2019年天津高考试题及答案(理科)

合集下载

(完整版)2019年天津卷理综化学高考试题及参考答案

2019年天津卷理科综合化学试题可能用到的相对原子质量：H-1 C-12 O-16 Zn-651．化学在人类社会发展中发挥着重要作用，下列事实不涉及．．．化学反应的是（）A．利用废弃的秸秆生产生物质燃料乙醇B．利用石油生产塑料、化纤等高分子材料C．利用基本的化学原料生产化学合成药物D．利用反渗透膜从海水中分离出淡水2．下列离子方程式能用来解释相应实验现象的是（）实验现象离子方程式A 向氢氧化镁悬浊液中滴加氯化铵溶液，沉淀溶解Mg(OH)2 + 2NH4+ = Mg2+ + 2NH3·H2OB 向沸水中滴加饱和氯化铁溶液得到红褐色液体Fe3+ + 3H2O = Fe(OH)3↓+ 3H+C 二氧化硫使酸性高锰酸钾溶液褪色3SO2+ 2MnO4-+ 4H+= 3SO42-+ 2Mn2++ 2H2OD 氧化亚铁溶于稀硝酸FeO + 2H+ = Fe2+ + H2O3．下列有关金属及其化合物的应用不合理．．．的是（）A．将废铁屑加入FeCl2溶液中，可用于除去工业废气中的Cl2B．铝中添加适量锂，制得低密度、高强度的铝合金，可用于航空工业C．盐碱地（含较多Na2CO3等）不利于作物生长，可施加熟石灰进行改良D．无水CoCl2呈蓝色，吸水会变为粉红色，可用于判断变色硅胶是否吸水4．下列实验操作或装置能达到目的的是（）A B C D混合浓硫酸和乙醇配制一定浓度的溶液收集NO2气体证明乙炔可使溴水褪色5．某温度下，HNO2和CH3COOH的电离常数分别为5.0×10-4和1.7×10-5。

将pH和体积均相同的两种酸溶液分别稀释，其pH随加水体积的变化如图所示。

下列叙述正确的是（）A．曲线Ⅰ代表HNO2溶液B．溶液中水的电离程度：b点＞c点C．从c点到d点，溶液中()()()HA OHAc cc--⋅保持不变（其中HA、A-分别代表相应的酸和酸根离子）D．相同体积a点的两溶液分别与NaOH恰好中和后，溶液中n(Na+)相同6．我国科学家研制了一种新型的高比能量锌-碘溴液流电池，其工作原理示意图如下。

2019年天津市高考物理试卷-含答案

2019年天津市高考物理试卷副标题一、单选题（本大题共5小题，共30.0分）1.2018年12月8日，肩负着亿万中华儿女探月飞天梦想的嫦娥四号探测器成功发射，“实现人类航天器首次在月球背面巡视探测，率先在月背刻上了中国足迹”。

已知月球的质量为M、半径为R，探测器的质量为m，引力常量为G，嫦娥四号探测器围绕月球做半径为r的匀速圆周运动时，探测器的（）A. 周期为B. 动能为C. 角速度为D. 向心加速度为2.2018年10月23日，港珠澳跨海大桥正式通车。

为保持以往船行习惯，在航道处建造了单面索（所有钢索均处在同一竖直面内）斜拉桥，其索塔与钢索如图所示。

下列说法正确的是（）A. 增加钢索的数量可减小索塔受到的向下的压力B. 为了减小钢索承受的拉力，可以适当降低索塔的高度C. 索塔两侧钢索对称且拉力大小相同时，钢索对索塔的合力竖直向下D. 为了使索塔受到钢索的合力竖直向下，索塔两侧的钢索必须对称分布3.如图所示，在水平向右的匀强电场中，质量为m的带电小球，以初速度v从M点竖直向上运动，通过N点时，速度大小为2v，方向与电场方向相反，则小球从M运动到N的过程（）A. 动能增加B. 机械能增加C. 重力势能增加D. 电势能增加4.笔记本电脑机身和显示屏对应部位分别有磁体和霍尔元件。

