2017年东北三省三校高三第一次联合模拟考试(三校一模)

东北三省三校2017届高三理综第一次联合模拟考试试题（扫描版）2017高三第一次联合模拟考试物理答案14. 以下说法正确的是（）A ．大量处于量子数n=4的氢原子向低能级跃迁时最多可以发出3种不同频率的光子B ．用不同频率的光照射同种金属发生的光电效应中，光电子的最大初动能与入射光的频率成正比C ．β衰变的实质是原子核内的一个中子转化成了一个质子和一个β粒子D ．原子核的质量大于组成它的核子的质量之和的现象叫做质量亏损 答案：C15. 某质点在x 轴上做直线运动的速度v 与时间t 的关系式为21v t t =+- （所有单位均取国际单位），则该质点（） A. 始终向x 轴正方向运动B.出发点的位置坐标一定为1m x =-C.正在做匀加速直线运动D.加速度逐渐增大 答案：D16. 如图所示，两个互相垂直的挡板甲和乙之间夹着一个光滑的小球。

最初小球静止，乙板与水平面间夹角为530，两板对球的支持力分别为N 甲、N 乙，现保持两板互相垂直的关系不变而将两板以O 点为轴在纸面内顺时针缓慢转过160，已知sin370.6o =、cos370.8o=，则（ ）A.转动之前N 甲<N 乙B.转动之后N 甲>N 乙C.转动过程中N 甲一直减小D.转动过程中N 甲一直增大 答案：C17. 将一个质量为1kg 的小球竖直向上抛出，从它开始运动到落回抛出点的过程中始终受到竖直向下的恒力F 与大小不变的空气阻力f 的作用，以小球的抛出点为坐标原点，以竖直向上为X 轴的正方向，整个过程中它的速度平方随位置坐标的变化的关系如图17中的①、②所示，210m /g s = ,则以下说法正确的是（）A.f=4NB.F=4N32m/sC.在小球上升过程中，它的加速度大小为232m/sD.在小球下降过程中，它的加速度大小为2答案：A18.2016年10月17日，我国载有两名航天员的神舟十一号飞船发射成功，10月19日凌晨，神舟11号与天宫二号完成自动交会对接。

东北三省三校2017届高三英语第一次联合模拟考试试题2017年东北三省三校模拟题（一）参考答案听力：1-5 BACBC 6-10 ACBAC 11—15 AABCC 16-20 BAABC阅读理解：21-23 CAC 24—27 CDAB 28—31 DBDD 32—35 CDBB36-40 BEDFA完形填空:41-45 DABCD 46—50 ADCBA 51—55 CDCBA 56-60 ACBDC语法填空：61. was convinced 62. changing 63。

what 64. to realize65。

happiness66。

a 67. from 68。

importantly 69. and 70。

ourselves短文改错：1。

that改成which 2. or改成and 3. coming前加the4.injuring改成injured5.stay改成stayed6. because改成why7. plant改成plants8.its 改成your9.去掉forgive后的for 10。

sincere改成sincerely参考范文:Dear Peter,How are you doing these days？ I am writing to invite you to participate in our campus book fair， which will be held on 23rd April, 2017—the World Book and Copyright Day。

Aimed to arouse the enthusiasm for reading, this activity is designed to provide a platform where the students can buy or exchange books and share their reading experience。

