鄂尔多斯市第二中学2017-2018高三第三次模拟考试

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鄂尔多斯市第二中学2018-2019学年高三上学期第三次月考试卷数学含答案

鄂尔多斯市第二中学2018-2019学年高三上学期第三次月考试卷数学含答案

鄂尔多斯市第二中学2018-2019学年高三上学期第三次月考试卷数学含答案 班级__________ 座号_____ 姓名__________ 分数__________一、选择题(本大题共12小题,每小题5分,共60分.每小题给出的四个选项中,只有一项是符合题目要求的.)1. 若函数21,1,()ln ,1,x x f x x x ⎧-≤=⎨>⎩则函数1()2y f x x =-+的零点个数为( ) A .1 B .2 C .3 D .42. 在ABC ∆中,60A =,1b =sin sin a b cA B C+++等于( )A .B .3C .3D .23. 执行如图所示程序框图,若使输出的结果不大于50,则输入的整数k 的最大值为( ) A .4B .5C .6D .74. 若函数1,0,()(2),0,x x f x f x x +≥⎧=⎨+<⎩则(3)f -的值为( )A .5B .1-C .7-D .25. 已知角的终边经过点()3P x ,()0x <且cos x θ=,则等于( )A .1-B .13- C .3- D .6. 已知数列{}n a 的首项为11a =,且满足11122n n n a a +=+,则此数列的第4项是( ) A .1 B .12 C. 34 D .587. 已知集合{2,1,1,2,4}A =--,2{|log ||1,}B y y x x A ==-∈,则A B =( )A .{2,1,1}--B .{1,1,2}-C .{1,1}-D .{2,1}-- 【命题意图】本题考查集合的交集运算,意在考查计算能力.8. 某校为了了解1500名学生对学校食堂的意见,从中抽取1个容量为50的样本,采用系统抽样法,则分段间隔为( )1111]A .10B .51C .20D .309. 某班设计了一个八边形的班徽(如图),它由腰长为1,顶角为α的四个等腰三角形,及其底边构成的正方形所组成,该八边形的面积为( )A .2sin 2cos 2αα-+B .sin 33αα-+ C. 3sin 31αα+ D .2sin cos 1αα-+ 10.某个几何体的三视图如图所示,该几何体的表面积为92+14π,则该几何体的体积为( ) A .80+20π B .40+20π C .60+10π D .80+10π11.单位正方体(棱长为1)被切去一部分,剩下部分几何体的三视图如图所示,则( )A.该几何体体积为 B.该几何体体积可能为 C.该几何体表面积应为+D .该几何体唯一12.设集合{}1234U =,,,,{}2540A x x x =∈-+<N ,则U C A 等于( ) A .{}12, B .{}14, C .{}24, D .{}134,,二、填空题(本大题共4小题,每小题5分,共20分.把答案填写在横线上)13.已知过双曲线22221(0,0)x y a b a b-=>>的右焦点2F 的直线交双曲线于,A B 两点,连结11,AF BF ,若1||||AB BF =,且190ABF ∠=︒,则双曲线的离心率为( )A.5- BC.6- D【命题意图】本题考查双曲线定义与几何性质,意要考查逻辑思维能力、运算求解能力,以及考查数形结合思想、方程思想、转化思想.14.设变量y x ,满足约束条件22022010x y x y x y --≤⎧⎪-+≥⎨⎪+-≥⎩,则22(1)3(1)z a x a y =+-+的最小值是20-,则实数a =______.【命题意图】本题考查线性规划问题,意在考查作图与识图能力、逻辑思维能力、运算求解能力. 15.已知△ABC 的面积为S ,三内角A ,B ,C 的对边分别为,,.若2224S a b c +=+, 则sin cos()4C B π-+取最大值时C = .16.等差数列{}n a 中,39||||a a =,公差0d <,则使前项和n S 取得最大值的自然数是________.三、解答题(本大共6小题,共70分。

文综 鄂尔多斯市2018年高三鄂尔多斯市第三次模拟考试

文综 鄂尔多斯市2018年高三鄂尔多斯市第三次模拟考试

文综鄂尔多斯市2018年高三鄂尔多斯市第三次模拟考试文综考试时间:____分钟题型单选题填空题总分得分单选题(本大题共35小题,每小题____分,共____分。

)2017年4月1日,中共中央、国务院印发通知,决定设立河北雄安新区。

其规划范围涵盖河北省雄县、容城、安新等3个小县及周边部分区域。

设立雄安新区对于集中疏解北京非首都功能,探索人口经济密集地区优化开发新模式、调整优化京津冀城市布局和空间结构等均具有重大现实意义和深远历史意义。

下图为2000年和2010年京津冀人口重心和经济重心分布图。

据此完成1~2题。

2000—2010年,京津冀经济重心空间位置变化明显,其原因可能是A. 河北南部乡镇企业迅速发展B. 保定、张家口资源加速开发C. 太行山区旅游业迅速发展D. 京津两核心城市经济发展势头强劲设立河北雄安新区一段时间后,京津冀人口重心总体移动方向为A. 向东B. 向西C. 向南D. 向北水闸是修建在河道和渠道上利用闸门控制流量和调节水位的低水头水工建筑物。

关闭闸门可以拦洪、挡潮或抬高上游水位,以满足灌溉、发电、航运、水产、环保、工业和生活用水等需要;开启闸门,可以宣泄洪水、涝水、弃水或废水,也可对下游河道或渠道供水。

下图是不同河段设置的水闸示意图。

据此完成3~4题。

图中甲处闸门的主要作用是A. 防止海水倒灌B. 提高水位,保障通航C. 拦蓄泥沙D. 保障供水在长江中下游地区,丙类闸门被迫开启的时间是A. 1月B. 5月C. 8月D. 11月电影《金蝉脱壳》中,西尔维斯特·史泰龙为了逃出自己所设计的轮船监狱,与阿诺德·施瓦辛格合作,最终逃出生天。

剧中史泰龙发现轮船位于不知名的海洋上,他与施瓦辛格的这段对话充满了地理的味道,“看到北极星视角高度约30度”,“这里是无风带,海面平静”,“昨天有个新囚犯进来,他说是9月10日”,“如果要兴建秘密监狱,国际海面、风平浪静、距陆地不远以取得补给品”,据此完成5~6题。

