi a -+=+-即2222)2()2(-+=+-b a b a ②……………………………………………………….…7分 由②得b a =代入①得2±=a ……………………………………………………….…9分 所以2==b a 或2-==b a ……………………………………………………….…10分 所以i z 22+=或i z 22--= ……………………………………………………….…12分 20.(Ⅰ)解:易知函数的定义域是{|0}x x ≠.,0()1?x e b e f x x==∴=⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯1+b 由题意知,f(1)=e 可得分2e e '()x x x f x x -==2(1)x e x x -……………………………………………………….…2分'()01f x x ≥≥令得: ;'()0001f x x x <<<<令得:或 ; 所以 ()y f x =的单调递增区间是: [1,)+∞,()y f x =的单调递减区间是:(,0),(0,1)-∞ ………………………………….…4分(Ⅱ)解:设2()e ()(0)x f x g x x x x ==>,则24e (2)'()x x x g x x-=. 令24e (2)'()0x x x g x x -==,解得2x =. …………6分 x 在(0,)+∞上变化时,'(),()g x g x 的变化情况如下表)故 当2x =时,()g x 取得最小值2e 4.………7分所以 当0x >时,()g x 的最小值是2e 4.……8分(Ⅲ)解:由题意可得:()()f x m g x x==, <易知x 0时,'()0g x >,所以g(x)在(,0)-∞单调递增,……9分 如图;原题转化为直线y=m 和曲线y=g(x)的交点个数问题。
2017—2018学年第二学期高二年级期中考试数学(理)试卷解析版
2017~2018学年第二学期高二年级期中考试数学(理)试卷一、选择题(本大题共12个小题,每小题5分,共60分,在每小题给出的四个选项中,只有一项是符合题目要求的)1.在复平面内,复数ii+310对应的点的坐标为( A )A .)3,1(B .)1,3(C .)3,1(-D .)1,3(-2.已知随机变量ξ服从正态分布),(2σμN ,若15.0)6()2(=>=<ξξP P ,则=<≤)42(ξP ( B )A .0.3B .0.35C .0.5D .0.7 3.设)(x f 在定义域内可导,其图象如图所示,则导函数)('x f 的图象可能是( B )4.用反证法证明命题:“若0)1)(1)(1(>---c b a ,则c b a ,,中至少有一个大于1”时,下列假设中正确的是( B )A .假设c b a ,,都大于1B .假设c b a ,,都不大于1C .假设c b a ,,至多有一个大于1D .假设c b a ,,至多有两个大于15.用数学归纳法证明3)12(12)1()1(2122222222+=+++-++-+++n n n n n 时,从)(*N k k n ∈=到1+=k n 时,等式左边应添加的式子是( B )A .222)1(k k +- B .22)1(k k ++ C .2)1(+k D.]1)1(2)[1(312+++k k6.3名志愿者完成4项工作,每人至少1项,每项由1人完成,则不同的安排方式共有( D )A .12种B .18种C .24种D .36种 7.在62)12(xx -的展开式中,含7x 的项的系数是( D ) A .60 B .160 C .180 D .2408.函数xe xf x2)(=的导函数是( C )A .xe xf 2'2)(= B .x e x f x 2'2)(= C .22')12()(x e x x f x -= D .22')1()(x e x x f x -=9.已知函数223)(a bx ax x x f +++=在1=x 处的极值为10,则数对),(b a 为( C )A .)3,3(-B .)4,11(-C .)11,4(-D .)3,3(-或)11,4(-10.若等差数列}{n a 公差为d ,前n 项和为n S ,则数列}{n S n 为等差数列,公差为2d.类似,若各项均为正数的等比数列}{n b 公比为q ,前n 项积为n T ,则等比数列}{n n T 公比为( C )A.2q B .2q C.q D.n q 11.将3颗骰子各掷一次,记事件A 表示“三个点数都不相同”,事件B 表示“至少出现一个3点”,则概率=)|(B A P ( C )A.21691 B.185 C.9160 D.2112.定义在R 上的偶函数)(x f 的导函数为)('x f ,若对任意实数x ,都有2)()(2'<+x xf x f 恒成立,则使1)1()(22-<-x f x f x 成立的实数x 的取值范围为( B )A .}1|{±≠x xB .),1()1,(+∞--∞C .)1,1(-D .)1,0()0,1( - 二、填空题(本大题共4小题,每小题5分,共20分)13.设),(~p n B ξ,若有4)(,12)(==ξξD E ,则=p 2/3 14.若函数32)1(21)(2'+--=x x f x f ,则=-)1('f -1 15.如图所示,阴影部分的面积是 32/316.已知函数)(x f 的定义域为]5,1[-,部分对应值如下表,)(x f 的导函数)('x f y =的图象如图所示,给出关于)(x f 的下列命题:②函数)(x f 在]1,0[是减函数,在]2,1[是增函数; ③当21<<a 时,函数a x f y -=)(有4个零点;④如果当],1[t x -∈时,)(x f 的最大值是2,那么t 的最小值为0. 其中所有正确命题是 ①③④ (写出正确命题的序号).三、解答题(本大题共6小题,共70分) 17.(本小题满分10分)设复数i m m m m z )23()32(22+++--=,试求实数m 的取值,使得 (1)z 是纯虚数; (2)z 对应的点位于复平面的第二象限. 解:(1)复数是一个纯虚数,实部等于零而虚部不等于0分5302303222 =∴⎪⎩⎪⎨⎧≠++=--m m m m m (2)当复数对应的点在第二象限时,分103102303222<<-∴⎪⎩⎪⎨⎧>++<--m m m m m 18.(本小题满分12分) 在数列}{n a 中,已知)(13,2*11N n a a a a n nn ∈+==+(1)计算432,,a a a 的值,并猜想出}{n a 的通项公式; (2)请用数学归纳法证明你的猜想. 解:(1)72123213112=+⨯=+=a a a ,19213,132********=+==+=a a a a a a于是猜想出分5562-=n a n (2)①当1=n 时,显然成立;②假设当)(*N k k n ∈=时,猜想成立,即562-=k a k 则当1+=k n 时,5)1(6216215623562131-+=+=+-⨯-=+=+k k k k a a a k k k , 即当1+=k n 时猜想也成立. 综合①②可知对于一切分12562,*-=∈n a N n n 19.(本小题满分12分)“莞马”活动中的α机器人一度成为新闻热点,为检测其质量,从一生产流水线上抽取20件该产品,其中合格产品有15件,不合格的产品有5件.(1)现从这20件产品中任意抽取2件,记不合格的产品数为X ,求X 的分布列及数学期望; (2)用频率估计概率,现从流水线中任意抽取三个机器人,记ξ为合格机器人与不合格机器人的件数差的绝对值,求ξ的分布列及数学期望. 解:(1)随机变量X 的可能取值为0,1,23821)0(22021505===C C C X P ,3815)1(22011515===C C C X P , 191)2(22001525===C C C X P , 所以随机变量X 的分布列为:分62192381380 =⨯+⨯+⨯=∴EX(2)合格机器人的件数可能是0,1,2,3,相应的不合格机器人的件数为3,2,1,0.所以ξ的可能取值为1,3,有题意知:1122213331319(1)()()()()444416P C C ξ==+=,3333331317(3)()()()()444416P C C ξ==+= 所以随机变量ξ的分布列为:分128163161)( =⨯+⨯=∴ξE 20.(本小题满分12分)编号为5,4,3,2,1的五位学生随意入座编号为5,4,3,2,1的五个座位,每位学生坐一个座位.设与座位编号相同的学生人数是X .(1)试求恰好有3个学生与座位编号相同的概率)3(=X P ; (2)求随机变量X 的分布列及均值.解:(1)恰好有3个学生与座位编号相同,这时另两个学生与座位编号不同,所以分412112010)3(5525 ====A C X P(2)随机变量X 的一切可能值为0,1,2,3,4,5. 且121)3(,00)4(,120112011)5(5555=========X P A X P A X P ; 83120459)1(,61120202)2(55155525========A C X P A C X P301112044)]5()4()3()2()1([1)0(===+=+=+=+=-==X P X P X P X P X P X P 随机变量X 的分布列为故分1211205041236281300)( =⨯+⨯+⨯+⨯+⨯+⨯=X E 21.(本小题满分12分)已知函数)(ln )(R a x ax x f ∈+=(1)若2=a ,求曲线)(x f y =在1=x 处的切线方程; (2)求)(x f 的单调区间;(3)设22)(2+-=x x x g ,若对任意),0(1+∞∈x ,均存在]1,0[2∈x ,使得)()(21x g x f <,求a 的取值范围. 解:(1)2),0(1)('=>+=a x x a x f )0(12)('>+=∴x xx f , 3)1('=∴f , 3=∴k又切点)2,1(,所以切线方程为)1(32-=-x y ,即:013=--y x 故曲线)(x f y =在1=x 处切线的切线方程为分4013 =--y x(2))0(11)('>+=+=x xax x a x f ①当0≥a 时,0)('>x f ,所以)(x f 的单调递增区间为分6),0( +∞②当0<a 时,由0)('=x f ,得ax 1-= 在区间)1,0(a -上0)('>x f ,在区间),1(+∞-a上,0)('<x f . 所以,函数)(x f 的单调递增区间为)1,0(a -,单调递减区间为分8),1( +∞-a(3)由已知,转化为]1,0[,1)1()(,)()(2max max ∈+-=<x x x g x g x f ,2)(max =∴x g 由(2)知,当0≥a 时,)(x f 在),0(+∞上单调递增,值域为R ,故不符合题意. (或者举出反例:存在23)(33>+=ae e f ,故不符合题意.)当0<a 时,)(x f 在)1,0(a -上单调递增,在),1(+∞-a上单调递减, 故)(x f 的极大值即为最大值,)ln(1)1()(max a af x f ---=-=, 所以2)ln(1<---a ,解得31e a -< 综上:分1213 ea -< 22.(本小题满分12分) 已知函数2()ln(1)f x ax x =++ (1)当14a =-时,求函数()f x 的极值; (2)若函数()f x 在区间[1)+∞,上为减函数,求实数a 的取值范围 (3)当[0)x ∈+∞,时,不等式()f x x ≤恒成立,求实数a 的取值范围. 解:(1))1()1(2)1)(2(1121)('->+-+-=++-=x x x x x x x f 令0)('>x f 得11<<-x ,令0)('<x f 得1>x .)(x f ∴在)1,1(-上是增函数,在),1(+∞上是减函数. 2ln 41)1()(+-==∴f x f 极大值,)(x f 无极小值分4(2)因为函数)(x f 在区间[1)+∞,上为减函数, 所以0112)('≤++=x ax x f 对任意的),1[+∞∈x 恒成立, 即)1(21+-≤x x a 对任意的),1[+∞∈x 恒成立,4121)211(2121)21(21)1(2122-=-+-≥-+-=+-x x x分841-≤∴a(3)因为当[0)x ∈+∞,时,不等式()f x x ≤恒成立, 即0)1ln(2≤-++x x ax 恒成立,令)0()1ln()(2≥-++=x x x ax x g , 转化为0)(max ≤x g 即可.1)]12(2[1112)('+-+=-++=x a ax x x ax x g 当0=a 时,1)('+-=x x x g ,0>x ,0)('<∴x g 即)(x g 在),0[+∞上单调递减,故0)0()(=≤g x g 成立. 当0>a 时,令0)('=x g 得,0=x 或121-=ax 若0121≤-a 即21≥a 时,),0(+∞∈x 有0)('>x g , 则)(x g 在),0[+∞上单调递增,0)0()(=≥g x g ,不满足题设; 若0121>-a 即210<<a 时,)121,0(-∈a x 有0)('<x g ,),121(+∞-∈ax 有0)('>x g , 则)(x g 在)121,0(-a 上单调递减,在),121(+∞-a上单调递增,无最大值,不满足题设; 当0<a 时,0>x ,0)('<∴x g即)(x g 在),0[+∞上单调递减,故0)0()(=≤g x g 成立. 综上:实数a 的取值范围为分12]0,( -∞。