当显示屏开启时磁体远离霍尔元件，电脑正常工作；当显示屏闭合时磁体靠近霍尔元件，屏幕熄灭，电脑进入休眠状态。

如图所示，一块宽为a、长为c的矩形半导体霍尔元件，元件内的导电粒子是电荷量为e的自由电子，通入方向向右的电流时，电子的定向移动速度为v。

当显示屏闭合时元件处于垂直于上表面、方向向下的匀强磁场中，于是元件的前、后表面间出现电压U，以此控制屏幕的熄灭。

则元件的（）A. 前表面的电势比后表面的低B. 前、后表面间的电压U与v无关C. 前、后表面间的电压U与c成正比D. 自由电子受到的洛伦兹力大小为5.如图为a、b、c三种光在同一光电效应装置中测得的光电流和电压的关系。

2019年高考天津卷化学试题含答案解析

2019年高考天津卷化学试题含答案解析以下数据供解题时参考：相对原子质量：H 1 C 12 N 14 O 161.化学在人类社会发展中发挥着重要作用，下列事实不涉及化学反应的是A.利用废弃的秸秆生产生物质燃料乙醇B.利用石油生产塑料，化纤等高分子材料C.利用基本的化学原料生产化学合成药物D.利用反渗透膜从海水中分离出淡水【试题解析】该题为生活中的化学常识考查。

A.废弃的秸秆为纤维素，经水解、分解生产生物质燃料乙醇，属于化学反应；B.石油的主要成分是烷烃、环烷烃、芳香烃，经过裂解得短链烯烃，用于生产塑料、化纤等高分子材料，为化学反应；C.利用基本化学原料生产化学合成药物，很明显是化学反应；D.利用反渗透膜，自然是物理方法。

【答案】：D2.下列离子方程式能用解释实验现象的是【试题解析】该题为化学基础知识的考查。

A. Mg(OH)2沉淀溶解原理是与NH4+ 反应生成NH3﹒H2O,正确；B.据题意生成液体故不能写沉淀符号，错误；C. 得失电子没有守恒，不要认为电荷和原子守恒就可以了，错误；D.硝酸无论浓稀均具有强氧化性,故应生成三价铁离子，错误。

【答案】：A3.下列有关金属及其化合物的应用不合理的是A.将废铁屑加入氯化亚铁溶液中，可用于除去工业废气中的氯气B.铝中添加适量锂，制得低密度、高强度的铝合金可用于航空工业C.盐碱地（含较多Na 2CO 3等）不利于作物生长，可施加熟石灰进行改良D.无水CoCl 2呈蓝色，吸水会变为粉红色，可用于判断变色硅胶是否吸水试题解析：该题检测金属及其化合物性质。

C 选项错误很明显，土壤已显碱性故不能加碱性物质，应该加些显酸性的物质，如可以施用铵肥。

【答案】C4.下列实验或操作能达到实验目的的是【试题解析】：该题考查学生基本实验操作。

A.有机化学制备乙酸乙酯时明确了试剂加入顺序，要先加乙醇再加浓硫酸，否则乙醇沸点低，加入浓硫酸中放热易挥发，错误；C.二氧化氮比空气重，排空气法收集应该长进短出，错误；D.电石中含有的杂质硫化钙、磷化钙与水产生的硫化氢、磷化氢也会使溴水褪色,故D 错误。

(完整版)2019年高考生物试题(天津卷)(解析版)

2019 年普通高等学校招生全国统一考试（天津卷）理科综合理科综合生物部分第I 卷1.用3H 标记胸腺嘧啶后合成脱氧核苷酸，注入真核细胞，可用于研究A. DNA 复制的场所B. mRNA 与核糖体的结合C. 分泌蛋白的运输D. 细胞膜脂质的流动【答案】A【解析】【分析】本题主要考查DNA 的复制、蛋白质的合成及运输及细胞膜的功能等有关知识。

DNA 复制需要脱氧核苷酸作为原料；mRNA 与核糖体结合，翻译形成蛋白质；分泌蛋白的运输需要内质网和高尔基体形成囊泡运输；细胞膜脂质的流动与物质跨膜运输有关。

【详解】DNA 复制需要DNA 模板、原料脱氧核苷酸、能量ATP 和DNA 聚合酶，A 正确；mRNA 与核糖体的结合，开始翻译mRNA 上的密码子，需要tRNA 运输氨基酸，不需要脱氧核苷酸，B 错误；分泌蛋白的需要内质网的加工，形成囊泡运到高尔基体，加工、分类和包装，形成分泌小泡，运到细胞膜，胞吐出去，与脱氧核苷酸无关，C 错误；细胞膜脂质的流动与物质跨膜运输有关，无需脱氧核苷酸，D 错误。