内蒙古鄂尔多斯市2017届高三模拟考试数学理试题 含答案 精品

内蒙古鄂尔多斯市2017届高三模拟考试数学理试题 含答案 精品

内蒙古鄂尔多斯市2017届高三模拟考试数学(理)试题第Ⅰ卷(共60分)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.若集合1{|0}2A x x x =><或,集合{|(1)(2)0}B x x x =+-<,则A B =( ) A .1{|2}2x x << B .1{|102}2x x x -<<<<或C .1{|1}2x x -<<D .1{|012}2x x x <<<<或2.设i 为虚数单位,2(34)(,)i a bi a b R -+=+∈,则下列判断正确的是( ) A .||5a bi += B .1a b += C .17a b -=- D .168ab =3.根据下边框图,当输入x 为2017时,输出的y 为( )A .43B .10C . 4D . 2 4.二项式*1()()n x n N x-∈的展开式中,存在常数项的一个充分条件是( )A .5n =B .6n = C. 7n = D .9n = 5.把函数sin(2)6y x π=-的图象向左平移6π个单位后,所得函数图象的一条对称轴为( )A .0x =B .2x π=C. 6x π=D .12x π=-6.《算术书》竹筒出土于上世纪八十年代,是我国现存最早的有系统的数学典籍,其中记载有求“囷(qun )盖”之术:置如其周,令相承也.又以高乘之,三十六成一.该术相当于给出了由圆锥的底面周长L 与高h ,计算其体积V 的近似公式2136V L h ≈. 它实际上是将圆锥体积公式中的圆周率π近似取为3,那么,近似公式2275V L h ≈相当于将圆锥体积公式中的π近似取为( ) A .227 B .258C. 15750 D .355113 7.如图所示,在边长为1的正方形OABC 内任取一点,用A 表示事件“点P 恰好在由曲线y =1x =及x 轴所围成的曲边梯形内”, B 表示事件“点P 恰好取自阴影部分内”,则(|)P B A =( )A .14 B . 15 C. 16 D .178.在等差数列{}n a 中,若4681012120a a a a a ++++=,则7513a a -的值为( ) A .8 B . 12 C. 16 D .729.某三棱锥的三视图如图所示,则该三棱锥的体积为( )A . 1B .13 C. 12 D .1610.函数1222x y x +=-的图象大致是( )A .B .C. D .11.设点12,F F 分别为双曲线:22221(0,0)x y a b a b-=>>的左、右焦点,若在双曲线左支上存在一点P ,满足112||||PF F F =,点1F 到直线2PF 的距离等于双曲线的实轴长,则该双曲线的离心率为( ) A.4B .43 C. 54 D .5312.已知223,20()1ln,021x x x f x x x ⎧-+-≤<⎪=⎨≤≤⎪+⎩,若()|()|g x f x ax a =--的图象与x 轴有3个不同的交点,则实数a 的取值范围为( ) A .ln 31[,)3e B .ln 31[,)32e C. 1(0,)e D .1(0,)2e第Ⅱ卷(共90分)二、填空题(每题5分,满分20分,将答案填在答题纸上)13.已知O 为坐标原点,点C 是线段AB 上一点,且(1,1)A 、(2,3)C ,||2||BC AC =,则向量OB 的坐标为 .14.已知实数,x y 满足2101230x y x x y -+≥⎧⎪≥⎨⎪-+≤⎩,则y x 的取值范围为 .15.在各项均为正数的等比数列{}n a 中,112(2)m m m a a a m -+=≥,数列{}n a 的前n 项积为n T ,若21512m T -=,则m 的值为 .16.过抛物线C :28y x =的焦点F 作直线与C 交于,A B 两点,线段AB 的垂直平分线交x 轴于点P ,则||||AB PF = . 三、解答题 (本大题共6小题,共70分.解答应写出文字说明、证明过程或演算步骤.)17. 在ABC ∆中,内角,,A B C 所对的边为,,a b c sin (2cos )0C c A -+=. (1)求角A 的大小;(2)若ABC ∆sin 2sin C B =,求最小边长.18. 为加快新能源汽车产业发展,推进节能减排,国家对消费者购买新能源汽车给予补贴,其中对纯电动乘车补贴标准如下表:某校研究性学习小组,从汽车市场上随机选取了M 辆纯电动乘用车,很据其续驶里程R (单次充电后能行驶的最大里程)作出了频率与频数的统计表:(1)求,,,x y z M 的值;(2)若从这M 辆纯电动乘用车中任选3辆,求选到的3辆车续驶里程都不低于180公里的概率;(3)如果以频率作为概率,若某家庭在某汽车销售公司购买了2辆纯电动乘用车,设该家庭获得的补贴为X (单位:万元),求X 的分布列和数学期望EX .19. 如图,在四面体ABOC 中,OC OA ⊥,OC OB ⊥,120AOB ∠=,且1OA OB OC ===.(1)设P 为AC 的中点,证明:在AB 上存在一点Q ,使PQ OA ⊥,并计算ABAQ的值; (2)求二面角O AC B --的平面角的余弦值.20. 已知动点(,)M x y 到直线:3l x =的距离是它到点(1,0)D. (1)求动点M 的轨迹C 的方程;(2)设轨迹C 上一动点T 满足:23OT OP OQ λμ=+,其中,P Q 是轨迹C 上的点,且直线OP 与OQ 的斜率之积为23-,若(,)N λμ为一动点,1(6F -,2(6F 为两定点,求12||||NF NF +的值. 21. 设()xxf x e ex -=--.(1)求()f x 的单调区间;(2)已知23()()(1)[()(1)](1)g x x f x x f x a x a x =+++-+-,若对所有0x ≥,都有()0g x ≥成立,求实数a 的取值范围.请考生在22、23两题中任选一题作答,如果多做,则按所做的第一题记分.22.选修4-4:坐标系与参数方程在平面直角坐标系xOy 中,以坐标原点O 为极点,以x 轴正半轴为极轴,建立极坐标系,若直线l 的参数方程为0cos sin x t y y t αα=⎧⎨=+⎩(t 为参数,α为l 的倾斜角),曲线E 的极坐标方程为4sin ρθ=,射线θβ=,4πθβ=+,4πθβ=-与曲线E 分别交于不同于极点的三点,,A B C .(1)求证:||||OB OC +;(2)当7=12πβ时,直线l 过,B C 两点,求0y 与α的值. 23.选修4-5:不等式选讲已知函数()|2|2|1|f x x x =-++的最小值为m . (1) 求m 的值;(2)若,,a b c R ∈,2222a b c m ++=,求()c a b +的最大值.绝密★启用前 试卷类型:A2017年鄂尔多斯市高考模拟考试 理数试题参考答案与评分标准一、选择题:(每小题5分,共60分)二、填空题:(每小题5分,共20分)13. (4,7) ; 14. 1(,3]2; 15. 5; 16. 2. 三、解答题:(共70分)17.解: (I)sin sin (2cos )0A C C A -+=, ∵sin 0C ≠,cos 2A A -=. ∴sin()16A π-=,且(0,)A π∈∴62A ππ-=, 23A π=(II) 易知a为最大边,故a =由sin 2sin C B =,得2c b =. ∴最小边为长b . 根据余弦定理,有2222cos a b c bc A =+-.∴222742b b b =++∴ 1.b = 即最小边长为1.18.解:(I)易求0.6x =,1y =,0.1z =,10.M =(II)373107.24C P C ==∴从这10辆纯电动乘用车中任选3辆,选到的3辆车续驶里程都不低于180公里的概率为7.24(III)X 所有可能的取值为5,6.5,8,8.5,10,12.