哈尔滨师大附中、东北师大附中、辽宁省实验中学2021 年高三第一次联合模拟考试文科数学试卷第一卷〔共 60 分〕一、选择题：本大题共12 个小题 , 每题 5 分 , 共 60 分. 在每题给出的四个选项中，只有一项为哪一项符合题目要求的.1. 集合x 1x |1 x 2 ，那么AIB 〔〕Ax | 0 ，Bx 2A．(1,2) B．(1,2] C．1,2 D．[ 1,2) 2. 设复数 z 满足 (1 i ) z 2i ，那么 z 〔〕A．1 i B．1 i C．1 i D．1 i3. r(1,2)r r r设向量 a ， b (m, m 1) ， a / /b ，那么实数 m 的值为〔〕A．1 B．1 C．1D．3 34. 双曲线的顶点到渐进线的距离等于虚轴长的1，那么此双曲线的离心率是〔〕B．3 4A．2 C．3 D．425. 一个四棱锥的底面为长方形，其三视图如下图，那么这个四棱锥的体积是〔〕A． 1 B． 2 C． 3 D． 46. 检测 600 个某产品的质量〔单位：g 〕，得到的直方图中，前三组的长方形的高度成等差数列，后三组所对应的长方形的高度成公比为的等比数列，检测的质量在的长方形高度为〔〕之间的产品数为 150，那么质量在A ．1B ．1C ．1D ．112306607. 数列 a n 是等差数列，满足 a 1 2a 2 S 5 ，以下结论中错误的选项是〔〕A ． S 9 0B ． S 5 最小C ． S 3 S 6D ． a 5 08. 函数 f ( x)sin( x ) 〔0 ，〕在区间 ( , ) 内是增函数，那么〔〕224 2A ． f ()1B ． f ( x) 的周期为 24D ． f (3C ． 的最大值为 4) 049. 如图是用二分法求方程 x 3 2 0 近似解的算法的程序框图，那么①②两处应依次填入〔 〕A ． a m ， b mB ． b m ， amC．a f (m)，b f ( m) D．b f ( m)，a f ( m)10. 过抛物线y2 2 px 〔 p 0 〕的焦点F作直线交抛物线于A， B，假设S OAF 4S OBF，那么直线 AB 的斜率为〔〕A．3B．4 3 45 5C．D．4 311. 四面体 A BCD 中，ABC 和BCD 都是边长为 6 的正三角形，那么当四面体的体积最大时，其外接球的外表积是〔〕A．60 B．30 C．20 D．1512. 定义在R 上的可导函数 f ( x)的导函数为 f '( x)，满足 f '( x) f ( x) ，且 f (0) 2 ，那么不等式f (x) 2 x 0的解集为〔〕eA．( 2, ) B．(0,) C．(1, ) D．(4,)第二卷〔共 90 分〕二、填空题〔每题 5 分，总分值 20 分，将答案填在答题纸上〕x y 4 0,13. 实数 x ，y满足x 3y 6 0, 那么z x 2 y 的最大值为．2x 3y 12 0,14. 假设0 a 2 ， 0 b2 ，那么函数 f(x) 1 x3 ax 2 2bx 3 存在极值的概率3为．15. 假设a 0 ，b 0 ，且2a b 1 ，且 2 ab 4a2 b2 的最大值是．16. 各项均为正数的数列a n 和 b n 满足： a n， b n， a n 1成等差数列，b n， a n 1， b n 1成等比数列，且 a1 1， a2 3 ，那么数列a n 的通项公式为．三、解答题〔本大题共6小题，共70分.解容许写出文字说明、证明过程或演算步骤.〕17. 在ABC 中内角 A ， B ， C 的对边分别为a， b ，c且ac sin A sin B ．a b sin( A B)〔Ⅰ〕求角 B 的值；〔Ⅱ〕假设ABC 的外接圆半径为1，求ABC 面积 S 的最大值．18.某市拟招商引资兴建一化工园区，新闻媒体对此进行了问卷调查，在所有参与调查的市民中，持“支持〞、“保存〞和“不支持〞态度的人数如表所示：支持保存不支持30 岁以下90012028030 岁以上〔含30 岁〕300260140〔Ⅰ〕在所有参与调查的人中，用分层抽样的方法抽取局部市民做进一步调研〔不同态度的群体中亦按年龄分层抽样〕，从“保存〞态度的人中抽取了19 人，那么在“支持〞态度的群体中，年龄在30 岁以上的人有多少人被抽取；〔Ⅱ〕在持“不支持〞态度的人中，用分层抽样的方法抽取 6 人做进一步的调研，将此 6人看作一个总体，在这 6 人中任意选取 2 人，求至少有 1 人在 30 岁以上的概率．19. 正三棱柱ABC A1B1C1中， AB AA1 2 ，点D为AC的中点，点 E 为AA1上．〔Ⅰ〕当AA1 4 AE 时，求证：DE 平面 BDC1；〔Ⅱ〕当 AA1 2 AE 时，求三棱锥 C1 EBD 的体积．20. 椭圆C：x2 y 2 1(a b 0) 的左、右顶点分别为 A ， B ，其离心率 e 1 ，点a2 b2 2P 为椭圆上的一个动点，PAB 面积的最大值为 2 3．〔Ⅰ〕求椭圆的标准方程；〔Ⅱ〕动直线 l 过椭圆的左焦点F1，且l与椭圆C交于M，N两点，试问在 x 轴上是否存uuuur uuurD 坐标并求出定值；假设不存在，请说在定点 D ，使得DM DN 为定值？假设存在，求出点明理由．21. 函数 f ( x) 2ln x x2ax 2( a R) ．〔Ⅰ〕讨论函数 f ( x) 的单调性；〔Ⅱ〕假设存在x0 (0,1] ，使得对任意的 a [ 2,0) ，不等式f ( x0 ) a2 3a 22me a (a 1) 〔其中e 是自然对数的底数〕都成立，求实数m 的取值范围．请考生在 22、 23 两题中任选一题作答，如果多做，那么按所做的第一题记分. 22.选修 4-4 ：坐标系与参数方程在直角坐标系xOy 中，直线 C1： y3x ，曲线 C2的参数方程是x3cos〔y 2 sin为参数〕，以坐标原点为极点，x 轴的正半轴为极轴建立极坐标系．〔Ⅰ〕求 C1的极坐标方程和C2的普通方程；〔Ⅱ〕把 C1绕坐标原点沿顺时针方向旋转得到直线C3，C3与C2交于A，B两点，求3|AB|．23.选修 4-5 ：不等式选讲a 0 ，b 0 ，函数 f ( x) | x a | | x b | 的最小值为4．〔Ⅰ〕求 a b 的值；〔Ⅱ〕求1 a24 1 b2的最小值．9哈尔滨师大附中、东北师大附中、辽宁省实验中学2021 年高三第一次联合模拟考试文科数学试卷答案一、选择题1-5:ACAAB6-10:DBCAD11 、 12： AB二、填空题1 2 1 16.n 2 n14.15.a n242三、解答题17. 解：〔Ⅰ〕 Q A B Csin( AB) sin C ，∴ac sin A sinB ， 由正弦定理得ac ab ，a b sin Ca bc即 b 2 a 2c 2 ac ，结合余弦定理，有 cos B1, B (0, )，∴ B．b 23〔Ⅱ〕 2R23 ，，解得 bsin3所以， b23 a2c 2 2accos2ac ac ac 〔当且仅当ac时取等〕，3所以 S1ac sin 33 3 ．2418. 解：〔 Ⅰ 〕设在 “支持 〞的群体中抽取 n 个人 ，其中年龄在 30 岁以下的人被抽取 x 人．由题意120 260 900n 300，得 n 60 ．那么 x 3 n 45 人．194所以在 “支 持 〞的群体中，年龄在 30 岁以下的人有 45 人被抽取．〔Ⅱ 〕设所选的人中，有 m 人年龄在30 岁以下．那么 280 2 m，∴ m 4 ．280 140 3 6即从 30 岁以下抽取 4 人，另一局部抽取 2 人．分别记作 A 1 , A 2 , A 3, A 4 ,B 1, B 2 ．