2017北京四中高二(下)期中数学(理)含答案
方法,转化为求函数最值处理.也可构造新函数然后利用导数来求解.注意利用数形结合的数学思想方法.
二、填空题:本大题共 6 小题,每小题 5 分,共 30 分.
11.【答案】
【解析】
.
12.【答案】
【解析】因为由题意得:所求封闭图形的面积为
。
13.【答案】(1,0)或(-1,-4)
【解析】函数求导,
,令
A. [-1,+∞) B. (-1,+∞) C. (-∞,-1] D. (-∞,-1)
20. 观察( )'=- ,(x3)'=3x2,(sinx)'=cosx,由归纳推理可得:若函数 f(x)在其定义域上满足 f(-x)=-f
(x),记 g(x)为 f(x)的导函数,则 g(-x)=
A. -f(x) B. f(x) C. g(x) D. -g(x)
3. 曲线 y=x·ex 在 x=1 处切线的斜率等于 A. 2e B. e C. 2 D. 1
Байду номын сангаас4.
等于
A. -21n 2 B. 21n 2 C. -ln 2 D. ln 2 5. 函数 f(x)=3+x lnx 的单调递增区间为
A. (0, ) B. (e,+∞) C. ( ,+∞)
D. ( ,e]
+∞)上恰有 2 个零点,求实数 a 的取值范围.
3 / 12
2017 北京四中高二(下)期中数学(文)
参考答案
一、选择题:本大题共 10 小题,每小题 5 分,共 50 分. 1.
【答案】D
【解析】
,故选 D.
2. 【答案】B
3. 【答案】A
【解析】
人教版2017高二(下学期)数学期中试题(理)附答案
一、单选题(每小题只有一个正确答案,共12小题,每题5分,共60分)1.已知函数f(x)=的定义域为M,g(x)=lnx的定义域为N,则M∩N=()A.{x|x>﹣1}B.{x|x<1}C.{x|0<x<1}D.∅2.复数z=(1+bi)(2+i)是纯虚数,则实数b=()A.﹣2 B.﹣ C.D.23.在等比数列{a n}中,已知a3=6,a3+a5+a7=78,则a5=()A.12 B.18 C.24 D.364.已知α为第二象限角,,则sin2α=()A.B.C.D.5.已知空间向量=(0,1,1),=(﹣1,0,1),则与的夹角为()A.B.C.D.6.已知m,n是空间中两条不同的直线,α、β是两个不同的平面,且m⊂α,n⊂β.有下列命题:①若α∥β,则m∥n;②若α∥β,则m∥β;③若α∩β=l,且m⊥l,n⊥l,则α⊥β;④若α∩β=l,且m⊥l,m⊥n,则α⊥β.其中真命题的个数是()A.0 B.1 C.2 D.37.若一个底面为正三角形、侧棱与底面垂直的棱柱的三视图如下图所示,则这个棱柱的体积为()A.B.C.D.68.阅读如图的程序框图.若输入n=5,则输出k的值为()A.2 B.3 C.4 D.59.双曲线﹣y=1的顶点到其渐近线的距离等于()A.B.C.D.10.已知实数x,y满足,则目标函数z=2x﹣y的最大值为()A.﹣3 B.C.5 D.611.根据如下样本数据得到的回归方程为=x+,则()A.>0,>0 B.>0,<0C.<0,>0 D.<0,<012.如图,设D是图中边长分别为1和2的矩形区域,E是D内位于函数图象下方的阴影部分区域,则阴影部分E的面积为()A.ln2 B.1﹣ln2 C.2﹣ln2 D.1+ln2二、填空题(共四小题,每小题5分,共20分)13.某高级中学共有450名学生,现用分层抽样的方法从该校学生中抽取1个容量为45的样本,其中高一年级抽20人,高三年级抽10人,则该校高二年级学生人数为..14.若曲线y=ax2﹣lnx在点(1,a)处的切线平行于x轴,则a=.15.若直线y=kx+3与圆x2+y2=1相切,则k=.16.设函数f(x)=(x>0),观察:f1(x)=f(x)=,f2(x)=f(f1(x))=,f3(x)=f(f2(x))=,f4(x)=f(f3(x))=,…根据以上事实,由归纳推理可得:(x))=.当n∈N*且n≥2时,f n(x)=f(f n﹣1三、解答题(共6小题,共70分)17.设锐角三角形ABC的内角A,B,C的对边分别为a,b,c,a=2bsinA (Ⅰ)求B的大小;(Ⅱ)若,c=5,求b.18.某超市为了解顾客的购物量及结算时间等信息,安排一名员工随机收集了在该超市购物的100位顾客的相关数据,如下表所示.已知这100位顾客中的一次购物量超过8件的顾客占55%.(Ⅰ)确定x,y的值,并估计顾客一次购物的结算时间的平均值;(Ⅱ)求一位顾客一次购物的结算时间不超过2分钟的概率.(将频率视为概率)19.已知椭圆C: +=1(a>b>0)的离心率为,其中左焦点为F(﹣2,0).(1)求椭圆C的方程;(2)若直线y=x+m与椭圆C交于不同的两点A,B,且线段A,B的中点M在圆x2+y2=1上,求m的值.20.已知f(x)=ax 3+bx2+cx(a≠0)在x=±1时取得极值,且f(1)=﹣1,(1)试求常数a、b、c的值;(2)试判断x=±1是函数的极大值还是极小值,并说明理由.21.如图,四棱锥P﹣ABCD中,底面ABCD为平行四边形,∠DAB=60°,AB=2,AD=1,PD⊥底面ABCD.(1)证明:PA⊥BD;(2)若PD=AD,求二面角A﹣PB﹣C的余弦值.22.某商场销售某种商品的经验表明,该商品每日的销售量y(单位:千克)与销售价格x(单位:元/千克)满足关系式y=+10(x﹣6)2,其中3<x<6,a 为常数.已知销售价格为5元/千克时,每日可售出该商品11千克.(Ⅰ)求a的值;(Ⅱ)若该商品的成品为3元/千克,试确定销售价格x的值,使商场每日销售该商品所获得的利润最大.参考答案一、CDBAA BBBCC BD二、13. 15014.15. ±216.三、17.【解答】解:(Ⅰ)由a=2bsinA,根据正弦定理得sinA=2sinBsinA,所以,由△ABC为锐角三角形得.(Ⅱ)根据余弦定理,得b2=a2+c2﹣2accosB=27+25﹣45=7.所以,.18.【解答】解:(Ⅰ)由已知得25+y+10=55,x+30=45,所以x=15,y=20;顾客一次购物的结算时间的平均值为=1.9(分钟);(Ⅱ)记A:一位顾客一次购物的结算时间不超过2分钟;A1:该顾客一次购物的结算时间为1分钟;A2:该顾客一次购物的结算时间为1.5分钟;A3:该顾客一次购物的结算时间为2分钟;将频率视为概率可得P(A1);P(A2)=;P(A3)=∴P(A)=P(A1)+P(A2)+P(A3)=0.15+0.3+0.25=0.7∴一位顾客一次购物的结算时间不超过2分钟的概率为0.7.19.【解答】解:(1)∵椭圆C: +=1(a>b>0)的离心率为,其中左焦点为F(﹣2,0),∴由题意得,解得a=2,b=2,∴椭圆C的方程为.(2)设点A(x1,y1),B(x2,y2),线段AB的中点为M(x0,y0),由,消去y得3x2+4mx+2m2﹣8=0,△=96﹣8m2>0,∴﹣2<m<2,∵x0==﹣,∴y0=x0+m=,∵点M(x0,y0)在圆x2+y2=1上,∴(﹣)2+()2=1,∴m=±.20.【解答】(1)解:由f′(1)=f′(﹣1)=0,得3a+2b+c=0,①3a﹣2b+c=0.②又f(1)=﹣1,∴a+b+c=﹣1.③由①②③解得a=,b=0,c=﹣.(2)解:f(x)=x3﹣x,∴f′(x)=x2﹣=(x﹣1)(x+1).当x<﹣1或x>1时,f′(x)>0;当﹣1<x<1时,f′(x)<0.∴x=﹣1时,f(x)有极大值;x=1时,f(x)有极小值.21.【解答】(1)证明:因为∠DAB=60°,AB=2,AD=1,由余弦定理得BD==,∴BD2+AD2=AB2,故BD⊥AD,∵PD⊥底面ABCD,BD⊂平面ABCD,∴BD⊥PD,又AD∩PD=D,∴BD⊥平面PAD,又PA⊂平面PAD,∴PA⊥BD.(2)解:以D为原点,DA为x轴,DB为y轴,DP为z轴,建立空间直角坐标系,由已知得A(1,0,0),P(0,0,1),B(0,,0),C(﹣1,,0),=(1,0,﹣1),=(0,,﹣1),=(﹣1,,﹣1),设平面APB的法向量=(x,y,z),则,取y=,得=(3,,3),设平面PBC的法向量=(a,b,c),则,取b=,得=(0,,3),设二面角A﹣PB﹣C的平面角为θ,由图象知θ为钝角,∴cosθ=﹣|cos<>|=﹣||=﹣||=﹣.∴二面角A﹣PB﹣C的余弦值为﹣.22.【解答】解:(Ⅰ)因为x=5时,y=11,所以+10=11,故a=2(Ⅱ)由(Ⅰ)可知,该商品每日的销售量y=所以商场每日销售该商品所获得的利润为从而,f′(x)=10[(x﹣6)2+2(x﹣3)(x﹣6)]=30(x﹣6)(x﹣4)于是,当x变化时,f(x)、f′(x)的变化情况如下表:由上表可得,x=4是函数f(x)在区间(3,6)内的极大值点,也是最大值点.所以,当x=4时,函数f(x)取得最大值,且最大值等于42答:当销售价格为4元/千克时,商场每日销售该商品所获得的利润最大.。
人教版2017高二(下学期)数学(理)期中试卷附答案
一、选择题(本题共12小题,每小题5分,共60分)1.“金导电、银导电、铜导电、锡导电,所以一切金属都导电”.此推理方法是()A.完全归纳推理B.归纳推理C.类比推理D.演绎推理2.已知曲线y=x2+2x﹣2在点M处的切线与x轴平行,则点M的坐标是()A.(﹣1,3)B.(﹣1,﹣3)C.(﹣2,﹣3)D.(﹣2,3)3.有一段“三段论”推理是这样的:对于可导函数f(x),如果f′(x0)=0,那么x=x0是函数f(x)的极值点,因为函数f(x)=x3在x=0处的导数值f′(0)=0,所以,x=0是函数f(x)=x3的极值点.以上推理中()A.大前提错误B.小前提错误C.推理形式错误D.结论正确4.已知二次函数f(x)的图象如图所示,则其导函数f′(x)的图象大致形状是()A. B. C.D.5.若复数z2+2=0,则z3等于()A.±2B.2 C.±2i D.﹣2i6.复数(3﹣i)m﹣(1+i)对应的点在第三象限内,则实数m的取值范围是()A.B.m<﹣1 C.D.或m<﹣17.由曲线y=x2,y=x3围成的封闭图形面积为()A.B.C.D.8.设f(x)=xlnx,若f′(x0)=2,则x0=()A.e2B.e C. D.ln29.由数字1,2,3,4,5组成没有重复数字的五位数,其中小于50000的偶数共有()A.60个B.48个C.36个D.24个10.n∈N且n<55,则乘积(55﹣n)(56﹣n)…(69﹣n)等于()A.B.C.D.11.在(1﹣x3)(1+x)10展开式中,x5的系数是()A.﹣297 B.﹣252 C.297 D.20712.在的展开式中的常数项是()A.7 B.﹣7 C.28 D.﹣28二、填空题(本题共4小题,每小题5分,共20分)13.过抛物线y=f(x)上一点A(1,0)的切线的倾斜角为45°则f′(1)=.14.在50件产品中有4件是次品,从中任意抽出5件,至少有3件是次品的抽法共种(用数字作答).15.已知集合S={﹣1,0,1},P={1,2,3,4},从集合S,P中各取一个元素作为点的坐标,可作出不同的点共有个.16.满足条件|z﹣i|=|1+i|的复数z在复平面上对应的点(x,y)的轨迹方程为.三、解答题(本题共70分,解答应写出文字说明、演算步骤或推证过程)17.(10分)(1)计算()2+;(2)复数z=x+yi(x,y∈R)满足z+2i=3+i求复数z.18.(12分)计算:(1);(2).19.