其中,2(5)0.30.09P X ===,12( 6.5)0.30.60.36P X C ==⨯=, 2(8)0.60.36P X ===,12(8.5)0.30.10.06P X C ==⨯=, 12(10)0.10.60.12P X C ==⨯=,2(12)0.10.01.P X ===∴X 的分布列为∴E (X )=5×0.09+6.5×0.36+8×0.36+8.5×0.06+10×0.12+12×0.01=7.519.解法一:(I)在平面OAB 内作ON ⊥OA 交AB 于点N ,连接NC . 又OC ⊥OA ,OA ∩ON =O , ∴OA ⊥平面ONC .∵NC ⊂平面ONC ,∴OA ⊥NC .取Q 为AN 的中点,连接PQ ,则PQ ∥NC , ∴PQ ⊥OA .在等腰△AOB 中,∠AOB =120O,∴∠OAB =∠OBA =30O. 在Rt △AON 中,∠OAN =30O,∴ON=12AN=AQ. 在△ONB 中,∠NOB =120O-90O=30O=∠NBO . ∴NB=ON=AQ ,∴3.ABAQ=(II)连接PN 、PO ,由已知得OC ⊥平面OAB ,又ON ⊂平面OAB , ∴OC ⊥ON ,又ON ⊥OA ,OA ∩OC =O ,∴ON ⊥平面AOC ,∴OP 是NP 在平面AOC 内的射影. 在等腰Rt △AOC 中,P 为AC 的中点, ∴AC ⊥OP ,则可知AC ⊥NP . ∴∠OPN 为二面角O-AC-B 的平面角.在等腰Rt △AOC 中,OC=OA =1,∴OP =2.在Rt △AON 中,ON =OA tan30o.在Rt △PON 中,PN =∴cos ∠OPN =PO PN =即二面角O-AC-B 的平面角的余弦值为5解法二:(I)过O 在平面OAB 内作OD ⊥OA ,交AB 于点D ,以O 点为原点,分别以OA 、OD 、OC 为x 轴、y 轴、z 轴,建立空间直角坐标系,如图所示.则A (1,0,0),C (0,0,1),B (12-,2,0),∴P (12,0,12). 设AQ QB λ=,则Q (1121λλ-+,21λ+,0),∴(1,0,0)OA =,122(1)21,,2(1)()PQ λλλ--++=.由OA ⊥PQ 得0OA PQ ⋅=,∴0122(1)λλ=-+,∴1.2λ=故存在Q (12,6,使得OA ⊥PQ . 此时,ABAQ3.AB AQ = (II)易求平面OAC 的一个法向量为(0,1,0)m =,而3(,22AB =-,(1,0,1)AC =-, 设平面ABC 的一个法向量为000()n x ,y ,z =,则00n AB n AC ⎧⋅=⎪⎨⋅=⎪⎩,∴00003020x y x z ⎧-=⎪⎨⎪-+=⎩, 解得(131)n ,,=.∴3cos ,m n mn m n⋅<>===即二面角O-AC-B20.解:(I)点(,)M x y 到直线3x =的距离是到点(1,0)F )0,2(T则3x -=,化简得22132x y += (II)设(,)T x y ,11(,)P x y ,22(,)Q x y ,则由23OT OP OQ λμ=+, 得1223x x x λμ=+,1223y y y λμ=+∵点T 、P 、Q 在椭圆22132x y +=上, ∴所以2211236x y +=,2222236x y +=,2223 6.x y += 故22221212232(23)3(23)x y x x y y λμλμ+=+++222222112212124(23)9(23)12(23)x y x y x x y y λμλμ=+++++ 221212245412(23)6x x y y λμλμ=+++=设OP OQ k k 、分别为直线OP 、OQ 的斜率,由题意知,121223OP OQ y y k k x x ⋅==-,因此1212230x x y y +=, ∴22491λμ+=.所以N 点是椭圆上22491λμ+=的点,而12F F 、恰为该椭圆的左、右焦点,由椭圆的定义,12 1.NF NF =+21.解:(I) ()1x xf x e e -'=+-12110≥=-=>,∴在()f x (,)-∞+∞上是增函数.(II)23()()(1)()(1)(1)(1)g x x f x x f x a x x a x =+++-++-23(1)()(1)[(1)]x x f x a x x x =+++-++ 2(1)[()(1)]x x f x x a =+++-显然210x x ++>,故若使()0g x ≥,只需()(1)0x xf x x a e e ax -+-=--≥ 即可.令()x x h x e e ax -=--,则()2.x x h x e e a a a -'=+-≥=- (i)当20a -≥即2a ≤时,()0h x '≥恒成立,∴()h x 在[)0,+∞内为增函数∴()(0)0h x h ≥=,即()0g x ≥在[)0,+∞上恒成立.(ii)当2a >时,则令()0h x '=,即0x x e e a -+-=,可化为2()10x x e ae -+=,解得2xa e =,∴两根1ln 02a x ==<(舍),20.x => 从而122()1()()()x x x x x x x xe ae e e e e h x e e -+--'==. 当20x x <<时,则12,x x x x e e e e ><,∴()0h x '<,∴()h x 在[]20,x 为减函数.又(0)0h =,∴2()0.h x <∴当2a >时,()0h x ≥不恒成立,即()0g x ≥不恒成立.综上所述,a 的取值范围为(],2.-∞22.解:(I)证明:依题意,||4sin OA β=,(II) B 点的极坐标为 C 点的极坐标为则直线l所以02y =,23.解:(I)由于3,1()+4,123,2x x f x x x x x -≤-⎧⎪=-≤<⎨⎪≥⎩ , 所以min ()|(1)3m f x f ==-=.(II) 由已知22232a cb ++=,有2222()()6a bc b +++=, 因为222a b ab +≥(当a b =取等号),222c b bc +≥(当c b =取等号),所以2222()()622a b c b ab bc +++=≥+,即3ab bc +≤ 故max [()] 3.b a c +=:A 绝密★启用前试卷类型2017年鄂尔多斯市高考模拟考试理数试题参考答案与评分标准一、选择题:(每小题5分,共60分)二、填空题:(每小题5分,共20分)13. (4,7) ; 14. 1(,3]2; 15. 5; 16. 2.三、解答题:(共70分)17.解: (I)sin sin (2cos )0A C C A -+=, ………………2分∵sin 0C ≠,cos 2A A -=. ∴sin()16A π-=,且(0,)A π∈……………………………………………4分∴62A ππ-=, 23A π=………………………………………………6分 (II) 易知a为最大边,故a =7分由sin 2sin C B =,得2c b =. ∴最小边为长b . ………………………9分 根据余弦定理,有2222cos a b c bc A =+-.∴222742b b b =++………………………………………………………11分 ∴ 1.b = 即最小边长为1.………………………………………12分18.解:(I)易求0.6x =,1y =,0.1z =,10.M =…………………………………3分(II)373107.24C P C ==∴从这10辆纯电动乘用车中任选3辆,选到的3辆车续驶里程都不低于180公里的概率为7.24……………………………………………………………6分 (III)X 所有可能的取值为5,6.5,8,8.5,10,12. …………………………7分其中,2(5)0.30.09P X ===,12( 6.5)0.30.60.36P X C ==⨯=,2(8)0.60.36P X ===,12(8.5)0.30.10.06P X C ==⨯=, 12(10)0.10.60.12P X C ==⨯=,2(12)0.10.01.P X ===∴X 的分布列为………………………10分∴E (X )=5×0.09+6.5×0.36+8×0.36+8.5×0.06+10×0.12+12×0.01=7.5…………………………………………………………………12分19.