那么从中任取2 人的所有根本领件为A 1, A 2〕〔,A 1, A 3〕〔, A 1 , A 4〕〔, A 1 ,B 1〕〔,A 1,B 2〕〔A 2, A 3〕〔,A 2 , A 4〕〔, A 2 , B 1〕〔,A 2 , B 2〕A 3 , A 4〕〔, A 3 ,B 1〕〔, A 3 , B 2〕〔, A 4 , B 1〕〔, A 4 , B 2〕,(B 1, B 2 ) ．共 15 个其中至少有 1人在 30 岁以上的根本领件有 9 个．分别是〔 A 1 , B 1〕〔, A 1 ,B 2〕〔 A 2 , B 1〕〔,A 2, B 2〕〔 A 3 , B 1〕〔,A 3, B 2〕〔,A 4 , B 1〕〔, A 4 , B 2〕,( B 1, B 2 ) ．所以在这 6 人中任意选取 2人，至少有130岁以上的概率为9 ．人在315 519. 〔Ⅰ〕证明： Q ABC 为正三角形，点 D 为 AC 的中点，∴ BDAC ，∴ BD 面 ACC 1 A 1 ，从而 BDDE ．连接 EC 1 ，Q AA 14AE ， AB AA 1 2 1 ，ED5，EC 122 9 5 ，，∴ EA224 2C 1D5 ，那么EC 12ED 2 C 1D 2 ，∴ EDC 1D ，又C 1DI BD D ，∴DE平面 BDC 1 ．〔Ⅱ〕 Q AA 12AE ，∴ ED2,C 1D C 1 E5，∴SC 1DE3，2由〔Ⅰ〕知 BD 面 ACC 1 A 1 ，所以 BD 为三棱锥 B C 1 DE 的高，所以V C 1EBDV B C 1DE1S C 1DE BD 1 3 33 ．33 2220. 解：〔 Ⅰ 〕由题意， ec 1, (S PAB ) max12ab ab2 3 ，且 a 2b 2c 2 .a22解得 a2,b3, c 1.∴椭圆的标准方程为x 2 y 2 1 ．43uuuur uuur〔Ⅱ 〕假设存在定点 D (m,0) ，使得向量 DM DN 为定值 n .①当直线 l 的斜率不为0 时，椭圆 C 左焦点 F 1( 1,0) ，设直线 l 的方程为 xty 1 .联立x 2 y 24 3 1，消去 x ，得 (3t 2 4) y2 6ty 9 0．x ty 1设 M ( x 1, y 1), N ( x 2 , y 2 ) ，那么y 1 y 26t , y 1 y 29．3t24 3t24uuuuruuurDM ( x 1 m, y 1), DN ( x 2 m, y 2 ) ，uuuur uuur (x m)( xm) y y x xm(x x ) m2y yDM DN 2 22111 2121(ty 1 1)(ty 2 1) m(t( y 1 y 2 ) 2) m 2 y 1 y 2(t 2 1)y 1 y 2 ( m 1)t ( y 1y 2 ) (m 1)29(t 2 1) 6t 2 (m 1) (m 1)2 ( 6m 15)t 2 9 (m 1)2 ．3t 2 4 3t 2 4 3t 2 4 uuuur uuur 6m 15 9 11 135假设 DM DN 为定值 n ，那么 ，即 m ，此时 n ．3 4 8 64uuurl11uuuur 527 135 ，亦 ②当直线 的斜率为 时， A( ， ， ,0), DM DN2 0), B(2 0), D (88864符合题意；∴存在点 D (11,0) ，使得向量 DMDN 为定值 n135 .86421. 解：〔 Ⅰ 〕 f (x)22xa2x 2ax 2 (x 0) ．xx令 h( x) 2x 2ax 2 ，a 2 16 ．①当 ②当a0 时， ax 0 ，∴ f ( x)h(x)) 上单调递增；0 ，函数 f ( x) 在 (0,x0 a 4 时， a 2 16 0 ，所以 h( x) 0 ，函数 f (x) 在 (0,) 上单调递增；③ 当 a4 时，a 2 16 0 ，令 h( x)aa 2 16aa 2 160 ，0 ，得 x 140, x 24f ' (x)0 x (0, x 1 ) U ( x 2 ,) ； f ' ( x) 0x (x 1, x 2 ) ．所以， f (x) 在 0, x 1 和 x 2，+上单调递增，在 ( x 1 , x 2 ) 单调递减．综上， 1o当 a1 时，函数 f (x) 在 (0,) 上单调递增；2o 当 a 1 时， f ( x) 在 0,x 1和 x 2 ,上单调递增，在 ( x 1, x 2 ) 单调递减．〔注： 如果在每种情况中已说明函数在哪个区间上的单调性，不写综上不扣分； 如果每种情况只解出不等式，最后没写综上 扣 1分〕〔Ⅱ 〕由〔 Ⅰ〕知， a [ 2,0) 时，函数 f ( x) 在区间 (0,1] 上单调递增 ，所以当 x (0,1] 时，函数 f (x) 的最大值是 f (1)3 a ，对任意的 a[ 2,0) ，都存在 x 0 (0,1] ，使得不等式 2me a (a 1) f ( x 0 ) a23a 2 成立，即对任意的 a [ 2,0) ， 2me a (a 1) f ( x 0 )max a 2 3a 2 都成立，即对任意的 a [ 2,0) ，不等式 2me a (a1) a 24a 1 0 都成立，记h( a)2me a (a 1)a 24a 1 ，那么 h( a) 2me a (a 2) 2a 42( a 2)( me a 1) ．Q a [2,0), a[ 1,1)，且 a 2 0 ．e2e①当 m1时， me a 1 0, h (a)0 ，即 a [ 2,0) 时， h( a) 单调递减 .∴ h( a) 0 ，只需 h(0)0 ，解得 m1 ，∴ m [ 1,1] ．2 2②当 m 1时，令 h ( a) 0 得 a 2 或 aln m ，因为 a [ 2,0) ，所以 2( a 2) 0 .〔ⅰ 〕当 1 m e 2 时， ln m [ 2,0) ，当 a( 2,ln m) 时， h '( a) 0 ；当 a( ln m,0) 时， h '(a) 0 ，∴ h(a) min h( ln m)ln 2 m 2ln m 3 0 ，解得 m(1, e 3) ，∴ m (1,e 2 ) ． e 1〔ⅱ 〕当 m e 2 时，因为 2 a 0 ，所以 e a1，所以 me a1 ，所以 h '(a) 0 ，e 2那么 h(a)在 [ 2,0) 上单调递增，得 h(2) 5 2me20 ，即 m 5e2，∴ m [ e 2 , 5e 2) .22综上， m 的取值范围是 [ 1 , 5e 2) ． 2 222. 解：〔Ⅰ〕直 线 C 1 ：sin3 cos2 (R) ，3曲线 C 2 的普通方程为 ( x 3) 2 ( y 2) 21 ．〔Ⅱ〕 C 3 ：( R) ，即 y 3x ．3东北三省三校2017届高三第一次联合模拟考试数学(文)试题Word 版含答案 11 / 11 圆 C 2 的圆 心到直线 C 3 的距离 d 3 2 1 ．2 2所以 AB 2 12 1 3 ． 423．解：〔Ⅰ〕 因为 f ( x) x a x b ( x a) ( x b) a b ， 当且仅当 a x b 时，等号成立，所以 f ( x) 的最小值为 a b 4 ． 〔Ⅱ〕由〔Ⅰ〕知 a b 4 ，由柯西不等式得 (1 a 2 1 b 2 )(4 9) ( a 2 b 3)2 16 ． 4 9 2 3 1 1 16 1 a 1b 16 36 即 ( a 2 b 2 ) ，当且仅当 2 3 ，即 a , b 时，等号成立． 4 9 13 2 3 13 13 所以， 1 a 2 1 b 2 的最小值为 16 ． 4 9 13另法：因为 a b 4 ，所以 b 4 a ，那么 1 a 2 1 b 2 a 2 (4 a)2 13a 2 32a 64 (0 a 4) 4 9 4 9 36 当 a 16 时， 1 a 2 1 b 2 取 最小值，最小值为 16 ． 13 4 9 13。