因此，本题答案选A。

【点睛】解答本题关键要熟悉细胞中不同的生理活动的具体过程，来判断是否需要脱氧核苷酸作为原料。

2.下列过程需ATP 水解提供能量的是A.唾液淀粉酶水解淀粉B.生长素的极性运输C.光反应阶段中水在光下分解D.乳酸菌无氧呼吸的第二阶段【答案】B【解析】【分析】本题主要考查细胞中生理活动的相关知识。

细胞中蛋白质、淀粉和多糖在酶的作用下水解，无需消耗能量；生长素在幼嫩组织中从形态学上端向形态学下端运输属于极性运输，其方式为主动运输，需要消耗能量；光合作用的光反应阶段，水在光下分解，不需要消耗ATP 的能量；无氧呼吸的第二阶段是丙酮酸分解成酒精和CO2或乳酸，都不消耗能量。

【详解】唾液淀粉酶水解淀粉，形成麦芽糖，不消耗能量，A 错误；.生长素的极性运输是以主动运输的方式，在幼嫩组织中从形态学上端运到形态学下端，需要ATP 提供能量，B 正确；光反应阶段中水在光下分解，需要光能，不需要ATP 功能，C 错误；乳酸菌无氧呼吸的第二阶段是丙酮酸变成乳酸，不需要ATP 供能，D 错误。

2019年天津市高考物理试卷以及答案解析

2019年普通高等学校招生全国统一考试（天津卷）地理答卷前，考生务必将自己的姓名、准考证填写在答题卡上，并在规定位置粘贴考试用条形码。

答卷时，考试务必将答案涂写在答题卡上，答在试卷上的无效。

考试结束后，将本试卷和答题卡一并交回。

祝各位考生考生顺利！一、单项选择题（每小题6分，共30分。

每小题给出的四个选项中，只有一个选项是正确的）1．（6分）2018年12月8日，肩负着亿万中华儿女探月飞天梦想的嫦娥四号探测器成功发射，“实现人类航天器首次在月球背面巡视探测，率先在月背刻上了中国足迹”。

已知月球的质量为M、半径为R，探测器的质量为m，引力常量为G，嫦娥四号探测器围绕月球做半径为r的匀速圆周运动时，探测器的（）A．周期为B．动能为C．角速度为D．向心加速度为2．（6分）2018年10月23日，港珠澳跨海大桥正式通车。

为保持以往船行习惯，在航道处建造了单面索（所有钢索均处在同一竖直面内）斜拉桥，其索塔与钢索如图所示。

下列说法正确的是（）A．增加钢索的数量可减小索塔受到的向下的压力B．为了减小钢索承受的拉力，可以适当降低索塔的高度C．索塔两侧钢索对称且拉力大小相同时，钢索对索塔的合力竖直向下D．为了使索塔受到钢索的合力竖直向下，索塔两侧的钢索必须对称分布3．（6分）如图所示，在水平向右的匀强电场中，质量为m的带电小球，以初速度v从M 点竖直向上运动，通过N点时，速度大小为2v，方向与电场方向相反，则小球从M运动到N的过程（）A．动能增加mv2B．机械能增加2mv2C．重力势能增加mv2D．电势能增加2mv24．（6分）笔记本电脑机身和显示屏对应部位分别有磁体和霍尔元件。

当显示屏开启时磁体远离霍尔元件，电脑正常工作；当显示屏闭合时磁体靠近霍尔元件，屏幕熄灭，电脑进入休眠状态。

如图所示，一块宽为a、长为c的矩形半导体霍尔元件，元件内的导电粒子是电荷量为e的自由电子，通入方向向右的电流时，电子的定向移动速度为v。

2019年天津市高考数学试卷(理科)(解析版)