解法一:(I)在平面OAB 内作ON ⊥OA 交AB 于点N ,连接NC .又OC ⊥OA ,OA ∩ON =O ,∴OA ⊥平面ONC . ………………………………2分 ∵NC ⊂平面ONC ,∴OA ⊥NC .取Q 为AN 的中点,连接PQ ,则PQ ∥NC , ∴PQ ⊥OA . ……………………………………3分 在等腰△AOB 中,∠AOB =120O ,∴∠OAB =∠OBA =30O . 在Rt △AON 中,∠OAN =30O ,∴ON=12AN=AQ. 在△ONB 中,∠NOB =120O -90O =30O =∠NBO . ∴NB=ON=AQ ,∴3.ABAQ=……………………5分 (II)连接PN 、PO ,由已知得OC ⊥平面OAB ,又ON ⊂平面OAB , ∴OC ⊥ON ,又ON ⊥OA ,OA ∩OC =O ,∴ON ⊥平面AOC ,∴OP 是NP 在平面AOC 内的射影.QCO BA P(第19题图)N在等腰Rt △AOC 中,P 为AC 的中点, ∴AC ⊥OP ,则可知AC ⊥NP .∴∠OPN 为二面角O-AC-B 的平面角.…………………………………………8分 在等腰Rt △AOC 中,OC=OA =1,∴OP=2. 在Rt △AON 中,ON =OA tan30o…………………………………………10分 在Rt △PON 中,PN=∴cos ∠OPN=5PO PN = ……………………………………………………11分 即二面角O-AC-B的平面角的余弦值为5……………………… ………12分 解法二:(I)过O 在平面OAB 内作OD ⊥OA ,交AB 于点D ,以O 点为原点,分别以OA 、OD 、OC 为x 轴、y 轴、z 轴,建立空间直角坐标系,如图所示.………1分 则A (1,0,0),C (0,0,1),B (12-,∴P (12,0,12). ……………………2分 设AQ QB λ=,则Q (1121λλ-+,21λ+,0),……………………………………3分∴(1,0,0)OA =,122(1)12()PQ λλ--+=.由OA ⊥PQ 得0OA PQ ⋅=,∴0122(1)λλ=-+,∴1.2λ=………… ……4分故存在Q (12,使得OA ⊥PQ .………………………………………5分 此时,ABAQ3.AB AQ = …………………………………6分 (II)易求平面OAC 的一个法向量为(0,1,0)m =而3(2AB =-,(1,0,1)AC =-,设平面ABC 的一个法向量为000()n x ,y ,z =,则00n AB n AC ⎧⋅=⎪⎨⋅=⎪⎩,∴000030220x y x z ⎧-+=⎪⎨⎪-+=⎩, 解得(131)n ,,=. ……………………………………………………………9分∴3cos ,1m n m n m n⋅<>===⨯ …………………………………11分 即二面角O-AC-B ……………………………12分 20.解:(I)点(,)M x y 到直线3x =的距离是到点(1,0)F )0,2(T则3x -=, …………………………………………2分化简得22132x y +=……………………………………………………………4分 (II)设(,)T x y ,11(,)P x y ,22(,)Q x y ,则由23OT OP OQ λμ=+,得1223x x x λμ=+,1223y y y λμ=+…………………………………………5分∵点T 、P 、Q 在椭圆22132x y +=上, ∴所以2211236x y +=,2222236x y +=,2223 6.x y +=…………………6分 故22221212232(23)3(23)x y x x y y λμλμ+=+++222222112212124(23)9(23)12(23)x y x y x x y y λμλμ=+++++ 221212245412(23)6x x y y λμλμ=+++= …………………8分设OP OQ k k 、分别为直线OP 、OQ 的斜率,由题意知,121223OP OQ y y k k x x ⋅==-,因此1212230x x y y +=,…………………………9分 ∴22491λμ+=. ……………………………………………………………10分 所以N 点是椭圆上22491λμ+=的点,而12F F 、恰为该椭圆的左、右焦点,由椭圆的定义,12 1.NF NF =+ ……………………………………………12分21.解:(I) ()1x x f x e e -'=+-……………………………………………………………2分12110≥=-=>, ……………………………………4分∴在()f x (,)-∞+∞上是增函数.……………………………………………5分(II)23()()(1)()(1)(1)(1)g x x f x x f x a x x a x =+++-++-23(1)()(1)[(1)]x x f x a x x x =+++-++2(1)[()(1)]x x f x x a =+++- ………………………………………………6分显然210x x ++>,故若使()0g x ≥,只需()(1)0x x f x x a e e ax -+-=--≥ 即可. ……………………………………………………………………………7分令()x x h x e e ax -=--,则()2.x x h x e e a a a -'=+-≥=-……………………………………………………………………………8分(i)当20a -≥即2a ≤时,()0h x '≥恒成立, ∴()h x 在[)0,+∞内为增函数∴()(0)0h x h ≥=,即()0g x ≥在[)0,+∞上恒成立.……………………………9分(ii)当2a >时,则令()0h x '=,即0x x e e a -+-=,可化为2()10x xe ae -+=,解得xe =,∴两根1ln 02a x ==<(舍),20.x => 从而122()1()()()x x x x x x x xe ae e e e e h x e e-+--'==. 当20x x <<时,则12,xxxxe e e e ><, ∴()0h x '<,∴()h x 在[]20,x 为减函数.又(0)0h =,∴2()0.h x <∴当2a >时,()0h x ≥不恒成立,即()0g x ≥不恒成立. ………………………11分 综上所述,a 的取值范围为(],2.-∞ ………………………………………………12分22.解:(I)证明:依题意,||4sin OA β=,…………………………………………2分…………………………………………5分(II)B 点的极坐标为 ……………………6分C 点的极坐标为7分8分 则直线l9分 所以02y =,10分23.解:(I)由于3,1()+4,123,2x x f x x x x x -≤-⎧⎪=-≤<⎨⎪≥⎩ ,…………………………………………… 3分 所以min ()|(1)3m f x f ==-=.………………………………………………5分(II) 由已知22232a cb ++=,有2222()()6a bc b +++=, ………………………………………………………7分因为222a b ab +≥(当a b =取等号),222c b bc +≥(当c b =取等号),……………………………………8分所以2222()()622a b c b ab bc +++=≥+,即3ab bc +≤故max [()] 3.b a c +=…………………………………………………………10分。