东北三省四市教研联合体2017届高三高考模拟试题（一）英语试题附参考答案第二部分阅读理解（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，从每题所给的四个选项（A、B、C、D）中，选出最佳选项，并在答题卡上将该项涂黑。

A“Her grades are fine; I’m not worried about that, but she just doesn’t seem to love learning any more.” Alice’s mum said.She’s absolutely right. I’d noticed the same thing about her daughter over the previous two or three years when I’d been Alice’s middle school English, Latin, and writing teacher, and I have an answer, right there on the tip of my tongue, for what has gone wrong. Yet I’m torn between my responsibility to help Alice, and the knowledge that what I have to say is a truth I’m not sure this mother is ready to hear.The truth- for this parent and so many others—is this: Her child has sacrificed her natural curiosity and love of learning at the altar(圣坛) of achievement, and it’s our fault. Alice’sparents, her teachers, society at large—we are all supposed to be blamed in this crime against learning. Form her first day of school, we pointed her toward that altar and trained her to measure her progress by means of points, scores, and awards. We taught Alice that her potential is tied to her intellect(智力) ,and that her intellect is more important than her character. We taught her to come home proudly bearing grades As, championship prizes, and college acceptances, and not on purpose, we taught her that we don’t really care how she obtains them. We taught her to protect her academic and extracurricular perfection at all costs and that it’s better to quit when things get challenging rather than risk losing that perfect record. Above all else, we taught her fear failure. That fear is what has destroyed her love of learning.21. Both the mother and the teacher have noticed that .A. Alice hates to tell them truthB. Alice’s test scores are not so goodC. Alice chooses several subjects to learnD. Alice doesn’t enjoy studying any more22. What dose the underlined wor d “torn ” mean in the passage?A. pained to chooseB. guilty to quitC. tired to manageD.certain to decide23. What dose the teacher think of the girl?A. Her academic perfection resulted form her characterB. She is a gifted girl with spirits of challenging.C. She is perfect in her character developmentD. Her academic performance is great.24. What leads to the girl’s change?A. A few failure experiencesB. Fear of failing getting good gradesC. Her mother’s over-protectionD. Her not being accepted by collegesBGraduate DeveloperWho You AreYou’re a creative person who’s brilliant at making smart decisions and even better at communicating them. You are passionate about software and not only like to learn about it at University, you like to get involved in your own projects at home too. You eat code for breakfast, lunch and dinner, and you’re always looking to learn and improve. You love working with like—minded peers and aspire to work with the best software developers in the industry.What you studyYou are a final-year/Honours/Masters/PHD student or a recent graduate(past two years)with a Distinction/High Distinction average in Computer Science, Computing or Software Engineering.What you will doYou’ll develop reliable and highly flexible systems in the market. The kind of software that ensures we buy and sell faster and more efficiently than anyone else. You will have full responsibility for designing and improving our systems, both technically and functionally to improve our trading success. At Optive we have no shortage of interesting and important work to do, so rest assured that you’ll be working on something that will add real value to the business.QualificationsYou are an Australian or New Zealand citizen or Permanent Resident of Australia.Salary$ 75,000 plus super plus performance driven profit share.25. You will be well-accepted if you are .A. creative and devoted to software EngineeringB. loving working with people from different majorsC. communicative and passionate in your university lifeD. enjoying eating code instead of delicious family dinners26. The Graduate Developer will probably be wanted as .A. a final-year Computing Engineering studentB. a master-degree graduate majoring in tradingC. a PHD student as a permanent resident in AustraliaD. a recent graduate with Honours in Computer Science27. You are not required to in the future job.A. design Optive systemB. ensure trade successC. work on something valuableD. develop the flexible market systemCThe news of Harper Lee’s second novel broke the internet. Go Set a Watchman would be published in July. Lee became a recluse (隐士)after the publication of her first book, To Kill a Mockingbird, in 1960. Although the author has been out of the public eye for more than 50 years, the wave that followed the announcement showed her novel’s beloved place in the American literary works.In 1991, The Book of the Month Club and the library of Congress conducted a survey that made readers identify books that had made a difference in their lives. Lee’s book followedonly The Bible. Alongside the works of Shakespeare and Twain, To Kill a Mockingbird remains one of the most widely taught books nationwide, reaching an estimated 70% of American public schools. What makes it such a perfect read for young people?English class is a place where young Americans come to know themselves. In the folds of dusty books, students can make contact with humanity beyond the superficial(肤浅的) conversations of the school hallways. Literature promises relief, a place to figure out one’s problems and get to know oneself better.To Kill a Mockingbird particularly distinguishes itself in this regard. It speaks in child’s voice without treating its readership as children. People have called it an ‘impossible’achievement, a children’s book penned in the prose(散文) of a well-educated adult—it’s unlikely that a child like Scout could exist in the real world. But that’s exactly what makes it such a charming read for young people. It alludes to (转弯抹角地说到)the consciousness of a well-educated adult going through difficult realities, but shows it through the light and playful voice of a curious little girl.28. In which year might Go Set a Watchman come out?A. 2015B. 1960C. 1910D. 199129. Which book leaves the deepest impression on American readers?A. Go Set a WatchmanB. To Kill a MockingbirdC. The BibleD. Shakespeare’s works30. Which of the following statements is right about To Killa Mockingbird?A. Its author may be a playful child.B. It’s a story about a well-educated adult.C. Young people can see themselves from it.D. It’s impossible for children like Scout to exist.31. What is the best title for the text?A. The Great Writer: Harper LeeB. The Great Young People NovelC. V oices from ChildrenD. Differences Harper Lee madeDLast week, Education Secretary Arne Duncan declared a war on paper textbooks. “Over the next few years,” he said in a speech at the National Press Club, “textbooks should be abandoned.” In their place would come a variety of digital-learning technologies, like e-readers and multi-media websites.Such technologies certainly have their place. But Secretary Duncan is threatening to light a fire to a tried-and-true technology—good old paper—that has been the foundation for one of the great educational systems on the planet. And while e-readers and multi-media may seem appealing, the idea of replacing an effective leaning platform with a widely hyped (炒作)but still unproven one is extremely dangerous.An expert on reading, Maryanne Wolf, has recently begun studying the effects of digital reading on learning, and so far the results are mixed. She worries that Internet reading, in particular, could be such a source of distractions(分散注意力) for the student that they may cancel out most other potential benefits of a web-linked, e-learning environment. and while it’s true that the high-tech industry has sponsored substantial amounts of research on the potential benefits of Web-based learning, not enough time has passed for longitudinal（纵向的）studies to demonstrate the full effects.In addition, digital-reading advocates claim that lightweight e-books benefit students’backs and save schools money. but the rolling backpack seems to have solved the weight problem, and the astounding costs to outfit every student with an e-reader, provide technical support and pay for regularsoftware updates promise to make the e-textbook a very pricey choice.As both a teacher who uses paper textbooks and a student of urban history, i can’t help but wonder what parallels exist between my own field and this sudden, wholesale abandonment of the technology of paper.32. Which of the following statements is true according to the passage?A. A tried and true technology is paper, long used in educational system.B. Digital learning technologies will replace the paper textbooks sooner or later.C. E-readers and multimedia websites are learning methods that are proved effective.D. Multi-media websites and a good old paper are kinds of digital-learning technologies.33. What are drawbacks of textbooks according to the passage?A. Its price.B. Its weight.C. Its content.D. Its appearance.34. What worries Maryanne Wolf is that .A. paper learning can provide more potential benefitsB. the results of digital reading effects are understandableC. students may not focus on learning by digital readingD. digital reading can’t provide potential benefits for users35. What is the author’s attitude towards digital-learning?A. disapprovingB. supportiveC. positiveD. objective第二节（共5小题；每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