(12分)(1)在(1+x)n的展开式中,若第3项与第6项系数相等,且n 等于多少?(2)(x+)n的展开式奇数项的二项式系数之和为128,则求展开式中二项式系数最大项.20.(12分)已知函数f(x)=3x3﹣9x+5.(Ⅰ)求函数f(x)的单调递增区间;(Ⅱ)求函数f(x)在[﹣2,2]上的最大值和最小值.21.(12分)5个排成一排,在下列情况下,各有多少种不同排法?(1)甲排头(2)甲不排头,也不排尾(3)甲、乙、丙三人必须在一起(4)甲、乙、丙三人两两不相邻(5)甲在乙的左边(不一定相邻)(6)甲不排头,乙不排当中.22.(12分)已知f(x)=ax3+bx2+cx(a≠0)在x=±1处取得极值,且f(1)=﹣1.(Ⅰ)求常数a,b,c的值;(Ⅱ)求f(x)的极值.参考答案一、BBABC CABCB DA二、13. 114. 418615. 2316. x2+(y﹣1)2=4.三、17.【解答】解:(1)原式==(2)∵z=x+yi且满足z+2i=3+i,∴(x+yi)+2i(x﹣yi)=3+i,即(x+2y)+(2x+y)i=3+i,由复数相等的定义可得.解得,∴z=﹣i.18.【解答】解:(1)原式=.(2)原式=.另一方法:=.19.【解答】解:(1)由已知得C n2=C n5⇒n=7(2)由已知得C n1+C n3+C n5+ (128)∴2n﹣1=128∴n=8,而展开式中二项式系数最大项是=70.20.【解答】解:(I)f′(x)=9x2﹣9.(2分)令9x2﹣9>0,(4分)解此不等式,得x<﹣1或x>1.因此,函数f(x)的单调增区间为(﹣∞,﹣1)和(1,+∞).((6分)(II)令9x2﹣9=0,得x=1或x=﹣1.(8分)当x变化时,f′(x),f(x)变化状态如下表:(10分)从表中可以看出,当x=﹣2或x=1时,函数f(x)取得最小值﹣1.当x=﹣1或x=2时,函数f(x)取得最大值11.(12分)21.【解答】解:(1)甲固定不动,其余全排列,故有A44=24种;(2)甲有中间3个位置供选择,故有C31A44=72种;(3)先排甲、乙、丙三人,再把该三人当成一个整体,再加上另2人,相当于3人的全排列,故有A33A33=36种;(4)先排甲、乙、丙之外的2人,形成了3个空,将甲、乙、丙三人排这3个空位,故有A22A33=12种;(5)不考虑限制条件有,甲在乙的左边(不一定相邻),占总数的一半,即A55=60种;(6)第一类,甲排列当中,有A44=24种,第二类,甲不排在当中,有A31A31A33=54种,故有24+54=78种22.【解答】解:(Ⅰ)由f(x)=ax3+bx2+cx,得f'(x)=3ax2+2bx+c,由已知有f'(1)=f'(﹣1)=0,f(1)=﹣1,即:⇒,解得:;(Ⅱ)由(Ⅰ)知,,∴.当x<﹣1时,或x>1时,f'(x)>0,当﹣1<x<1时,f'(x)<0.∴f(x)在(﹣∞,﹣1)和(1,+∞)内分别为增函数;在(﹣1,1)内是减函数.因此,当x=﹣1时,函数f(x)取得极大值f(﹣1)==1;当x=1时,函数f(x)取得极小值f(1)==﹣1.。
2017北京四中高二(下)期中数学(文)含答案
2017北京四中高二(下)期中数学(文)卷(I)一、选择题:本大题共10小题,每小题5分,共50分。
1. 复数=A. +iB. +iC. 1-iD. 1+i2. 下列求导正确的是A. (3x2-2)'=3xB. (log2x)'=C. (cosx)'=sinxD. ()'=x3. 曲线y=x·e x在x=1处切线的斜率等于A. 2eB. eC. 2D. 14. 设a>0,b>0,则“a>b”是“lna>lnb”的A. 充分不必要条件B. 必要不充分条件C. 既不充分也不必要条件D. 充要条件5. 函数f(x)=3+xlnx的单调递增区间为A. (0,)B. (e,+∞)C. (,+∞)D. (,e)6. 在复平面内,复数(i是虚数单位)的共轭复数对应的点位于A. 第四象限B. 第三象限C. 第二象限D. 第一象限7. 命题“x0∈(0,+∞),lnx0=x0-1”的否定是A. x0∈(0,+∞),lnx0≠x0-1B. x0(0,+∞),1nx0=x0-1C. x∈(0,+∞),lnx≠x-1D. x(0,+∞),lnx=x-18. 已知f(x)=1+(1+x)+(1+x)2+(1+x)3+…+(1+x)n,则f'(0)=A. nB. n-1C.D.9. 函数f(x)=x3+ax2+(a+6)x+1有极大值和极小值,则实数a的取值范围是A. (-1,2)B. (-3,6)C. (-∞,-3)∪(6,+∞)D. (-∞,-1)∪(2,+∞)10. 方程x2=xsinx+cosx的实数解个数是A. 3B. 0C. 2D. 1二、填空题:本大题共6小题,每小题5分,共30分。
11. 复数(2+i)·i的模为___________.12. 命题“若a-b=0,则(a-b)(a+b)=0”的逆否命题为___________.13. 若曲线y=x3+x-2上的在点P0处的切线平行于直线y=4x-1,则P0坐标为__________.14. 函数f(x)=在区间[0,3]的最大值为___________.15. 若命题“x{x|x2-5 x+4>0}”是假命题,则x的取值范围是___________.16. 对于函数y=f(x),x D,若对于任意x1D,存在唯一的x2D,使得,则称函数f(x)在D上的几何平均数为M. 那么函数f(x)=x3-x2+1,在x=[1,2]上的几何平均数M=____________.三、解答题:本大题共2小题,共20分.17. 设函数f(x)=lnx-x2+x.(I)求f(x)的单调区间;(II)求f(x)在区间[,e]上的最大值.18. 已知函数f(x)=,其中a∈R.(I)当a=1时,求曲线y=f(x)在原点处的切线方程;(II)求f(x)的极值.卷(II)一、选择题:本大题共3小题,每小题5分,共15分。
2016-2017学年度第二学期高二年级期中考试试题
2016-2017学年度第二学期高二年级期中考试试题AI’m sure you know the song “Happy Birthday”. But do you kno w who wrote the song and for whom it was written?The retired professor, Archibald A. Hill in Lucasville, USA coul d tell us the story. Ninety-seven years ago, two of Mr. Archibald Hill’s aunts, Miss Patty Hill and Miss Mildred Hill were asked to write some songs for a book called “Song Storied for the Sunday Morning”. Miss Pa tty Hill and Miss Mildred Hill were both kindergarten teachers t hen. They loved children very much and wrote many beautiful songs for the book. One of them was the famous “Good Morn ing to You.” The song said, “Good morning to you, good morni ng to you, dear children, good morning to all.” This song was very popular at that time among kindergarten children. But not many grown-ups knew it. A few years later little Archibald was born. As his aunt, Miss Patty Hill sang the song “Happy Birthday” to the m elody(曲调)of “Good Morning to you” to her little nephew (侄子). She sang the song like this: Happy birthday to you! Hap py birthday to you! Happy birthday, dear Archie! Happy birthd ay to you!Miss Patty Hill and her sister had never expected that this son g would become so popular, but it really did. People all over t he world like the song because of its simplicity in tune(曲调简单)and friendliness in words.1. Archibald A.Hill was __________.A. EnglishB. RussianC. AmericanD. Australian2. Professor Archibald A. Hill was Miss Mildred Hill’s ________ __.A. sonB. studentC. brotherD. nephew3. The song “Happy Birthday” has a history of __________ yea rs.A. 97B. about 90C. more than 90D. less than 904. Why did Miss Hiss wrote songs?A. Because they are musicians.B. Because they love children.C. Because they were asked to write for a book.D. Because their nephew asked them to do so.【答案】1. C 2. D 3. B 4. C【解析】我肯定你知道那首歌“生日快乐”。
17学年下学期高二期中考试英语试题(附答案)(4)
南康中学2016—2017学年度第二学期高二期中考试英语试卷第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A.B、C三个选项中选出最佳选项并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1.Where does the conversation probably take place?A.At a TV store. B.In a restaurant. C.In a hotel.2.What’s the most probable relationship between the speakers?A.Customer and waiter. B.Teacher and student. C.Doctor and patient. 3.What does the man say about turtles?A.They’re ugly and lazy. B.They’re cute and smart. C.They’re noisy and dirty. 4.What does the woman mean?A.The man doesn’t have to move at all.B.The man’s new apartment is near his school.C.She once lived on Victoria Street.5.What does the woman advise the man to do?A.Buy an iPad 4 this week. B.Play games on his iPad 4.C.Surf the Internet together.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。
集宁201704高二下学期期中数学(理)试题及答案