内蒙古鄂尔多斯市高三化学模拟考试试题(扫描版)

内蒙古鄂尔多斯市高三化学模拟考试试题(扫描版)

绝密★启用前试卷类型:A2017年鄂尔多斯市高考模拟考试化学试题参考答案与评分标准第I 卷选择题(每题6分,共42分)7.C 8.B 9.A 10.D 11.B 12.C 13.D第II 卷(一)必考题26.(15分)(1) ①bde (2分,漏选一个扣一分,错选不得分)②增大 (2分)③0.005mol·L —1·min -1(2分) 25%(2分)是(1分) 因为10min 时的Qc=K (2分)(2) CH 4(g)+N 2O 4(g) ==N 2(g) +2H 2O(l ) + CO 2(g) △H = —898.1kJ/mol (2分)(3) 14(2分)27.(14分) (1) (2分)(2)NaCN+NaClO = NaCNO+NaCl (2分)2CNO -+2H ++3ClO -= N 2↑+2CO 2↑+3Cl -+H 2O (2分)(3)除去氯气 (2分)(4)过滤、洗涤、干燥、称量戊中沉淀的质量,计算出CN -被处理的百分率(2分)防止空气中的二氧化碳进入戊装置影响沉淀测定的准确度(2分)(5)CO 2+H 2O+NaCN = HCN+NaHCO 3 (2分)28.(14分) N ︰]- Na +[︰C(1)2FeCl3 + 10NaOH + 3NaClO = 2Na2FeO4 + 9NaCl + 5 H2O (2分) NaClO(1分)(2)K2FeO4的溶解度小于Na2FeO4的溶解度(2分)(3)碱性(1分)c(Na+)〉 c(ClO—)> c(OH—)> c(H+) (2分)(4)Fe+8OH—-6e—= FeO42-+4H2O (2分)(5)正(2分);(6)本身具有强氧化性可杀菌消毒,在反应中被还原成的三价铁离子水解生成氢氧化铁胶体吸附水中的杂质而起到净水作用。

(答出一点的得一分)(2分)(二)选考题35.[化学—选修3:物质结构与性质](15分)(1)1s22s22p63s23p63d24s2或[Ar]3d24s2(2分)三周期 VIIA族 (2分)(2)V形(1分) sp3杂化(2分)(3)分子晶体(2分)分子间作用力(范德华力)和共价键(2分)(4)CaTiO(2分) (2分)336.[化学—选修5:有机化学基础](15分)(1)醛基 (1分) (2)①③④(2分)(3)(2分)(2分)(4)(2分,任写一种给2分) 5mol (2分)(5)2种(2分)或 (2分)尊敬的读者:本文由我和我的同事在百忙中收集整编出来,本文稿在发布之前我们对内容进行仔细校对,但是难免会有不尽如人意之处,如有疏漏之处请指正,希望本文能为您解开疑惑,引发思考。

内蒙古鄂尔多斯市第二中学2017-2018学年高三上学期期中考试化学试题 Word版含答案

内蒙古鄂尔多斯市第二中学2017-2018学年高三上学期期中考试化学试题 Word版含答案

鄂尔多斯市第二中学2017-2018学年高三第一学期期中考试化学试题可能用到的相对原子质量:H-1 C-12 N-14 O-16 Na-23 Mg-24 Al-27 S_32Cl-35.5 K-39 Fe-56 Cu-64 Cl-35.5 Ba-137第Ⅰ卷 (共50分)一.选择题(每小题只有一个选项符合题意,每小题2.5分)1.下列“化学与生活”的说法不正确的是A.硫酸钡可用于钡餐透视B.盐卤可用于制豆腐C.明矾可用于水的消毒、杀菌D.醋可用于除去暖水瓶中的水垢2.海水是巨大的资源宝库,下列说法正确的是A.碘元素又称“海洋元素” B.镁是海水中含量最多的金属元素C.从海水中提取食盐和碘的过程都是物理变化 D.氯碱工业和纯碱工业的原料都有食盐3.下列有关贮存方法的叙述错误的是A.浓硝酸用棕色瓶盛放,贮存在阴凉处 B.少量液溴可用水封存,防止溴挥发C.氢氧化钠溶液贮存在带磨口玻璃塞的玻璃试剂瓶中 D.实验室中少量的金属钠保存在煤油中4.NA为阿伏伽德罗常数的值。

下列说法正确的是A.18gD2O和18gH2O中含有的质子数均为10N AB.50ml 12mol/L盐酸与足量MnO2共热,转移的电子数为0.3N AC.过氧化钠与水反应时,生成0.1mol氧气转移的电子数为0.2N AD.密闭容器中2molNO与1molO2充分反应,产物的分子数为2N A5.化学与社会、生活密切相关。

对下列现象或事实的解释正确的是6.下列有关铝及其化合物的叙述正确的是A.铝粉与氧化镁混合共热可制得金属镁B.足量铝分别与含1 mol HCl、1 mol NaOH的酸、碱溶液反应,产生相同量的氢气C.工业上用电解熔融Al2O3的方法冶炼铝D.铝在等体积的浓硫酸、浓硝酸的混合液中钝化7、下列叙述正确的是A.将CO2通入BaCl2溶液中至饱和,无沉淀产生;再通入SO2,产生沉淀B.在稀硫酸中加入铜粉,铜粉不溶解;再加入Cu(NO3)2固体,铜粉仍不溶解C.向AlCl3溶液中滴加氨水,产生白色沉淀;再加入过量NaHSO4溶液,沉淀消失D.纯锌与稀硫酸反应产生氢气的速率较慢;再加入少量CuSO4固体,速率不改变8.比较锌和二氧化锰分别与盐酸的反应,下列叙述错误的是A.若Zn和MnO2都是足量的,则前一反应中盐酸全部被还原,后一反应中盐酸部分被氧化 B.两个都是氧化还原反应 C.盐酸都是被还原D.等物质的量的Zn和MnO2分别与足量的浓盐酸反应,转移的电子数相等9.下列指定反应的离子方程式正确的是A.氯气溶于水:Cl2+H2O=2H++Cl-+ClO-B.Na2CO3溶液中CO32-的水解:CO32-+H2O=HCO3-+OH-C.酸性溶液中KIO3与KI反应生成I2:IO3-+I-+6H+=I2+3H2OD.NaHCO3溶液中加足量Ba(OH)2溶液:HCO3-+Ba2++OH-=BaCO3↓+H2O10.11.用FeCl3溶液腐蚀印刷电路板上的铜,所得的溶液中加入铁粉。

内蒙古鄂尔多斯市2017届高三物理模拟考试试题 精

内蒙古鄂尔多斯市2017届高三物理模拟考试试题 精

绝密★启用前 试卷类型:A2017年鄂尔多斯市高考模拟考试 物理试题参考答案与评分标准二、选择题(每题6分,共48分)14.A 15.B 16.D 17.C 18.D 19.ACD 20.BD 21.AC三、非选择题(共62分) (一)必考题(共47分) 22.(6分)(1)CD ; (2)1.50; 2.00(1.99~2.01); 0.50(0.49~0.51) 评分标准:本题6分,其中(1)2分,(2)第一空、第二空各1分,第三空2分。