哈尔滨师大附中、东北师大附中、辽宁省实验中学2017年高三第一次联合模拟考试文科数学试卷第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.已知集合1|02x A x x +⎧⎫=<⎨⎬-⎩⎭，{}|12B x x =<≤，则A B =I （ ） A ．(1,2)B ．(1,2]C ．[]1,2-D ．[1,2)-2.设复数z 满足(1)2i z i +=，则z =（ ） A ．1i -+B ．1i --C ．1i +D ．1i -3.设向量(1,2)a =r ，(,1)b m m =+r，//a b r r ，则实数m 的值为（ ）A ．1B ．1-C ．13-D ．3-4.双曲线的顶点到渐进线的距离等于虚轴长的14，则此双曲线的离心率是（ ） A ．2B ．32C ．3D ．45.一个四棱锥的底面为长方形，其三视图如图所示，则这个四棱锥的体积是（ ）A ．1B ．2C ．3D ．46.检测600个某产品的质量（单位：g ），得到的直方图中，前三组的长方形的高度成等差数列，后三组所对应的长方形的高度成公比为0.5的等比数列，已知检测的质量在100.5~105.5之间的产品数为150，则质量在115.5~120.5的长方形高度为（ ）A ．112B ．130C ．16D ．1607.已知数列{}n a 是等差数列，满足1252a a S +=，下列结论中错误的是（ ） A ．90S =B ．5S 最小C ．36S S =D ．50a =8.函数()sin()f x x ωϕ=+（0ω>，22ππϕ-<<）在区间(,)42ππ内是增函数，则（ ）A ．()14f π=-B ．()f x 的周期为2πC ．ω的最大值为4D ．3()04f π=9.如图是用二分法求方程320x -=近似解的算法的程序框图，则①②两处应依次填入（ ）A ．a m =，b m =B ．b m =，a m =C ．()a f m =，()b f m =D ．()b f m =，()a f m =10.过抛物线22y px =（0p >）的焦点F 作直线交抛物线于A ，B ，若4OAF OBF S S ∆∆=，则直线AB 的斜率为（ ） A ．35±B ．45±C ．34±D ．43±11.已知四面体A BCD -中，ABC ∆和BCD ∆都是边长为6的正三角形，则当四面体的体积最大时，其外接球的表面积是（ ） A ．60πB ．30πC ．20πD ．15π12.已知定义在R 上的可导函数()f x 的导函数为'()f x ，满足'()()f x f x <，且(0)2f =，则不等式()20xf x e -<的解集为（ ） A ．(2,)-+∞B ．(0,)+∞C ．(1,)+∞D ．(4,)+∞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.已知实数x ，y 满足40,360,23120,x y x y x y -+≥⎧⎪--≤⎨⎪+-≤⎩则2z x y =+的最大值为 ．14.若02a <<，02b <<，则函数321()233f x x bx =+-存在极值的概率为 ．15.若0a >，0b >，且21a b +=，且224a b -的最大值是 ． 16.各项均为正数的数列{}n a 和{}n b 满足：n a ，n b ，1n a +成等差数列，n b ，1n a +，1n b +成等比数列，且11a =，23a =，则数列{}n a 的通项公式为 ．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.已知在ABC ∆中内角A ，B ，C 的对边分别为a ，b ，c 且sin sin sin()a c A Ba b A B -+=-+． （Ⅰ）求角B 的值；（Ⅱ）若ABC ∆的外接圆半径为1，求ABC ∆面积S 的最大值．18.某市拟招商引资兴建一化工园区，新闻媒体对此进行了问卷调查，在所有参与调查的市民中，持“支持”、“保留”和“不支持”态度的人数如表所示：支持 保留 不支持 30岁以下 900 120 280 30岁以上（含30岁）300260140（Ⅰ）在所有参与调查的人中，用分层抽样的方法抽取部分市民做进一步调研（不同态度的群体中亦按年龄分层抽样），已知从“保留”态度的人中抽取了19人，则在“支持”态度的群体中，年龄在30岁以上的人有多少人被抽取；（Ⅱ）在持“不支持”态度的人中，用分层抽样的方法抽取6人做进一步的调研，将此6人看作一个总体，在这6人中任意选取2人，求至少有1人在30岁以上的概率．19.已知正三棱柱111ABC A B C -中，12AB AA ==，点D 为AC 的中点，点E 为1AA 上．（Ⅰ）当14AA AE =时，求证：DE ⊥平面1BDC ； （Ⅱ）当12AA AE =时，求三棱锥1C EBD -的体积．20.已知椭圆C ：22221(0)x y a b a b +=>>的左、右顶点分别为A ，B ，其离心率12e =，点P 为椭圆上的一个动点，PAB ∆面积的最大值为23（Ⅰ）求椭圆的标准方程；（Ⅱ）动直线l 过椭圆的左焦点1F ，且l 与椭圆C 交于M ，N 两点，试问在x 轴上是否存在定点D ，使得DM DN ⋅u u u u r u u u r为定值？若存在，求出点D 坐标并求出定值；若不存在，请说明理由．21.已知函数2()2ln 2()f x x x ax a R =+-+∈． （Ⅰ）讨论函数()f x 的单调性；（Ⅱ）若存在0(0,1]x ∈，使得对任意的[2,0)a ∈-，不等式20()322(1)a f x a a me a >++-+（其中e 是自然对数的底数）都成立，求实数m 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直角坐标系xOy 中，直线1C ：y =，曲线2C 的参数方程是cos 2sin x y ϕϕ⎧=⎪⎨=-+⎪⎩（ϕ为参数），以坐标原点为极点，x 轴的正半轴为极轴建立极坐标系． （Ⅰ）求1C 的极坐标方程和2C 的普通方程； （Ⅱ）把1C 绕坐标原点沿顺时针方向旋转3π得到直线3C ，3C 与2C 交于A ，B 两点，求||AB ．23.选修4-5：不等式选讲已知0a >，0b >，函数()||||f x x a x b =++-的最小值为4． （Ⅰ）求a b +的值； （Ⅱ）求221149a b +的最小值．哈尔滨师大附中、东北师大附中、辽宁省实验中学2017年高三第一次联合模拟考试文科数学试卷答案一、选择题1-5:ACAAB 6-10:DBCAD 11、12：AB二、填空题13.8 14.1415.212- 16.22n n n a +=三、解答题17.解：（Ⅰ）sin()sin A B C A B C π++=∴+=Q ，∴sin sinBsin a c A a b C-+=-， 由正弦定理得a c a ba b c-+=-， 即222b a c ac =+-， 结合余弦定理，有1cos ,(0,)2B B π=∈，∴3B π=． （Ⅱ）22sin3b R π==，解得3b =，所以，22232cos23b ac ac ac ac ac π==+-≥-=（当且仅当a c =时取等）， 所以133sin 234S ac π=≤． 18.解：（Ⅰ）设在“支持”的群体中抽取n 个人，其中年龄在30岁以下的人被抽取x 人． 由题意n 30090019260120+=+，得60=n ．则4543==n x 人．所以在“支持”的群体中，年龄在30岁以下的人有45人被抽取． （Ⅱ）设所选的人中，有m 人年龄在30岁以下．则632140280280m==+，∴4m =．