集宁第二学期期中考试 高二年级理科数学试题本试卷满分为150分,考试时间为120分钟第Ⅰ卷(选择题,共60分)一、选择题(本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的)1.极坐标方程(ρ-1)·(θπ-)=0(ρ≥0)表示的图形是( )(A )两个圆 (B )两条直线(C )一个圆和一条射线 (D )一条直线和一条射线 2.将曲线y =sin 2x 按照伸缩变换⎩⎨⎧x ′=2x y ′=3y 后得到的曲线方程为( )A .y =3sin xB .y =3sin 2xC .y =3sin 12xD .y =13sin 2x 3. 若复数ii a 213++(i R a ,∈为虚数单位)是纯虚数,则实数a 的值为( )A. 6-B. 2-C. 4D. 64.六把椅子摆成一排,3人随机就座,任何两人不相邻的坐法种数为( ) A .144 B .120 C .72 D .245.盒中有10只螺丝钉,其中有3只是坏的,现从中随机地抽取4只,那么310为( ) A.恰有1只坏的概率 B.恰有2只好的概率 C .4只全是好的概率 D.至多2只坏的概率6. 某人射击一次击中目标的概率为0.6,经过3次射击,设X 表示击中目标的次数,则(2)P X ≥等于( ) A.81125B.54125C.36125D.271257.设~(100.8)X B ,,则(21)D X +等于( ) A.1.6 B.3.2 C.6.4 D.12.88.设随机变量X 的分布列如下表,且 1.6EX =,则a b -=( ) A.0.2B.0.1C.0.2-D.0.4-9. 已知x 、y 取值如下表:x0 1 45 6 y 1.3m3m5.67.4画散点图分析可知:y 与x 线性相关,且求得回归方程为ˆ1yx =+,则m 的值(精确到0.1)为( ) A.1.5 B.1.6 C.1.7 D.1.810.如果随机变量ξ~N (-1,σ2),且P (-3≤ξ≤-1)=0.4,则P (ξ≥1)=( ) A .0.2 B.0.3 C .0.4 D .0.111. 用数学归纳法证明12+22+…+(n -1)2+n 2+(n -1)2+…+22+12=n (2n 2+1)3时,从n =k 到n =k +1时,等式左边应添加的式子是( )A .(k -1)2+2k 2B .(k +1)2+k 2C .(k +1)2D.13(k +1)[2(k +1)2+1]12.调查某医院某段时间内婴儿出生的时间与性别的关系,得到下面的数据表:晚上 白天 合计 男婴 24 31 55 女婴 8 26 34 合计325789 你认为婴儿的性别与出生时间有关系的把握为( )A .80%B .90%C .95%D .99%参考公式及数据:))()()(()(22d b c a d c b a bc ad n K ++++-=X 0123 P0.1ab0.1P (k k 2)0.25 0.15 0.10 0.05 0.025 k1.3232.0722.7063.8415.024第Ⅱ卷(非选择题 共90分)二、填空题:(把答案填在答题纸对应的横线上,每小题5分,共20分。
湖北省七校2016_2017学年高二英语下学期期中联考试题201704240323
湖北省七校 2016-2017 学年高二英语放学期期中联考试题考试时间: 2017 年 4 月 21 日下午 2:30 — 4:30第 I 卷第一部分听力(共两节,满分30 分)做题时,先将答案标在试卷上。
录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节 ( 共 5小题;每题分,满分分)听下边 5段对话。
每段对话后有一个小题,从题中所给的 A 、B、C 三个选项中选出最正确选项,并标在试卷的相应地点。
听完每段对话后,你都有10秒钟的时间往返答相关小题和阅读下一小题。
每段对话仅读一遍。
例: How much is the shirt?A.£19.15.B.£9.15.C. £ 9.18.答案是 B。
1. What will the woman do today?A. Visit a friend.B. Attend a lecture.C. Do an experiment.2.What does the man find difficult?A.Understanding the instructions.B.Putting together the folding table.C.Fixing a toy train.3. Where is the woman ’s cell phone?A. In the classroom.B. In the dining hall.C. In her bag.4. When does the woman need the book?A. On April 1st.B. On April 2nd.C. On April 3rd.5. What does the man mean?A. Most readers don ’t agree with him.B. The woman can ’t convince him.C. Few people read his article.第二节 ( 共 15小题;每题分,满分分)听下边 5 段对话或独白。
2016-2017北京海淀高二下期中【文】
海淀区高二年级第二学期期中考试数 学 (文科) 2017.4学校 班级 姓名 成绩本试卷共100分.考试时间90分钟.一、选择题:本大题共8小题,每小题4分,共32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.下列各数中,是纯虚数的是( ).A . 2iB . πC.1+ D.(12.函数2()cos f x x x =+的导数()f x '为( ).A .sin x x -B .2sin x x -C .sin x x +D .2sin x x +3.函数3211()32f x x x =+的单调递增区间是( ).A .(,1)-∞-,(0,)+∞B .(,1)(0,)-∞-+∞UC .(1,0)-D .(,0)-∞,(1,)+∞4.若复数1z ,2z 在复平面内的对应点关于虚轴对称,且11i z =+,则2z =( ).A .1i +B .1i -C .1i --D .1i -+5.定义在R 上的函数()f x 和()g x ,其各自导函数()f x '和()g x '的图象如图所示,则函数()()()F x f x g x =-极值点的情况是( ).A .只有三个极大值点,无极小值点B .有两个极大值点,一个极小值点C .有一个极大值点,两个极小值点D .无极大值点,只有三个极小值点6.函数()ln f x x =与函数2()g x ax a =-的图象在点(1,0)的切线相同,则实数a 的值为( ).A .1B .12-C .12D .12或12-7.函数e (21)x y x =-的大致图象是( ).x ()A .B .C .D .8.为弘扬中国传统文化,某校在高中三个年级中抽取甲、乙、丙三名同学进行问卷调查.调查结果显示这三名同学来自不同的年级,加入了不同的三个社团:“楹联社”、“书法社”、“汉服社”,还满足如下条件:(1)甲同学没有加入“楹联社”; (2)乙同学没有加入“汉服社”;(3)加入“楹联社”的那名同学不在高二年级; (4)加入“汉服社”的那名同学在高一年级; (5)乙同学不在高三年级. 试问:甲同学所在的社团是( ).A .楹联社B .书法社C .汉服社D .条件不足无法判断二、填空题:本大题共6小题,每小题4分,共24分.9.已知复数(23i)i z =+,在复平面内与复数z 对应的点的坐标为__________.10.设函数()f x ,()g x 在区间(0,5)内导数存在,且有以下数据:2x =处的导数值是__________;曲线()f x 在点(1,(1))f 处的切线方程是__________.11.如图,函数()f x 的图象经过(0,0),(4,8),(8,0),(12,8)四个点,试用“>,=,<”填空:(1)(4)(2)2f f -__________(12)(8)4f f -; (2)(6)f '__________(10)f '.12.已知函数321()233f x x ax x =-++在(,)-∞+∞上单调递增,则实数a 的取值范围是__________.13.若数列{}n a 满足12312()()n n a a a a a a n ++++=+L ,则数列{}n a 是等差数列. 类比上述结论,可以猜想:若数列{}n b 满足__________,则数列{}n b 是等比数列.14.函数()e ln x f x a x =-(其中a ∈R ,e 为自然常数) ①a ∃∈R ,使得直线e y x =为函数()f x 的一条切线; ② 对0a ∀<,函数()f x 的导函数()f x '无零点; ③ 对0a ∀<,函数()f x 总存在零点;则上述结论正确的是__________.(写出所有正确的结论的序号)三、解答题:本大题共4小题,共44分.解答应写出文字说明,证明过程或演算步骤. 15.(本小题满分10分)计算:(Ⅰ)(12)(34)(2)i i i -+-+(Ⅱ)(12)(34)i i +÷-16.(本小题满分10分)已知函数32()392f x x x x =--+,求函数()f x 在区间[]2,2-上的最小值. 17.(本小题满分12分)已知数列{}n a 满足11a =,1n n a a ++*n ∈N . (Ⅰ)求2a ,3a ,4a ,并猜想数列{}n a 的通项公式.(Ⅱ)设数列{}n a 的前n 项和为n S ,求证:数列{}n S 不是等差数列. 18.(本小题满分12分)已知函数()(1)ln af x x a x x=-+-,其中a ∈R .(Ⅰ)求证:当1a =时,函数()y f x =没有极值点. (Ⅱ)求函数()y f x =的单调增区间.海淀区高二年级第二学期期中参考答案 2017.4数 学(文科)阅卷须知:1.评分参考中所注分数,表示考生正确做到此步应得的累加分数. 2.其它正确解法可以参照评分标准按相应步骤给分. 一、选择题:本大题共8小题,每小题4分,共32分.1.D 2.B 3.A 4.D 5.C 6.C 7.A8.C二、填空题:本大题共6小题,每小题4分,共24分.(有两空的小题每空2分) 9.(3,2)- 10.16;31y x =-11.(1)>(2)< 12.[ 13.21231()()n n n b b b b b b =L 14.①②③ 三、解答题: 本大题共4小题,共44分. 15.解:(Ⅰ)(12)(34)(2)i i i -+-+ (112)(2)i i =--+2015i =-+ (Ⅱ)(12)(34)i i +÷-12(12)(34)34(34)(34)i i i i i i +++==--+ 510122555i i -+==-+ 16.解:2()369f x x x '=-- 23(23)3(1)(3)x x x x =--=+-∵[]2,2x ∈-,令()0f x '=,得1x =-.当x 变化时,()f x ,()f x '在区间[]2,2-上的变化状态如下:所以函数(f 在区间上的最小值为20-.17.解:(Ⅰ)21a =,3a =,4a =归纳猜想n a(Ⅱ)由(Ⅰ)得111S a ==,212S a a =+=323S S a =+= 假设数列{}n S 是等差数列, 则1S ,2S ,3S 成等差数列, 所以1322S S S +=,即1=,两边平方得48+,2,这显然不成立,所以假设错误,所以数列{}n S 不是等差数列. 18.(Ⅰ)证明:函数()y f x =的定义域是(0,)+∞.当1a =时,1()2ln f x x x x=--, 函数221()1f x x x '=-+,222221(1)0x x x x x -+-==≥,所以函数()y f x =在定义域()0,+∞上单调递增. 所以当1a =时,函数()y f x =没有极值点.(Ⅱ)21()1a af x x x+'=-+, ()0,x ∈+∞, 222(1)(1)()x a x a x x a x x -++--==,令()0f x '=,得11x =,2x a =,0a ≤时, 由()0f x '>可得1x >,所以函数()y f x =的增区间是(1,)+∞; ①当01a <<时,由()0f x '>可得0x a <<,或1x >,所以函数()y f x =的增区间是(0,)a ,(1,)+∞; ②当1a >时,由()0f x '>可得01x <<,或x a >,所以函数()y f x =的增区间是(0,1),(,)a +∞; ③当1a =时,由(Ⅰ)可知函数()y f x =在定义域(0,)+∞上单调递增. 综上所述,当0a ≤时,函数()y f x =的增区间是(1,)+∞; 当01a <<时,所以函数()y f x =的增区间是(0,)a ,(1,)+∞; 当1a =时,函数()y f x =在定义域(0,)+∞上单调递增; 当1a >时,所以函数()y f x =的增区间是(0,1),(,)a +∞.。