23.(9分)(1)A (2)5、6 (3)2.60 ,0.27 (4)偏大 评分标准:本题共9分,前四空每空2分,第五空1分。

24.(13分)解:(1)滑块在圆弧轨道上下滑过程中,由机械能守恒定律得: 2021mv mgR =① 解得: m/s 30=v②在B 点,由牛顿第二定律得:Rv m mg F N 20=-③代入数据得,轨道对滑块的支持力N 60=N F④(2)设滑块最终和小车共同速度为v ,由动量守恒定律得:()v M m mv +=0⑤ 解得: m/s 1=v⑥设小车前进的距离为x ,对小车由动能定理得:221Mv mgx =μ ⑦代入数据解得:m 25.0=x⑧评分标准:本题共13分。

其中⑦式3分,①③④⑤⑧每式2分。

25.(19分)解:(1)粒子在磁场中运动时,根据牛顿第二定律得:rm B q 20v v =① 而mRqB 0=v ,所以 R r = ②即粒子运动轨迹与弧ab 的交点A 及a 、O 组成一正三角形。

故A 点的横、纵坐标分别为: ︒=30cos 1R x③ ︒=30sin 1R y④解得: R x 231= ⑤ R y 211=⑥ (2)粒子在I 区域的轨迹弧长对应的圆心角为3π,进入II 区域后,偏转方向改变,轨道半径不变。

轨迹与弧cd 的交点C 与A 、O 在一条直线上,运动轨迹如图2。

粒子在I 、II 两区域的运动轨迹对应的圆心角相等,故运动时间相等⑦粒子从O 到C 的运动时间vrt 32π⨯=⑧ 整理得: 032qB mt π=⑨(3)粒子经过C 点时速度沿+x 方向,由几何关系可知C 点的坐标值yxOa c eb d fA 图1C yxOa c eb dfA 图2R x 32=⑩R y =2⑪若III 区域的磁场方向向外,粒子恰好从f 点经过时,轨迹如图3。

2018届内蒙古鄂尔多斯市高三下学期模拟考试理科数学试题及答案

2018届内蒙古鄂尔多斯市高三下学期模拟考试理科数学试题及答案

绝密★启用并使用完毕前2017年鄂尔多斯市高考模拟考试理 科 数 学注意事项:1.本试卷分第I 卷(选择题)和第II 卷(非选择题)两部分。

答卷前,考生务必将自己的姓名、座位号、考生号、试卷类型(A 或B )涂写在答题卡上。

本试卷满分150分,考试时间120分钟。

2.回答第I 卷时,选出每小题答案后,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。

写在本试卷上无效。

3.回答第II 卷时,将答案写在答题卡上,写在本试卷上无效。

4.考试结束后,将本试卷和答题卡一并交回。

第I 卷(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.已知全集{1,2,3,4,5}U =,集合{1,3,4}A =,集合{2,4}B =,则()U C A B为 A .{2,4,5} B .{1,3,4} C .{1,2,4} D .{2,3,4,5}试卷类型:A2.已知复数12312z bi z i =-=-,,若12z z 是实数,则实数b 的值为 A .0 B .32- C .6 D .6- 3. 已知{}n a 是由正数组成的等比数列,n S 表示{}n a 的前n 项的和,若13a =,24144a a =,则5S 的值是A .692B .69C .93D .1894. 设αβγ、、为平面,l n m 、、为直线,以下四组条件,可以作为β⊥m 的一个充分条件的是 A .,,l m lαβαβ⊥=⊥B .,,m αγαγβγ=⊥⊥C .,,m αγβγα⊥⊥⊥D .,,n n m αβα⊥⊥⊥5.在5()a x x-的展开式中3x 的系数等于5-,则该展开 式各项的系数中最大值为A .5B .10C .15D .206.已知数列{}n a 的各项均为正数,执行程序框图(如右 图),当4k =时,31=S ,则2014a = A .2012 B .2013 C .2017 D .20187.已知某几何体的三视图如图所示,其中俯视图是圆,且 该几何体的体积为1V ;直径为2的球的体积为2V . 则12:V V = A .1:4 B .1:2 C .1:1D. 2:1 8. 若ln 2ln 3ln 5,,235a b c ===,则 A .a b c << B .c a b << C .c b a << D .b a c <<9.已知实数x 、y 满足约束条件22,24,41x y x y x y +≥+≤-≥-⎧⎪⎨⎪⎩.若),(y x a =,)1,3(-=b ,设z 表示向量在方向上的投影,则z 的取值范围是 A .3,62⎡⎤-⎢⎥⎣⎦B .[]1,6- C.[ D.[10. 已知0a ≥,函数2()(2)x f x x ax e =-,若()f x 在[1,1]-上是单调减函数,则a 的取值范围是A .304a << B .1324a << C .34a ≥D .102a <<11.已知直线)2(+=x k y (k >0)与抛物线2:8C y x =相交于A 、B 两点,F为C 的焦点,若||2||FA FB =,则k 的值为A .13B.3C .23 D.312. 设函数()1f x xx=-+ )(R x ∈,集合{},),(M x x f y y N ∈==其中[]b a M ,=a (<)b ,则使N M =成立的实数对),(b a 有A .0个B .1个C .2个D .无数多个第II 卷(共90分)本卷包括必考题和选考题两部分,第13题~第21题为必考题,每个试题考生都必须作答。

内蒙古鄂尔多斯市高考数学三模试卷(理科)

内蒙古鄂尔多斯市高考数学三模试卷(理科)