即从30岁以下抽取4人，另一部分抽取2人．分别记作214321,,,,,B B A A A A ． 则从中任取2人的所有基本事件为）（）（）（）（）（2111413121,,,,,,,,,B A B A A A A A A A ）（）（）（）（22124232,,,,,,,B A B A A A A A ),(,,,,,,,,,,212414231343B B B A B A B A B A A A ）（）（）（）（）（．共15个其中至少有1人在30岁以上的基本事件有9个．分别是）（）（2111,,,B A B A ）（）（2212,,,B A B A ),(,,,,,,,,2124142313B B B A B A B A B A ）（）（）（）（． 所以在这6人中任意选取2人，至少有1人在30岁以上的概率为53159=． 19.（Ⅰ）证明：ABC ∆Q 为正三角形，点D 为AC 的中点， ∴BD AC ⊥，∴BD ⊥面11ACC A ，从而BD DE ⊥．连接1EC ，Q 14AA AE =，12AB AA ==，∴12EA =，52ED =，2195242EC =+=，15C D =则22211EC ED C D =+，∴1ED C D ⊥，又1C D BD D =I ，∴DE ⊥平面1BDC ．（Ⅱ）Q 12AA AE =，∴112,5ED C D C E ==132C DE S ∆=， 由（Ⅰ）知BD ⊥面11ACC A ，所以BD 为三棱锥1B C DE -的高， 所以111113333322C EBD B C DE C DE V V S BD --∆==⋅=⨯=． 20. 解：（Ⅰ）由题意，max 11,()22322PAB c e S ab ab a ∆===⨯==222a b c =+. 解得2,3,1a b c ===.∴椭圆的标准方程为22143x y +=． （Ⅱ）假设存在定点(,0)D m ，使得向量DM DN ⋅u u u u r u u u r为定值n .①当直线l 的斜率不为0时，椭圆C 左焦点1(1,0)F -，设直线l 的方程为1x ty =-.联立221431x y x ty ⎧+=⎪⎨⎪=-⎩，消去x ，得22(34)690t y ty +--=．设1122(,),(,)M x y N x y ，则12122269,3434t y y y y t t -+==++． 1122(,),(,)DM x m y DN x m y =-=-u u u u r u u u r，21212121212()()()DM DN x m x m y y x x m x x m y y ⋅=--+=-+++u u u u r u u u r2121212(1)(1)(()2)ty ty m t y y m y y =---+-++221212(1)(1)()(1)t y y m t y y m =+-++++222222229(1)6(1)(615)9(1)(1)343434t t m m t m m t t t -++---=-++=+++++． 若DM DN ⋅u u u u r u u u r 为定值n ，则615934m ---=，即118m =-，此时13564n =-． ②当直线l 的斜率为0时，11527135(20),(20),(,0),88864A B D DM DN --⋅=-⨯=-u u u u r u u u r ，，，亦符合题意； ∴存在点)0,811(-D ，使得向量DN DM ⋅为定值64135-=n . 21. 解：（Ⅰ）2222()2(0)x ax f x x a x x x-+'=+-=>． 令2()22h x x ax =-+，216a ∆=-．①当0a ≤时，0ax -≥，∴()()0h x f x x'=>，函数()f x 在(0,)+∞上单调递增； ②当04a <≤时，2160a ∆=-≤，所以()0h x ≥，函数()f x 在(0,)+∞上单调递增； ③当4a >时，2160a ∆=->，令()0h x =，得221216160,0a a a a x x --+-=>=>， '12()0(0,)(,)f x x x x >⇒∈+∞U ；'12()0(,)f x x x x <⇒∈．所以，()f x 在()10,x 和()2+x ∞，上单调递增，在12(,)x x 单调递减． 综上，1o当1a ≤时，函数()f x 在(0,)+∞上单调递增；2o 当1a >时，()f x 在()10,x 和()2,x +∞上单调递增，在12(,)x x 单调递减．（注：如果在每种情况中已说明函数在哪个区间上的单调性，不写综上不扣分；如果每种情况只解出不等式，最后没写综上扣1分）（Ⅱ）由（Ⅰ）知，[2,0)a ∈-时，函数()f x 在区间(0,1]上单调递增，所以当(0,1]x ∈时，函数()f x 的最大值是(1)3f a =-，对任意的[2,0)a ∈-，都存在0(0,1]x ∈，使得不等式202(1)()32a me a f x a a ++>++成立， 即对任意的[2,0)a ∈-，20max 2(1)()32a me a f x a a ++>++都成立，即对任意的[2,0)a ∈-，不等式22(1)410a me a a a +--+>都成立，记2()2(1)41ah a me a a a =+--+，则()2(2)242(2)(1)aah a me a a a me '=+--=+-．21[2,0),[,1)a a e e ∈-∴∈Q ，且20a +≥． ①当1m ≤时，10,()0ame h a '-<∴≤，即[2,0)a ∈-时，()h a 单调递减. ∴()0h a >，只需(0)0h ≥，解得12m ≥-，∴1[,1]2m ∈-． ②当1m >时，令()0h a '=得2a =-或ln a m =-，因为[2,0)a ∈-，所以2(2)0a +≥. （ⅰ）当21m e <<时，ln [2,0)m -∈-，当(2,ln )a m ∈--时，'()0h a <；当(ln ,0)a m ∈-时，'()0h a >，∴2min ()(ln )ln 2ln 30h a h m m m =-=-++>，解得31(,)m e e∈ ，∴2(1,)m e ∈．（ⅱ）当2m e ≥时，因为20a -≤<，所以211a e e≤<，所以1a me ≥，所以'()0h a ≥，则()h a在[2,0)-上单调递增，得2(2)520h me --=->，即252e m <，∴225[,)2e m e ∈. 综上，m 的取值范围是215[,)22e -． 22. 解：（Ⅰ）直线1C ： 2sin 3cos ()3R πρθρθθρ=-⇒=∈， 曲线2C 的普通方程为22(3)(2)1x y ++=． （Ⅱ）3C ： ()3R πθρ=∈，即3y x =．圆2C 的圆心到直线3C 的距离32122d -+==． 所以212134AB =-=． 23．解：（Ⅰ）因为()()()f x x a x b x a x b a b =++-≥+--=+， 当且仅当a x b -≤≤时，等号成立，所以()f x 的最小值为4a b +=． （Ⅱ）由（Ⅰ）知4a b +=，由柯西不等式得22211()(49)(23)164923a ba b ++≥⨯+⨯=．即221116()4913a b +≥，当且仅当113223ba =，即1636,1313ab ==时，等号成立．所以，221149a b +的最小值为1613．另法：因为4a b +=，所以4b a =-，则2222211(4)133264(04)494936a a a a ab a --++=+=<< 当1613a =时，221149a b +取最小值，最小值为1613．。