2016-2017第二学期高二化学期中考试题
青海油田一中2016---2017年第二学期期中试题高二化学本试卷分第I 卷(选择题)和第II 卷(非选择题)两部分。
满分为150分,答题时间为120分钟。
所需相对原子质量:H 1 O 16 C 12 N 14 P 31 Cu:64一、选择题(每小题只一个正确答案,请将答案填在第Ⅱ卷上表格中,每小题4分) 1. 关于强、弱电解质及非电解质的组合完全正确的是( )2. 用pH 试纸测某溶液的pH 值时,规范的操作是( ) A.将pH 试纸放入溶液,观察其颜色变化,跟标准比色卡比较 B.用干燥的洁净玻璃棒蘸取溶液,点在pH 试纸上,跟标准比色卡比较C.用蒸馏水湿润pH 试纸,用玻璃棒蘸取溶液,滴在pH 试纸上,跟标准比色卡比较D.在试管内放入少量溶液煮沸,把pH 试纸放在试管口上观察,跟标准比色卡比较 3. 0.1 mol•L -1的醋酸溶液中存在电离平衡:CH 3COOHCH 3COO -+H +,对于该平衡,下列叙述正确的是( )A .加入少量0.1 mol•L -1的盐酸,溶液中c (H +)减小B .加入少量NaOH 固体,平衡向正反应方向移动 C.入少量醋酸钠固体,CH 3COOH 的电离程度增大D .升高温度,溶液pH 增大4. 用水稀释0.1mol/L 氨水时,溶液中随着水量的增加而减小的是( )A .)O H NH ()OH (23⋅-c c B .)OH ()O H NH (23-⋅c cC.c(H+)和c(OH-)的乘积D.OH-的物质的量5.有一支50mL酸式滴定管中盛盐酸,液面恰好在amL刻度处,把管内液体全部放出,盛入量筒内,所得液体体积一定是()A. a mLB.(50―a)mLC. 大于(50―a)mLD. 大于a mL6. 某学生的实验报告所列出的下列数据中合理的是()A.用10mL量筒量取7.13mL稀盐酸B.用托盘天平称量25.20g NaCl C.用广泛pH试纸测得某溶液的pH为2.3D.用25mL滴定管做中和滴定时,用去某浓度的碱溶液21.70mL7.下列金属防腐的措施中,属于牺牲阳极的阴极保护法的是()A.水中的钢闸门连接电源的负极B.金属护拦表面涂漆C.汽水底盘喷涂高分子膜D.地下钢管连接锌板8.将足量的BaCO3粉末分别加入下列溶液中,充分溶解至溶液饱和,各溶液中Ba2+的浓度最小的为()A.10 mL 0.2 mol/LNa2CO3溶液 B.40 mL水C.50 mL 0.01 mol/L 氯化钡溶液 D.100 mL 0.01 mol/L盐酸9.为了除去氯化镁酸性溶液中的Fe3+离子,可在加热的条件下加入一种试剂,过滤后再加入适量盐酸,这种试剂是( )A. 碳酸钠B. 氢氧化钠C.氨水D. 碳酸镁10.现有浓度为1 mol/L的五种溶液:①HCI,②H2SO4,③CH3COOH,④NH4Cl,⑤NaOH,由水电离出的C(H+)大小关系正确的是()A.④>③>①=⑤>② B.①=②>③>④>⑤C.②>①>③>④>⑤ D.④>③>①>⑤>②11.温度相同浓度均为0.1 mol·L-1的:①(NH4)2CO3,②(NH4)2Fe(SO4)2,③氨水,④NH4NO3,⑤NH4HSO4溶液,它们中的c(NH+4)由大到小的排列顺序是()A.②⑤④①③B.③⑤②④①C.②①⑤④③D.⑤②①④③12.氢氧燃料电池已用于航天飞机。
2017学年高二第二学期数学期中测试卷(复数+立体几何)—教师版
1智立方2017学年高二第二学期数学期中测试卷(A 卷)考试时间:90分钟 满分100分一.填空题(每小题3分,共36分)1. 计算:2(12)(32)1i i i+-++= 。
【答案】8+3i2. 已知复数1z =3+4i ,2z =t +i ,且21z z ⋅是实数,则实数t 等于 __________。
【答案】343. 若正四棱锥的底面边长为23cm ,体积为4cm 3,则它的侧面与底面所成的二面角的大小是__________。
【答案】30°4. 如右图所示:一副三角板以平面ABC ⊥平面 ACD 的方式放置,若,公共边AC=22则异面直线 AB 与CD 的距离是_______。
【答案】25. 设E 为正方体1111ABCD-A B C D 的棱1CC 的中点,则平面1AB E 和底面1111A B C D 所成角的余弦值为_________。
【答案】326. 若一个球的体积为π34,则它的表面积为_______。
【答案】 12π7. 当母线长为1的圆锥的侧面积与底面积的差达到最大值时,其侧面展开图圆心角等于_________。
【答案】 2π8. 有下列五个命题:下面关于四棱柱的四个命题:9. ①若有两个侧面垂直于底面,则该四棱柱为直四棱柱;10. ②若两个过相对侧棱的截面都垂直于底面,则该四棱柱为直四棱柱; 11. ③若四个侧面两两全等,则该四棱柱为直四棱柱;12. ④若四棱柱的四条对角线两两相等,则该四棱柱为直四棱柱. 13. 其中,真命题的编号是 。
【答案】 ②④ 14. 4+R,z-2=2,z=z∈若复数z 满足z 且则___________。
【答案】 1±3i15. 2z z 1z 1z =-+若复数满足,则的最大值为__________。
【答案】 316. 若直线m 与平面α所成角为3π,直线a 在平面α上,且与m 异面,则直线m 与直线a 所成角的范围为__________。
人教版2017高二(下学期)(理科)期中数学试题附答案
一.选择题:本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,有且只有一项是符合题目要求的.1.设集合M={x2﹣2x<0},N={x|x≤1},则M∩N=()A.(0,1)B.(1,2)C.(0,2)D.(0,1]2.已知复数(i为虚数单位),那么z的共轭复数为()A.B.C.D.3.等差数列{a n}中,a7+a9=16,a4=1,则a12=()A.15 B.30 C.31 D.644.在平面内的动点(x,y)满足不等式,则z=2x+y的最大值是()A.6 B.4 C.2 D.05.“a=2”是“直线y=﹣ax+2与y=垂直”的()A.充分不必要条件 B.必要不充分条件C.充要条件D.既不充分也不必要条件6.如图,在长方体ABCD﹣A1B1C1D1中,AB=BC=2,AA1=1,则BC1与平面BB1D1D 所成角的正弦值为()A.B.C.D.7.已知△ABC的边BC上有一点D满足=3,则可表示为()A.=﹣2+3 B.=+C.=+D.=+ 8.如图曲线y=x2和直线x=0,x=1,y=所围成的图形(阴影部分)的面积为()A.B.C.D.9.执行如图的程序框图,则输出的n为()A.9 B.11 C.13 D.1510.已知x>0,y>0,x+y+=2,则x+y的最小值是()A.B.1 C.D.11.函数f(x)在定义域(0,+∞)内恒满足:①f(x)>0;②2f(x)<xf′(x)<3f(x),其中f′(x)为f(x)的导函数,则()A.<<B.<<C.<<D.<<12.过双曲线(a>0,b>0)的右焦点F(c,0)作圆x2+y2=a2的切线,切点为M.直线FM交抛物线y2=﹣4cx于点N,若(O为坐标原点),则双曲线的离心率为()A.B.C.D.二.填空题:本大题共4小题,每小题5分,共20分.13.若,则cos2θ=.14.将全体正整数排成如图的一个三角形数阵,按照此排列规律,第10行从左向右的第5个数为.15.如图是一个正方体被切掉部分后所得几何体的三视图,则该几何体的体积为.16.已知函数f(x)=﹣kx(e为自然对数的底数)有且只有一个零点,则实数k的取值范围是.三、解答题:本大题共6小题,共70分.解答应给出文字说明、证明过程及演算步骤.17.(10分)设锐角三角形ABC的内角A,B,C的对边分别为a,b,c,a=2bsinA (Ⅰ)求B的大小;(Ⅱ)若,c=5,求b.=2a n﹣118.(12分)已知数列{a n}满足a1=2,a n+1(1)求证数列{a n﹣1}是等比数列(2)设b n=n•(a n﹣1),求数列{b n}的前n项和S n.19.(12分)已知函数f(x)=x3﹣ax2﹣3x(1)若x=﹣是f(x)的极值点,求f(x)在[﹣1,a]上的最大值和最小值.(2)若f(x)在区间上[1,+∞)是增函数,求实数a的取值范围.20.(12分)在如图所示的五面体中,面ABCD为直角梯形,∠BAD=∠ADC=,平面ADE⊥平面ABCD,EF=2DC=4AB=4,△ADE是边长为2的正三角形.(Ⅰ)证明:BE⊥平面ACF;(Ⅱ)求二面角A﹣BC﹣F的余弦值.21.(12分)已知椭圆+=1(a>b>0)和直线l:﹣=1,椭圆的离心率e=,坐标原点到直线l的距离为.(Ⅰ)求椭圆的方程;(Ⅱ)已知定点E(﹣1,0),若直线m过点P(0,2)且与椭圆相交于C,D 两点,试判断是否存在直线m,使以CD为直径的圆过点E?若存在,求出直线m的方程;若不存在,请说明理由.22.(12分)已知函数f(x)=(m,n为常数,e=2.71828…是自然对数的底数),曲线y=f(x)在点(1,f(1))处的切线方程是y=;(Ⅰ)求m,n的值;(Ⅱ)求f(x)的单调区间;(Ⅲ)设g(x)=f′(x)•(其中f'(x)为f(x)的导函数),证明:对任意x>0,g(x)<1+e﹣2.参考答案一、DBAAA DBDCC DB二、13.14. 5015.16. (0,)三、17.【解答】解:(Ⅰ)由a=2bsinA,根据正弦定理得sinA=2sinBsinA,所以,由△ABC为锐角三角形得.(Ⅱ)根据余弦定理,得b2=a2+c2﹣2accosB=27+25﹣45=7.所以,.18.【解答】解:(1)证明:∵a n+1=2a n﹣1,变形为:a n+1﹣1=2(a n﹣1),∴数列{a n﹣1}是等比数列,首项为1,公比为2,∴a n﹣1=2n﹣1,即a n=1+2n﹣1.(2)b n=n•(a n﹣1)=n•2n﹣1,∴数列{b n}的前n项和S n=1+2×2+3×22+…+n×2n﹣1,①∴2S n=2+2×22+…+(n﹣1)×2n﹣1+n•2n,②由①﹣②,得﹣S n=1+2+22+…+2n﹣1﹣n•2n=﹣n•2n=(1﹣n)•2n﹣1.∴S n=(n﹣1)•2n+1.19.【解答】解:(1)f′(x)=3x2﹣2ax﹣3,x=﹣是f(x)的极值点,则f′(﹣)=3×+2a×﹣3=0,解得a=4,f(x)=x3﹣4x2﹣12,f′(x)=3x2﹣8x﹣3=(x﹣3)(3x+1)=0,解得x=﹣,3,x,f(x),f′(x )变化如下表:(﹣1﹣)﹣(﹣)所以f(x)max=f(﹣)=,f(x)min=f(3)=18(2)函数f(x)=x3﹣ax2﹣3x求导得f′(x)=3x2﹣2ax﹣3,f(x)在区间上[1,+∞)是增函数,则f′(x)=3x2﹣2ax﹣3≥0在[1,+∞)恒成立,即a在[1,+∞)恒成立,a,y=x﹣在[1,+∞)为增函数,则(x﹣)min=0∴a≤0,∴实数a的取值范围为(﹣∞,0]20.【解答】证明:(Ⅰ)取AD中点O,以O为原点,OA为x轴,过O作AB的平行线为y轴,OE为z轴,建立空间直角坐标系,则B(1,1,0),E(0,0,),A(1,0,0),C(﹣1,2,0),F(0,4,),=(﹣1,﹣1,),=(﹣1,4,),=(﹣2,2,0),=1﹣4+3=0,=2﹣2=0,∴BE⊥AF,BE⊥AC,又AF∩AC=A,∴BE⊥平面ACF.解:(Ⅱ)=(﹣2,1,0),=(﹣1,3,),设平面BCF的法向量=(x,y,z),则,取x=1,得=(1,2,﹣),平面ABC的法向量=(0,0,1),设二面角A﹣BC﹣F的平面角为θ,则cosθ===.∴二面角A﹣BC﹣F的余弦值为.21.【解答】解:(Ⅰ)由直线,∴,即4a2b2=3a2+3b2﹣﹣①又由,得,即,又∵a2=b2+c2,∴﹣﹣②将②代入①得,即,∴a2=3,b2=2,c2=1,∴所求椭圆方程是;(Ⅱ)①当直线m的斜率不存在时,直线m方程为x=0,则直线m与椭圆的交点为(0,±1),又∵E(﹣1,0),∴∠CED=90°,即以CD为直径的圆过点E;②当直线m的斜率存在时,设直线m方程为y=kx+2,C(x1,y1),D(x2,y2),由,得(1+3k2)x2+12kx+9=0,由△=144k2﹣4×9(1+3k2)=36k2﹣36>0,得k>1或k<﹣1,∴,,∴y1y2=(kx1+2)(kx2+2)=k2x1x2+2k(x1+x2)+4∵以CD为直径的圆过点E,∴EC⊥ED,即,由,,得(x1+1)(x2+1)+y1y2=0,∴(1+k2)x1x2+(2k+1)(x1+x2)+5=0,∴,解得,即;综上所述,当以CD为直径的圆过定点E时,直线m的方程为x=0或.22.【解答】解:(Ⅰ)由得(x>0).由已知得,解得m=n.又,即n=2,∴m=n=2.…(Ⅱ)由(Ⅰ)得,令p(x)=1﹣x﹣xlnx,x∈(0,+∞),当x∈(0,1)时,p(x)>0;当x∈(1,+∞)时,p(x)<0,又e x>0,所以当x∈(0,1)时,f'(x)>0;当x∈(1,+∞)时,f'(x)<0,∴f(x)的单调增区间是(0,1),f(x)的单调减区间是(1,+∞).…(8分)(Ⅲ)证明:由已知有,x∈(0,+∞),于是对任意x>0,g(x)<1+e﹣2等价于,由(Ⅱ)知p(x)=1﹣x﹣xlnx,x∈(0,+∞),∴p'(x)=﹣lnx﹣2=﹣(lnx﹣lne﹣2),x∈(0,+∞).易得当x∈(0,e﹣2)时,p'(x)>0,即p(x)单调递增;当x∈(e﹣2,+∞)时,p'(x)<0,即p(x)单调递减.所以p(x)的最大值为p(e﹣2)=1+e﹣2,故1﹣x﹣xlnx≤1+e﹣2.设q(x)=x﹣ln(1+x),则,因此,当x∈(0,+∞)时,q(x)单调递增,q(x)>q(0)=0.故当x∈(0,+∞)时,q(x)=x﹣ln(1+x)>0,即.∴1﹣x﹣xlnx≤1+e﹣2<.∴对任意x>0,g(x)<1+e﹣2.…(14分)。