内蒙古鄂尔多斯市高考数学三模试卷(理科)姓名:________ 班级:________ 成绩:________一、选择题 (共12题;共24分)1. (2分)已知集合,,则()A .B .C .D .2. (2分)复数(i是虚数单位)在复平面上对应的点位于()A . 第一象限B . 第二象限C . 第三象限D . 第四象限3. (2分)设是等差数列的前n项和,若,则等于()A .B .C .D .4. (2分) AD,BE分别是的中线,若,且与的夹角为,则()A .B .C .D .5. (2分)已知抛物线与双曲线有相同的焦点F,点A是两曲线的一个交点,且轴,则双曲线的离心率为()A . 2B .C .D .6. (2分)(2016·绵阳模拟) 某四棱锥的三视图如图所示,则该四棱锥的体积是()A . 36B . 30C . 27D . 127. (2分) (2016高一上·荆州期中) 碘﹣131经常被用于对甲状腺的研究,它的半衰期大约是8天(即经过8天的时间,有一半的碘﹣131会衰变为其他元素).今年10月1日凌晨,在一容器中放入一定量的碘﹣131,到10月25日凌晨,测得该容器内还剩有2毫克的碘﹣131,则10月1日凌晨,放人该容器的碘﹣131的含量是()A . 8毫克B . 16毫克C . 32毫克D . 64毫克8. (2分)执行如图所示的程序框图.若输出的结果为3,则可输入的实数x的个数为()A . 1B . 2C . 3D . 49. (2分)某公共汽车上有10名乘客,沿途有5个车站,乘客下车的可能方式有()A . 种B . 种C . 50种D . 10种10. (2分)已知f(x)=sin(x+),若sinα=(<α<π),则f(α+)=()A .B . -C .D . -11. (2分)在区间内任取两个数a,b,则使方程的两个根分别作为椭圆与双曲线的离心率的概率为()A .B .C .D .12. (2分) (2019高一上·海林期中) 设函数的最小值是1,则实数的取值范围是()A .B .C .D .二、填空题 (共4题;共4分)13. (1分) (2016高二下·丹阳期中) 在(x2﹣)6的展开式中,常数项等于________.14. (1分)已知数列{an}和{bn}的通项公式分别是,,其中a、b是实常数,若,,且a,b,c成等差数列,则c的值是________15. (1分) (2016高二上·灌云期中) 设实数x,y满足,则z=|x﹣1|+|y+2|的取值范围为________.16. (1分)(2018·潍坊模拟) 一个几何体的三视图如图所示,则该几何体的外接球的体积为________.三、解答题 (共7题;共70分)17. (10分) (2019高三上·德州期中) 如图,在四边形中,,,,连接,.(1)求的值;(2)若,,求的面积最大值.18. (10分)(2016·中山模拟) 有一批货物需要用汽车从生产商所在城市甲运至销售商所在城市乙,已知从城市甲到城市乙只有两条公路,且通过这两条公路所用的时间互不影响.据调查统计,通过这两条公路从城市甲到城市乙的200辆汽车所用时间的频数分布如表:所用的时间(天数)10111213通过公路l的频数20402020通过公路2的频数10404010假设汽车A只能在约定日期(某月某日)的前11天出发,汽车B只能在约定日期的前12天出发(将频率视为概率).(1)为了尽最大可能在各自允许的时间内将货物运往城市乙,估计汽车A和汽车B应如何选择各自的路径;(2)若通过公路l、公路2的“一次性费用”分别为3.2万元、1.6万元(其他费用忽略不计),此项费用由生产商承担.如果生产商恰能在约定日期当天将货物送到,则销售商一次性支付给生产商40万元,若在约定日期前送到;每提前一天销售商将多支付给生产商2万元;若在约定日期后送到,每迟到一天,生产商将支付给销售商2万元.如果汽车A,B按(I)中所选路径运输货物,试比较哪辆汽车为生产商获得的毛利润更大.所以汽车A选择公路1.汽车B选择公路219. (10分) (2017高一上·珠海期末) 如图,DE∥BC,BC=2DE,CA⊥CB,CA⊥CD,CB⊥CD,F、G分别是AC、BC中点.(1)求证:平面DFG∥平面ABE;(2)若AC=2BC=2CD=4,求二面角E﹣AB﹣C的正切值.20. (5分)(2017·长春模拟) 如图,在矩形ABCD中,|AB|=4,|AD|=2,O为AB中点,P,Q分别是AD和CD上的点,且满足① = ,②直线AQ与BP的交点在椭圆E: + =1(a>b>0)上.(Ⅰ)求椭圆E的方程;(Ⅱ)设R为椭圆E的右顶点,M为椭圆E第一象限部分上一点,作MN垂直于y轴,垂足为N,求梯形ORMN 面积的最大值.21. (10分)(2019·江门模拟) 设函数,是自然对数的底数,是常数.(1)若,求的单调递增区间;(2)讨论曲线与公共点的个数.22. (10分)(2017·甘肃模拟) 选修4-4:坐标系与参数方程在直角坐标系xOy中,曲线C1的参数方程为(α为参数,﹣π<α<0),曲线C2的参数方程为(t为参数),以O为极点,x轴的正半轴为极轴建立极坐标系.(1)求曲线C1的极坐标方程和曲线C2的普通方程;(2)射线θ=﹣与曲线C1的交点为P,与曲线C2的交点为Q,求线段PQ的长.23. (15分) (2016高二下·长安期中) 设函数f(x)=|ex﹣a|+| ﹣1|,其中a,x∈R,e是自然对数的底数,e=2.71828…(1)当a=0时,解不等式f(x)<2;(2)求函数f(x)的单调增区间;(3)设a≥ ,讨论关于x的方程f(f(x))= 的解的个数.参考答案一、选择题 (共12题;共24分)1-1、2-1、3-1、4-1、5-1、6-1、7-1、8-1、9-1、10-1、11-1、12-1、二、填空题 (共4题;共4分)13-1、14-1、15-1、16-1、三、解答题 (共7题;共70分) 17-1、17-2、18-1、18-2、19-1、19-2、20-1、21-1、21-2、22-1、22-2、23-1、23-2、23-3、。

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第二部分阅读读理解(共两节,满分40分第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的A、B,C和D四个选项中,选出最佳选项。