10.用下列装置进行实验，不能达到相应实验目的的是（）D．装置丁：可制得金属锰C H O且含有（）结构的同分异构体共有（）11．分子式为48212．X 、Y 、Z 、W 是原子序数依次增大的三个不同短周期主族元素。

X 、W 均为所在周期数和族序数相等的元素。

Y 、Z 易形成质量比为7：8和7：16的两种常见化合物。

下列说法正确的是（ ）A ．原子半径：Y<Z<WB ．原子最外层电子数：W>Z>YC ．X 、Y 可以形成既有极性键也有非极性键的化合物D ．由X 、Y 、Z 三种元素形成的化合物的水溶液一定显酸性13．下列各项关系中正确的是（ ）A ．230.1 mol/LNa SO 溶液中：+2--3324c Na =2c SO +c HSO +c H SO =0.2mol/L （）（）（）（） B ．常温下pH=a 的稀24H SO 与pH=b 的氨水等体积混合后恰好完全反应，则a+b<14C ．30.2 mol/LCH COOH 溶液与0.1 mol/LNaOH 溶液等体积混合：+--33c(H )c(OH )=c(CH COO )c(CH COOH)--D ． p H 相同的三种溶液①4NH Cl ②424(NH )SO ③44NH HSO ，+4c(NH )大小关系为：>>①②③14．以下说法正确的是（ ）A ．大量处于量子数n=4的氢原子向低能级跃迁时最多可以发出3种不同频率的光子B ．用不同频率的光照射同种金属发生的光电效应中，光电子的最大初动能与入射光的频率成正比C ．β衰变的实质是原子核内的一个中子转化成了一个质子和一个β粒子D ． 原子核的质量大于组成它的核子的质量之和的现象叫做质量亏损15．某质点在x 轴上做直线运动的速度v 与时间t 的关系式为2v t t 1=+-（所有单位均取国际单位），则该质点（ ）A ．始终向x 轴正方向运动B ．出发点的位置坐标一定为x= 1 m -C ．正在做匀加速直线运动D ． 加速度逐渐增大16．如图所示，两个互相垂直的挡板甲和乙之间夹着一个光滑的小球。

哈尔滨师大附中、东北师大附中、辽宁省实验中学2017年高三第一次联合模拟考试文科数学试卷第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. 已知集合错误！未找到引用源。

，错误！未找到引用源。

，则错误！未找到引用源。

（）A. 错误！未找到引用源。

B. 错误！未找到引用源。

C. 错误！未找到引用源。

D. 错误！未找到引用源。

【答案】A【解析】因错误！未找到引用源。

，错误！未找到引用源。

，故错误！未找到引用源。

，应选答案A。

2. 设复数错误！未找到引用源。

满足错误！未找到引用源。

，则错误！未找到引用源。

（）A. 错误！未找到引用源。

B. 错误！未找到引用源。

C. 错误！未找到引用源。

D. 错误！未找到引用源。

【答案】C【解析】试题分析：因为错误！未找到引用源。

，所以错误！未找到引用源。

故选C．3. 设向量错误！未找到引用源。

，错误！未找到引用源。

，错误！未找到引用源。

，则实数错误！未找到引用源。

的值为（）A. 错误！未找到引用源。

B. 错误！未找到引用源。

C. 错误！未找到引用源。

D. 错误！未找到引用源。

【答案】A【解析】因错误！未找到引用源。

，故错误！未找到引用源。

，应选答案A。

4. 双曲线的顶点到渐进线的距离等于虚轴长的错误！未找到引用源。

，则此双曲线的离心率是（）A. 错误！未找到引用源。

B. 错误！未找到引用源。

C. 错误！未找到引用源。

D. 错误！未找到引用源。

【答案】A【解析】不妨设双曲线的方程为错误！未找到引用源。

，则其中一个顶点为错误！未找到引用源。

，一条渐近线为错误！未找到引用源。

，则顶点到渐进线的距离错误！未找到引用源。

，由题意错误！未找到引用源。

，故该双曲线的离心率错误！未找到引用源。

，应选答案A。

5. 一个四棱锥的底面为长方形，其三视图如图所示，则这个四棱锥的体积是（）A. 1B. 2C. 3D. 4【答案】B【解析】试题分析：由三视图可知该几何体为四棱锥，底面为矩形，边长为2,1，棱锥的高为3，所以其体积为错误！未找到引用源。