【点睛】圆柱的底面半径是棱锥底面对角线长度的一半、不是底边棱长的一半。
12.设，直线和圆（为参数）相切，则的值为____.
【答案】
【解析】
【分析】
根据圆的参数方程确定圆的半径和圆心坐标，再根据直线与圆相切的条件得出满足的方程，解之解得。
【详解】圆化为普通方程为，
圆心坐标为，圆的半径为，
2019年普通高等学校招生全国统一考试（天津卷）
数学（理工类）
第Ⅰ卷
注意事项：
1.每小题选出答案后，用铅笔将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。
2.本卷共8小题。
参考公式：
·如果事件、互斥，那么 .
·如果事件、相互独立，那么 .
·圆柱的体积公式，其中表示圆柱的底面面积，表示圆柱的高.
4.阅读右边的程序框图，运行相应的程序，输出的值为
A.5B.8C.24D.29
【答案】B
【解析】
【分析】
根据程序框图，逐步写出运算结果。
【详解】，
结束循环，故输出。
故选B。
【点睛】解答本题要注意要明确循环体终止的条件是什么，会判断什么时候终止循环体．
5.已知抛物线的焦点为，准线为 .若与双曲线的两条渐近线分别交于点A和点B，且（为原点），则双曲线的离心率为
由直线与圆相切，则有，解得。
【点睛】直线与圆的位置关系可以使用判别式法，但一般是根据圆心到直线的距离与圆的半径的大小作出判断。
13.设，则的最小值为______.
【答案】
【解析】
分析】
把分子展开化为，再利用基本不等式求最值。
【详解】
，

(完整版)2019天津高考化学解析

2019年普遍高等学校招生全国统一考试（天津卷）理科综合化学部分第I卷1.化学在人类社会发展中发挥着重要作用，下列事实不涉及化学反应的是A.利用废弃的秸秆生产生物质燃料乙醇B.利用石油生成塑料、化纤等高分子材料C.利用基本的化学原料生产化学合成药物D.利用反渗析膜从海水中分离出淡水解析：本次考查内容为化学反应实质，是否有新物质生成。

考查内容相对基础。

同时也从题目中的四个选项中发现，考查点相对细致和琐碎。

选项中ABC三个选项都是涉及化学变化，D选项考查渗析原理，没有涉及化学变化。

答案：选D2.下列离子方程式能用来解释相应实验现象的是B沸水中加入饱和溶液，生成胶体；C选项离子方程式左右两边带电量不对；D选项二价铁遇到硝酸发生氧化还原反应。

答案：选择A。

3.下列有关金属及其化合物的应用不合理的是A.将废铁屑加入FeCl2溶液中，可用于除去工业废气中的Cl2B.铝中添加适量锂，制的低密度，高强度的铝合金，可以用于航空工业C.盐碱地（含较多Na2CO3等）不利于作物生长，可施加熟石灰进行改良D.无水CoCl2呈蓝色，吸水会变成粉红色，可用于判断变色硅胶是否吸水解析：选项A，氯气与二价铁反应，铁单质与三价铁反应最终将氯气吸收；选项B，考查的是合金的性质；选项C，Na2CO3+Ca(OH)2=2NaOH+CaCO3↓生成物质还是碱性物质，并没有改良。

选项D，物质颜色变化判断性质。

答案：C选项。

4.下列实验操作或者装置能达到目的的是解析：选项A密度先小后大，硫酸加入到乙醇中；选项B，读取刻度数看凹液面；选项C，NO2密度大于空气，长近短出；选项D，没有除去杂质。

答案：选B5.某温度下，HNO2和CH3COOH的电离常数分别为5.0×10-4和1.7×10-5.将PH和体积均为相同的两种酸溶液分别稀释，其PH随加水体积的变化如图所示。

下列叙述正确的是A.曲线1代表HNO2溶液B.溶液中加水的电离程度：b点>c点C.从c点到d点，溶液中)()(c) (c---⋅A c OHHA保持不变（其中HA、A-分别代表相应的酸和酸根离子）D.相同体积a点的两溶液分别与NaOH恰好中和后，溶液中n（Na+）相同解析：HNO2和CH3COOH相比，HNO2酸性更强，加水稀释，越稀越电离，醋酸电离程度大，曲线1为醋酸，选项A错误；b点PH大，氢离子浓度大，对水抑制程度大，水电离程度小，选项B 错误；C选项表达式只和温度有关，加水稀释过程中温度没变，保持不变，选项c正确；因为HNO2和CH3COOH酸性不一样，在PH相同的情况下，浓度不一样，中和需要的NaOH也不一样。