2017学年第二学期高二数学期中考试试卷答案
考试答案一、填空题:w W w .x K b 1.c o M1、异面、平行;2、;3、;4、;5、垂直;6、;7、;8、;9、;10、③④;11、;12、取中点R,P地轨迹即为线段RC.二、选择题:13、A;14、D;15、A;16、A;17、A;18、C三、解答题:19、<1)由………3分故:两根为所以:………6分新课标第一网<2)证明:假设直线与共面,设该平面为.………2分可知直线与在平面上,所以……………4分即即直线为共面直线,与已知为异面直线矛盾.故原假设不成立,则直线与为异面直线.……………6分20、解:<1)………3分<2)………4分..........6分………8分21、解:(1>,将代入,得....3分<2)设,中点..........6分新课标第一网将代入得:AB中点轨迹为8分22、<1)延长DB与交于点P,P即为所求点.<图略)……………4分<2)过N点作交AB于点E,连结CN,CE.可知即为异面直线AM、CN所成角.......6分.,可求得.......9分则……………………10分X k B 1 . c o m23、<1)结论:上述直线上所有地点都是“点”………2分由题意得:直线……………3分设,由A为BP中点,可知由A、B两点在抛物线上,则:w W w .x K b 1.c o M化简得关于地方程:<*)…………5分其判别式恒成立,可知对方程<*)恒有解.即对直线上所有地点P,存在过P点地直线交抛物线于A、B两点,使得A为BP中点.…………8分<2)设直线地斜率为,直线,直线与抛物线地交点,…………2分斜率和为定值0……………4分如存在满足条件地点M,使得为定值仅当,即时,……………8分申明:所有资料为本人收集整理,仅限个人学习使用,勿做商业用途.。
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高二年级第二学期期中练习2017 4一、. 单项填空1. The mayor of Beijing says that all construction work by the end of next year.A. has been completedB. has completedC. will have been completedD. will have completed2. At this time tomorrow over the Atlantic.A. we’re going to flyB. we’ll be flyingC. we’ll flyD. we’re to fly3. He will have learned English for eight years by the time he from the university next year.A. will graduateB. will have graduatedC. graduatesD. is to graduate4. Selecting a mobile phone for personal use is no easy task because technology so rapidly.A. is changingB. has changedC. will have changedD. will change5. In some western counties, demand for graduates from MBA courses has .A. turned downB. turned overC. fallen downD. fallen over6. —When will you come to see me, Dad?—I will go to see you when you the training course.A. will have finishedB. will finishC. are finishingD. finish7. The forest guards often find campfires that have not been completely.A. turned downB. put outC. put awayD. turned over8. The idea puzzled me so much that I stopped for a few seconds to try to .A. make it outB. make it offC. make it upD. make it over9. I’ve won a holiday for two weeks to Florida. I my mum.A. am takingB. have takenC. takeD. will have taken10. —Do you think I should get a good guidebook?—Yes, of course. , you also need a good camera and comfortable shoes.A. What’s moreB. In other wordsC. By the wayD. All in all11. —Can I join your club, Dad? —You can when you a bit older.A. getB. will getC. are gettingD. will have got12. Lakers Phoenix Suns 114 to 106 last Tuesday.A. conqueredB. beatC. wonD. struck13. They will provide the poor students with textbooks, dictionaries, schoolbags, reference(c参考)books, pens,exercise-books. All the following answers are right except . in another way.A. not to mentionB. let aloneC. without mentioningD. to put it14. My most famous relative of all, who really left his mark on America, was Reb Sussel, mygreat-grandfather. A. one B. the one C. he D. some15. The rise in world population will mean greater competition for natural resources, which could easilybecome .A. quarrelB. argumentC. conflictD. war16. If I can help , I don’t like workin g late into the night.A. soB. thatC. itD. them17. We thought there were 20 students in the dining hall, , in fact, there were 40.A. whileB. whetherC. whatD. which18. —What’s the name?—Khulaifi. I spell that for you?A. ShallB. WouldC. CanD. Might19. —I’m thinking of the test tomorrow. I’m afraid I can’t pass this time. —! I’m sure you’ll make it.A. Go aheadB. Good luckC. No problemD. Cheer up20. Everywhere man has cut down forests in order to grow crops, or to use wood as fuel or as buildingmaterial. A. the; the B. the; / C. /; the D. /; /二、完形填空A Shelter for the HomelessLast summer I was a volunteer (志愿者) at a shelter for the homeless, a place for homeless people to sleep at night. I wasn’t working that summer and was 21only two classes in summer school, so I had some 22time. Three nights a week, I 23in the kitchen of the shelter along with four other volunteers. We planned and 24 for 45 people hot meals 25vegetables, chicken, fish and fruit. The homeless people 26this good food because many of them usually didn’t eat well.I 27this volunteer work, making 28with the four volunteers in the kitchen. One was a very nice elderly housewife, one a movie actor, another a young teacher, and the other a college student, 29me.I talked to a lot of the homeless people at the shelter. Their life stories 30me with sympathy (同情). Some of them had 31with alcohol (酒) or drugs while others only had bad 32. One woman worked for almost 15 years for a small company, and 33she lost her job. She looked for a 34job, but couldn’t find one, for she was too 35. She