东北三省三校2017-2018年高三第一次联合模拟考试参考答案第一部分：1. C2. A3. B4. B5. A6. C7. B8. A 9. C 10. B11. C 12. A 13. A 14. B 15. C 16. C 17. A 18. B 19. A 20. C第二部分：21. A 22. B 23. A 24. C 25. D 26. A 27. C28. B 29. D 30. B 31. C 32. A 33. C 34. A 35. C 36. G 37. C 38. A 39. D 40. F第三部分：41. B 42. C 43. A 44. D 45. B 46. B 47. C48. D 49. B 50. A 51. C 52. D 53. B 54. A 55. C 56. B 57. A 58. C 59. D 60. A61. a 62. at 63. screaming 64. Impatient65. bothered66. pieces 67. until / before 68. have finished 69. there 70. It/ That/This第四部分：第一节：Of all my teachers, Mr. Smith is the one who impresses me most. Because he is 54,Though/Althoughhe looks very young at his age. And he’s one of the most popular teachers in ourforschool. Compared with other teacher, Mr. Smith pays more attention to his way ofteachers teaching. He tries various ways to make his class lively and interestingly. In hisinteresting opinion, we would not only know “what”, but also understand “why”. So, instead ofshouldgiving us answers immediately, he encourages us to think by themselves wheneverourselves he puts forward questions, With his help, we /\learned how to analyze and settlehavedown problems. What a wonderful world of “what” he leads us to! He is such a去掉 whylearning person that we all admire him very much.learned第二节：One Possible Version:Dear Editor,We took part in a voluntary activity organized by our school last Sunday. We gathered at the Culture Square, passing leaflets to passers-by from which they can learn what the earthquake is and how to protect themselves against the earthquake. The leaflets drew the attention of the public to the damage that an earthquake can cause and told them to take action in a right and efficient way. Some made a public speech, telling people about the best place they can hide when an earthquake occurs.We have learned much from the voluntary activity, which offers us an opportunity to contribute to society. We told the public about the situation because we thought everyone should be concerned about each other and take responsibility for promoting a public cause.Best wishes!ours,i Hua听力部分录音稿1. W: John, why don’t you watch NBA games on TV?M: Oh, I’m studying for a Chinese test tomorrow.2. M: Where is Adam？ He was supposed to meet us here half an hour ago, but he hasn’t come yet.W: I’ve just phoned and we’ve had a quarrel. He was busy with the contract and didn’t notice the time.3. M: May I help you?W: Yes, I’d like to apply for a loan. We need the loan to build an addition to our house.M: All right. I’ll give you an application form. You can fill it out here, or take it home if you like.4. M: Hello, Kate. Do you know how many students will go to the mountain?W: Our monitor said that the whole class will go except Tom.He has something urgent to do. That is, 20 students.M: Oh, 18 students have been there by now.5. W: Chris, did you know about Gemma Hudson? Have you heard that she’s leaving?M: Really? How do you know that?W: She told me. She’s leaving next month.M: Actually, I’m not surprised. I didn’t think she was very happy working here.6. W: What is the tall building over there?M: It’s the library, which houses over half a million books.It also has computer labs and quiet study spaces.W: And the building next to it?M: That’s the computer center. The IT facilities are really advanced there. Some of its classrooms are open 24 hoursa day.W: Is there a sports center on campus?M: Of course. The sport facilities are excellent. A stadium is on the northeast side of the campus. A 50-metre swimming pool is just beside it. There is a football match this afternoon. Let’s go there and have a look.7. W: So, do you like watching the World Cup?M: Um, no. I can’t stand it.W: Mmm, I hate it too. But I want to watch the World Cup Final this weekend. But our TV won’t work.M: I can’t play, that’s my problem. If I’m not very good at a sport, I don’t like watching it.W: So, what do you like, then?M: I play golf.W: Oh, right. My husband plays too. I’m not interested in it, though. I go horse riding.M: I’d love to have a go at riding if I have time. I bet it’s good fun.8. M: So, Sally, we should probably start thinking about our winter vacation plans.W: Yeah, I think we should fly down to Florida for a couple of weeks.M: Uh, I’d really prefer not to. Those holiday travel packages cost way too much.W: Well, how about skiing up at Sky Valley Resort for a week or so?M: I’d rather not. We did it last year. Remember?W: And we had a really good time. Do you have any suggestions? M: Umm… you know, I never have any good ideas.W: OK. Here’s something different. How about going to the New Year’s Eve party in New York City? We’ve always wanted to do that.M: But I don’t really like big parties. They’re always crowded. W: You’re driving me crazy, Tom! All right. Let’s just stay home for two weeks!M: No way! That sounds too boring! Can’t you think of anything else?9. W: Andy, are you busy at the moment?M: Oh, hi. Well, actually I’m expecting an important call from the office any minute but go ahead.W: I’m sorry, but it’s really important. It’s about the car. It was broken into last night…M: Oh, no!W: I want to know if you saw or heard anything.M: Did I see anything? Where?W: In the road. Did you see anybody looking suspicious?M: No, I don’t think I looked out of the window.W: Can you remember if you heard anything? They can’t have done it completely silently. I came back late from a party and I parked it at about 11:30, and then I had a shower and went to bed.M: No, I’m sorry. What have they taken?W: Well, this morning I found the driver’s window smashed.There’s glass everywhere, and the CD player has gone. M: How annoying! Is anything else missing?W: Well, it could have been worse. Luckily, there wasn’t much in the car anyway.10. M: In today’s lecture we’ll begin with Doxey’s model forthe socio-cultural impact of tourism. He identifies fourstages.He identifies the first stage as happiness. Because initially the tourists are regarded as a new and interesting thing.As tourist development begins to increase, some sectionsof the local population become involved with tourists while others don’t, and it is increasingly the case that commercial rather than social factors are influencing relationships between tourists and the host community.People are less interested in the tourists for their own sake. This is identified as the second stage by Doxey. If development continues to increase, the hosts get annoyed with the tourists, and this is the third stage. Well, development of the tourist area may start to get out of control, which is going to make life difficult for local people. So the policy makers, the government, the local authorities, and so on, provide more roads, more car parks and so on to try to help cope with the coming of tourists.But the lives of the local people are made increasingly difficult and in the final stage of the model, the hosts show unkindness to the tourists.。

2017年东北三省三校高三第一次联合模拟考试-生物试题1．有关生物体内水和无机盐的叙述，正确的是A．镁是所有光合色素的组成成分B．植物蒸腾作用失去的水是自由水C．秋冬季节，植物体内结合水与自由水比例下降D．骨骼和牙齿中的钙主要以离子形式存在2．水绵是常用的生物实验材料，下列说法正确的是A．恩格尔曼利用水绵和好氧细菌设计实验时，自变量是氧气的有无B．水绵没有叶绿体，但细胞内含有色素和相关的酶，因此可以进行光合作用C．在水绵进行无性繁殖的过程中，会出现核膜的消失、重建过程D．利用光学显微镜观察有丝分裂间期的水绵，能观察到染色体和中心体3．基因在表达过程中如有异常mRNA会被细胞分解，下图是s基因的表达过程，则下列有关叙述正确的是A．异常mRNA的出现是基因突变的结果B．图中所示的①为转录，②为翻译过程C．图中②过程使用的酶是反转录酶D．S基因中存在不能翻译成多肽链的片段4．现有一株基因型为AaBbCc的豌豆，三对基因独立遗传且完全显性，自然状态下产生子代中重组类型的比例是A．1／8 B．1／4 C．37／64 D．27／2565．角膜移植手术是对角膜病治疗的主要手段。

虽然角膜因其无血管、无淋巴管的特性，属于相对免疫赦免区，但仍不能完全避免免疫排斥这一器官移植所面对的困扰。

据此判断下列说法正确的是A．移植后角膜脱落属于非特异性免疫B．使用免疫抑制剂可提高人体免疫系统的敏感性C．对于接受者来说，移植的角膜中含有抗原类物质D．角膜属于相对免疫赦免区，说明人体的免疫系统存在一定的缺陷6．浮萍(一种被子植物)在环境污染治理和能源生产方面具有巨大潜力。

研究发现，浮萍不但能净化水质，还能分泌有关物质促进水中藻类叶绿素的分解，并覆盖在水体表面，使藻类处于遮光状态，从而降低水华的发生。

以下推论正确的是A．浮萍能分解水中有机污染，属于分解者B．浮萍通过影响叶绿体的生理功能抑制蓝藻生长C．浮萍与藻类是竞争关系D．浮萍可用于能源生产，体现了其在生态系统中的间接价值29．(10分)图一为夏日晴朗天气，光合作用强度的日变化曲线。