AOfficial Historical Tour of HarvardStudent-led Public Walking TourPlease note:Due to ongoing construction,our tour schedule will be limited,Please ch eck our updated tour times on the calendar. In the event of bad weather,please check the calendar or regarding a scheduled tour time.The tour is student-led and includes an outdoor walk(please dress appropriately) through Harvard Yard,providing a history of the university,general information, and a unique view on the student's individual experience. The tour is around one hour long and completely free of charge. There is no pre-registration for general visitors, and we begin registration for each tour one hour before departure time.Currently we are only able to accommodate groups of up-to14 people on our free,public tour. If you have a group of15or more,reservations are required (please see below).Group Tours and ReservationsReservations are required for tours of 15 or more people, with an absolute maximum of 125 people. We request 2 weeks notice to schedule a tour Please note that we cannot accommodate groups that arrive without a reservation.A Self-Guided Walking Tour of Harvard YardIf you are interested in touring the campus on your own, you may want a self-guided walking tour. These tour pamphlets(册), offered in nine different languages, can be purchased from our office for a minimal charge.You may also download audio files of historical information on each of the buildings in Harvard Yard. These files office together including an Audio Tour, can also be obtained here with a simple click.21. What will you do on a student-led public walking tour?A. Wear whatever you like.B. Pay for admission and reservation.C. Go at any time and don't have to register.D. Learn about a student's personal experience.22. What should a group of 20 do to visit the Harvard campus?A. Register on arrival.B. Register before leaving.C. Reserve at least a fortnight beforehand.D. Make a reservation one week in advance23. Where is the text probably fromA .In a newspaper. B. In a textbook.C. On the Internet.D. Ina magazine.BLast April, on a Sunday, we took one of our“ nowhere” drive. My husband was quietly driving along a back road. I was occupied in the front passenger seat watching the scenery.I noticed out of the corner of my eye that my husband was struggling to look out of my window. This frightened me, since his eyes should be on the road in front ofhim. I asked him what he was looking at out of the windows, and he quietly replied, nothing.After a few minutes, I looked over at my husband and noticed a tear running down his check. I asked him what was wrong. This time he told me, “I was just thinking about Pop and a story he had once told me.” It had something to do with Pop, his friend from childhood, and I wanted to know the story, so I asked him to share it with me.He said, "When I was about 8 years old, Pop and I were out fishing and he told me that the pine trees know when it is Easter.”I had no idea what he meant by that, so I pressed him for more information, He continued on. "The pine trees start their new growth in the weeks before Easter because spring is drawing near. If you looks at the tops of the pine trees, you will see the yellow shoots. As the days get closer to Easter Sunday, the tallest shoot will branch off and form a cross. By the time Easter Sunday comes around, you will see that most of the pine trees will have small yellow crosses on all of the tallest shoots.”I turned to look out of the window and I couldn't believe my eyes. It was a week before Easter, and you could see all of the trees with the tall yellow shoots stretching to the sky.The tallest ones shone in the sunlight like rows of tiny golden crosses. May you find your Easter season filled with beautiful golden crosses!24. How did the author feel when she found her husband looking outward while driving?A. CuriousB. CalmC. AfraidD. Excited25. Why did the husband cry?A. He saw something strange out of the window.B. He was scolded by his wife for driving carelessly.C.Ye He was too frightened to drive the car on the roadD. He thought of an old friend and was moved by his story.26. What can we infer from the text?A.The author knew pop's story very well.B. Pop was traveling together with the couple.C. The author was amazed at the appearance of pine trees.D. Pine trees have something to do with the origin of Easter27. What does the author intend to tell us?A. We should be careful while driving.B. Pine trees are more beautiful before EasterC. Easter should be decorated with golden crosses.D. Easter promises hope with the coming of spring.CEvery time a new year is coming, people set out to better themselves. They promise they will lose weight, find a new job, or maybe even take that vacation they've always talked about. But why do we make the promises to ourselves, and where did this tradition come from? Why doe this tradition live on when so many people fail to keep the resolutions they made? Well we can start by blaming the ancient BabyloniansAround 4, 000 years ago in Babylon, the earliest recorded celebration honoring the coming of a new year was held. Calendars weren't as they are today, so the Babylonians did it in late March during the first new moon after the Spring Equinox (春分). The festivities were meant for the rebirth of the sun god, but the Babylonians made promises in order to please their gods. They felt this would help them start the new year off well.Resolutions continued on with the Romans. When the early Roman calendar no longer synced(同步) up with the sun, Julius Caesar decided to make a change. He consulted with the best astronomers and mathematicians of the time and introduced the Julian calendar, which more closely represents the modern calendar we use today. Caesar declared January I the first day of the year to honor the god of new beginnings, Janus, The Romans celebrated the New Year by offering sacrifices to Janus.To this day, the traditions of the ancient Babylonians and Romans continue on around the world. So much that Google launched a Resolution Map in 2019 where people could add resolutions and see others adding theirs in real time. However, no matter how many people participated in Google's project, the numbers are bleak when it comes to the amount of people who maintain their resolutions -only eight percent of people are successful in sticking them out.28. Why did the ancient Babylonians make promises to their gods?A. To honor the god of new beginnings.B. To observe the rebirth of the sun god and please him.C. To honor the coming of a new year and satisfy their gods.D. To please their gods and hope for a good start of the year.29. What can we know about the tradition of the New year's resolutionsA. The tradition of making New Years resolutions will soon die.B A lot of people don't feel like making New Years resolutionsC. Julius Caesar made January I the first day of a year on his owny The history of making New Years resolutions is not well known.30. What does the underlined word"bleak"in the last paragraph mean?A. LargeB. Unpleasant.C. Encouraging.D.Hopeful31. Which of the following can be the best title for the text?A. How Making New Year’s Resolutions Came Into BeingB. The Ancient Babylonians and the RomansC The Change of Roman CalendarsD How People Better Themselves33. What can prevent the users' information from being let out?A. The application of users' fingerprintB. The advice from top fitness experts.C. The use of voice and face recognition.D. Automatically stored analysis results34. What does the underlined part "The people behind the mirror refer to?A. The producer.B. The userC. The salesmanD.The beautician35. What helps prepare people for various occasions about the mirror?A . Expert skin and fitness analysis. B. A variety of lighting sit unions. C. A custom built skincare routine D. Several make-ups for you to try on第二节七选五There is one food that takes center stage during autumn the pumpkin!__36_____But we also use it in many foods that are popular during the fall season, like pumpkin bread, pumpkin coffee, pumpkin cookies and pumpkin pie. But perhaps we should try to eat pumpkin all year long, and not just in the fall. Pumpkins are packed with important nutrients and vitamins.__37_____ Pumpkin is good for your eyesight.The nutrient beta carotene(胡萝ト素)and Vitamin A are necessary for maintaining good eyesight.___38____In fact, one nutrition website says that one ct, of cooked, mashed pumpkin can give you more than 200 percent of your daily recommended amount of Vitamin A.___39_____The National Cancer Institute in the United States says on its website that antioxidants (抗氧化物)may help reduce a person’s risk of getting cancer. That is because antioxidants in pumpkins kill something called free radicals in the body. The Institute defines free radicals as"highly reactive chemicals that have the potential to harm cells”.Pumpkin helps your skinBeta-carotene also helps to keep your skin healthy and young looking. Yet Another good reason to eat pumpkin!Pumpkin may be good for weight loss.___40___Fiber slows down digestion of food. So, when you eat pumpkin you feel full and eat less.So, if it is available where you live, try to find ways to include pumpkin in your next meal.A. Pumpkin may help your heart.B. Pumpkin also has a lot of fiber.C. Pumpkin are a great source of both.D. Pumpkins are the king of vegetables.E. Pumpkin may help reduce your risk of getting cancer.F. Many people use Pumpkins as a decoration at Halloween.G. If cooked in a healthy way, pumpkin is a powerhouse of health benefits第二节语法填空I’m an Indian, speak, read and write Chinese, and have lived in Singapore for 8 years, Mainland China for 7, Hong Kong foe 3 and Taiwan for 2 yeas. I ____(marry) to a lady in Taiwan and I am back in Shanghai for 4th period of living on the mainland.Mainland China provides lots of _______(opportunity). things can be as expensive or cheap as you want. Big cities are clean and efficient. People are very friendly if you speak even ______little Chinese. Also, contrary to the popular ______(believe), there is a great degree of personal freedom in this country, _____it’s relatively easy to set up a business and it’s _______(extreme) safe. I wouldn’t retire here, but in my active working hours, I keep ______(come) back to Mainland China because it’s just amazing Shanghai, in particular, is very international. While it’s more expensive ____most other cities in Mainland China, you can still live here cheaply if you need to.I had to do that for about 2 years when I was in a startup and managed ______(live) quite well on very little money, which might not have worked in Singapore or HL.So personally, it would be _____(good) to live in Mainland China than in any other place in the world.改错I’m sorry to hear that you didn’t do good in your exam and that you are upsetting and worried bout your study. I quite understand what you are feeling now. As your friend, I have some advices for you.First of all, I want to say a failure is the mother of success. So never give it up though you fail this time. You should analyze the cause then take a lesson from it, which can help you making up for the gap in the knowledge and skills in these aspects. The next is to build up your confidence. For it is the key for your improvement. I am sure we will do better next time in an exam and gradually gain your confidence back.书面表达假定你是李华,你班同学准备在“六一儿童节”那天到当地一所幼儿园看望孩子们。

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