(完整word)2019年高考试题汇编理科数学--数列,推荐文档

解答: 13，设等比数列公比为q3、25•- (ag )ag••• q 3• S 121 …S 53（1）证明:a nb n 是等比数列，a n b n 是等差数列;(2 )求a n 和b n 的通项公式. 答案： (1) 见解析 1 x n 11 x n 1(2)a n () n，b n () n2222解析：(1)将 4a n 1 3a n b n 4 , 4b n 1 3b n a n 4 相加可得 4a n1 4b n 1 3a n 3b n a n b n ,11 整理可得a n 1 b n 1丄(a n b n )，又玄1 Q 1，故a . b n 是首项为1,公比为1的等比数列22将 4a n 1 3a n b n 4, 4b n 13b n a n 4 作差可得 4a n14b n13a n 3b n a . b n 8,整理可得a n 1 b n 1a nb n 2，又a 1 Q 1，故a .b n 是首项为1，公差为2的等差数列1 1A. a n 2n 5B.3n 3n 10 CS2n 28nD.S n■In 2 2n 2答案：A解析：S 4 4冃 6d 0a 1 3 5, S n2依题意有可得 a nn 4n .3S 31 4d 5 d 2 n（2019全国1理）9•记S n 为等差数列 a n 的前n 项和•已知S 40 , a 5 5，则（2（2019全国1理）14.记S n 为等比数列 a n 的前 n 项和,a 436，则 S5答案: S 51213 2019全国2理)19.已知数列a n 和b n满足a 10 , 4a n 1 3a n b n 4, 4b n 1 3b n a n 4.-31 2 3436(2)由a n b n是首项为1 ,公比为?的等比数列可得a n b n ()"①；由a n bn 是首项为1公差为2的等差数列可得a n b n 2n 1②;【解析】【分析】首先确定公差，然后由通项公式可得 a 5的值，进一步研究数列中正项？负项的变化规律,得到和的最小值.【详解】等差数列 a n 中，8s 5a 3 10,得a 3 2& 3,公差da 3 a ?1, a§% 2d 0,由等差数列a n 的性质得n 5时，a n 0, n 6时，a n 大于0,所以S n 的最小值为S 4或S 5,即为10.①②相加化简得a n（!）n n 1，①②相减化简得b n 2 2（2019全国3理）5.已知各项均为正数的等比数列的前4项和为15，且a s 3a 3 4印，则a ?（）A. 16B. 8 答案： C解答：C. 4D.设该等比数列的首项 a i ，公比由已知得,4a©3dq 24a i ，因为a 0且q 0，则可解得2，又因为 a i (1q 3) 15,即可解得c 1，则4.（2019全国3理）14.记S n 为等差数列 a n 的前n 项和，若q0, a 2 3a ，则 3°S 5答案:4解析:设该等差数列的公差为d 2a 1 a 1 0,d 0 ,10 a 1 a 10S 0____________2S 55 a 1 a 522 2a 1 9d3 4.2a 1 4d 5d（2019北京理）10.设等差数列的前n 项和为S n,若a 2=-3 ,S s =-10，则a s = ,S n 的最小值为【答案】 (1). 0. (2). -10.【点睛】本题考查等差数列的通项公式？求和公式？等差数列的性质，难度不大，注重重要知识？基础知识？基本运算能力的考查a i (2019北京理)20.已知数列｛a n｝，从中选取第i1项、第i2项、…、第i m项(i l<i2<・・Vm)，若a h a2则称新数列a h, a i2, , a m为｛a n｝的长度为m的递增子列•规定：数列｛a n｝的任意一项都是｛a n｝的长度为1的递增子列．(I)写出数列1 , 8, 3, 7, 5, 6, 9的一个长度为4的递增子列；(H)已知数列｛a n｝的长度为p的递增子列的末项的最小值为a m o，长度为q的递增子列的末项的最小值为a n0.