could do nothing but sell her furniture—sofas, chairs and tables so that she could pay for her food. The woman went on job hunting, but she still couldn’t find one. She had no money for her flat and had to sleep in her car. Then she had to sell her car. Alone, afraid, and homeless, she finally came to the shelter.21. A. studying B. planning C. holding D. taking22. A. full B. free C. extra D. long23. A. helped B. waited C. appeared D. washed24. A. ordered B. bought C. cooked D. finished25. A. with B. from C. by D. in26. A. demanded B. needed C. collected D. requested27. A. wanted B. had C. owned D. enjoyed28. A. decisions B. wishes C. friends D. changes29. A. as B. like C. except D. besides30. A. struck B. filled C. seized D. provided31. A. habits B. worries C. difficulties D. problems32. A. luck B. pain C. experience D. loss33. A. so B. again C. then D. thus34. A. good B. fresh C. skilled D. new35. A. old B. poor C. sick D. dull第二部分:阅读理解ADiscoverNews magazine of science devoted to the wonders and stories of modern science, written for the educated general reader, published (出版) by Disney Magazine Publishing Co., Discover tells many of the same stories that professionals (专业人员) read in Scientific American. A truly delightful family science magazine, each issue (每期) brings to light new and newsworthy topics to make dinnertime and water-cooler conversations interesting.Cover Prices $59. 88Price: $4. 95($1. 66/issue)You Save: $39. 93(67%)Issues: 12 issues/12 monthsSelfPublished by Conde Nast Publications Inc., Self is a handbook devoted to women’s overall physical and mental health. Every issue contains usable articles such as “Style Lab” , in which wearable clothes are mixed and matched on non-models and the “Eat-Righ t Road Map” , with tips on how to eat properly.Cover Price: $35. 86Price: $15. 00($2. 5/issue)You Save: $5. 86(58%)Issues: 6 issues/12 monthsInStyleInStyle is a guide to the lives and lifestyles of the world’s famous people. The magazine cov ers the choices people make about their homes, their clothes and their free time activities. With photos and articles, it opens the door to these people’s homes, families, parties and weddings, offering ideas about beauty, fitness and in general, lifestyle s. Publisher: The Time Inc. Magazine Company.Cover Price: $47. 88Price: $23. 88($2. 38/issue)You Save: $24. 00(50%)Issues: 10 issues/12 monthsWiredThis magazine is designed for leaders in the field of information engineering including top managers and professionals in the computer, business, design and education industries. Published by Conde Nast Publications Inc. Wired often carries articles on how technology changes people’s lives.Cover Price: $59. 40Price: $10. 00($1. 00/issue)You Save: $49. 40(83%)Issues: 10 issues/12 months36. Which of the following magazines is published monthly?A. In styleB. SelfC. DiscoverD. Wired37. Which two magazines are published by the same publisher?A. Wired and DiscoverB. Discover and In styleC. Self and DiscoverD. Self and Wired38. Which magazine offers the biggest price cut?A. InstyleB. WiredC. DiscoverD. Self39. The “Style Lab” in Self provides readers with articles which .A. offer advice to ordinary women on clothesB. show how a woman can become famousC. introduce places with the best foodD. discuss ways of training modelsBThe year is 594. It has been announced that a comet (彗星) is heading towards the Earth. Most of it will miss our planet, but two pieces will probably hit the southern half of the Earth.On 2 July, a piece, four kilometres wide, enters the Earth’s atmosphere (大气层) with a massive explosion. About half of the piece is destroyed, but the remaining part hits the South Atlantic at 50 times the speed of sound. The sea boils and a huge hole is made in the seabed. Huge waves are created and speed outwards from the hole. The wall of water, a kilometer high, rushes towards southern Africa at 800 kilometres an hour. Cities on the African coast are totally destroyed and millions of people are drowned.Before the waves reach South America, the second piece of the comet lands in Argentina. Earthquakes and volcanoes are set off in the Andes Mountains. The shock waves move north into California and all around the Pacific Ocean. The cities of Los Angeles, San Francisco and Tokyo are completely destroyed by earthquakes. Millions of people in the southern half of the earth are already dead, but the north won’t escape for long. Because of the explosions, the sun is hidden by clouds of dust, temperatures around the world fall to almost zero. Crops are ruined. The sun won’t be seen again for many years. Wars break out as countries fight for food. A year later, no more than 10 million people remain alive.Could it really happen? In fact, it has already happened more than once in the history of the Earth. The dinosaurs were on Earth for over 10 million years. Then 65 million years ago they suddenly disappeared. Many scientists believe that the Earth was hit by a piece of object in space. The dinosaurs couldn’t live through the cold climate that followed and they died out. Will we meet the same end?40. What is mainly described in the passage?A. Historic discovery.B. An event of imagination.C. A research on space.D. A scientific adventure.41. When the first piece hits the South Atlantic, it causes.A. an earthquakeB. damages to citiesC. an Earth explosionD. huge waves42. Why can’t the northern half of the earth escape for long?A. Because the land is covered with water.B. Because the light and heat from the sun cannot reach the Earth.C. Because people there cannot live at the temperature of zero.D. Because wars break out among countries.CWould you eat a ready meal from the fridge rather than cook from scratch? Have you been doing Internet shopping rather than going to the stores? What can’t you be bother ed to do?A study into how lazy British people are has found more than half of adults are so lazy that they’d catch the lift rather than climb two flights of stairs.Just over 2, 000 people were quizzed by independent researchers at Nuffield Health, Brita in’s largest health charity. The results were astonishing.About one in six people surveyed said if their remote control was broken, they would continue watching the same channel rather than get up.More than one third of those questioned said they would not run to catch a bus. Worryingly, of the 654 respondents with children, 64% said they were often too tired to play with them.This led the report to conclude that it’s no wonder that one in six children in the UK are classified as obese (very fat) before they start school.Dr. Sarah Dauncey, medical director of Nuffield Health, said, “People need to get fitter, not just for their own sake, but for the sake of their families, friends and evidently their pets too.”“If we don’t start to take control of t his problem, a whole generation will become too unfit to perform even the most rudimentary tasks.”And Scotland’s largest city, Glasgow, was shamed as the most inactive city in the UK, with 75% surveyed admitting they do not get enough exercise, followed closely by Birmingham and Southampton, both with 67%.The results cause serious challenges for the National Health Service, where obesity-related illnesses such as heart disease and cancer have been on a steady increase for the past 40 years and are costing billions of pounds every year.43. What causes children in the UK to be obese?A. Eating ready meal.B. Watching TV.C. Doing Internet shopping.D. Being lazy.44. What can be inferred from the passage?A. British people are too lazy to do anything.B. Internet shopping will do harm to your health.C. Many people would not run to catch a bus in Britain.D. People should be more active and take regular exercise to keep fit.45. What does the underlined word “rudimentary” (in the eighth paragraph) most probably mean?A. basicB. seriousC. vitalD. hard46. All the following are among the most lazy cities in the UK except .A. GlasgowB. BirminghamC. NuffieldD. SouthamptonDOur professor introduced himself and challenged us to get to know someone. I stood up to look around when a gentle hand touched my shoulder. I turned around to find a wrinkled, very old lady looking at me with a smile that lit up her entire being.“Why are you in college at such a young, innocent age?” I asked. She replied, “I’m here to meet a rich husband, get married, have a couple of children, and then retire and travel.” “No seriously,” I said. “I always dreamed of having a college education and now I’m getting one!” She told me. We became instant friends.One day we invited Rose to speak at our football banquet. She was introduced and stepped up. “We do not stop playing because we are old; we grow old because we stop playing. There are some secrets to staying young, being happy, andachieving success. You have to laugh and find humour every day. You’ve got to have a dream. When you lose your dreams, you die. We have so many people walking around who are dead, and they don’t even know it!” She said. “T here is a huge difference between growing older and growing up. If you are nineteen years old and lie in bed for one full year and don’t do one productive thing, you will turn 5 years old. If I am 87 years old, and stay in bed for a year, and never do anyt hing, I will turn 88. Anybody can grow older. That doesn’t take any talent or ability. The idea is to grow up by always finding the opportunity in change. Have no regrets. The elderly usually don’t have regrets for what we did, but rather for things we did not do.”Rose finished the college degree she had begun all those years ago. One week after graduation Rose died peacefully in her sleep. Over two thousand college students attended her funeral.47. The author intends to tell us that .A. it’s n ever too late to be all you can possibly beB. it’s a poor heart not to enjoy lifeC. it’s right that knowledge is long, life is shortD. both old friends and old wine are best48. The best title of the text is .A. Finding the Opportunity In ChangeB. Growing Up or Growing OldC. Friendship Despite Great Difference in AgeD. Rose, a Hardworking Old Woman49. The huge difference between growing older and growing up lies in .A. whether to be smarterB. whether to go to collegeC. whether to make friends with youthsD. whether to find the new opportunity50. Which of the following sentences is seriously said?A. Why are you in college at such a young, innocent age?B. I’m here to meet a rich husband, get married, have a couple of children, and then retire and travel.C. We have so many people walking around who are dead.D. I always dreamed of having a college education.第三部分: 短文填词阅读下面短文, 根据以下提示:1) 汉语提示, 2) 首字母提示, 3) 语境提示, 在每个空格内填入一个适当的英语单词, 并将该词完整地写在右边相对应的横线上。