若p<q,求证：a m°<a n°；(川)设无穷数列｛a n｝的各项均为正整数，且任意两项均不相等若｛ a n｝的长度为s的递增子列末项的最小值为2s -, 且长度为S末项为2s-1的递增子列恰有2s-1个(s=1 , 2,…)，求数列｛a n｝的通项公式.【答案】(I )1,3,5,6.(n )见解析; (川)见解析.【解析】【分析】(I )由题意结合新定义的知识给出一个满足题意的递增子列即可；(n )利用数列的性质和递增子列的定义证明题中的结论即可；(川)观察所要求解数列的特征给出一个满足题意的通项公式，然后证明通项公式满足题中所有的条件即可•【详解】(I )满足题意的一个长度为4的递增子列为：1,3,5,6.(n)对于每一个长度为q的递增子列a n a2丄a q，都能从其中找到若干个长度为p的递增子列色总丄a p，此时a p a q ，设所有长度为q的子列的末项分别为：a q, ,a q2,a q3 ,L ,所有长度为p的子列的末项分别为：a p1,a p2,a p3,L ,则a n0 min a q1,a q2,a q3,L ，注意到长度为P的子列可能无法进一步找到长度为q的子列，故a m0 min a p1,a p2,a p3,L ，据此可得：a m0a n0n 1, n为偶数(川)满足题意的一个数列的通项公式可以是a n 斗才来朴2,1,4,3,6,5,8,7,L ,n 1,n为奇数面说明此数列满足题意很明显数列为无穷数列，且各项均为正整数，任意两项均不相等.长度为s 的递增子列末项的最小值为2s-1，下面用数学归纳法证明长度为s 末项为2s-1 的递增子列恰有2s 1个s 1,2,L ：当n 1 时命题显然成立，假设当n k时命题成立，即长度为k末项为2k-1的递增子列恰有21个，则当n k 1时，对于n k 时得到的每一个子列a s1,a s2,L ,a s k 1,2k 1，可构造：aq,a s2丄，a s「2k 1,2 k 1 1和a5^,a S2,L ,a^l,2k,2 k 1 1两个满足题意的递增子列，则长度为k+1 末项为2k+1 的递增子列恰有 2 2k 12k2k 1 1个，n 1, n为偶数综上可得，数列a n、，卄沁.2,1,4,3,6,5,8,7,L是一个满足题意的数列的通项公式•n 1, n为奇数注：当s 3时，所有满足题意的数列为：2,3,5 , 1,3,5 , 2,4,5 , 1,4,5 ，当s 4 时，数列2,3,5 对应的两个递增子列为：2,3,5,7 和2,3,6,7 .【点睛】“新定义”主要是指即时定义新概念、新公式、新定理、新法则、新运算五种，然后根据此新定义去解决问题，有时还需要用类比的方法去理解新的定义，这样有助于对新定义的透彻理解.但是，透过现象看本质，它们考查的还是基础数学知识，所以说“新题”不一定是“难题”，掌握好三基，以不变应万变才是制胜法宝.2019天津理) 19.设a n 是等差数列，b n 是等比数列.已知a1 4,b1 6，b2 2a2 2,b3 2a3 4.(I)求a n和b n的通项公式;（n）设数列q满足G 1,c n X 2 J 2「其中k Nn 1 n b k,n 2k ,i )求数列a2n c2n1 的通项公式;2nii ）求a i c i n Ni1答案】（I ）a n 3n 1 ; b n 3 2n（n ）（i ）a2n c2n 1 9 4n1 （ii ）* 2n 1n 1 *aqnN 27 25 2 n 12 nNi 1【解析】【分析】(I )由题意首先求得公比和公差，然后确定数列的通项公式即可; (n )结合(I )中的结论可得数列a 2n c 2n 1的通项公式，结合所得的通项公式对所求的数列通项公式进行等2n价变形，结合等比数列前n 项和公式可得aG 的值.i 12 4 d 26 2d，解得2 4 2d 4 12 4d故a n 4 (n 1) 33n1 ,b n6 2n13 2n.所以,a n的通项公式为 a n 3n 1 , b n的通项公式为b n3 2n (n )( i ) a 2n C 2n 1 a ?n b n 1 3 2n 1 3 2n 19 4n 1所以,数列 a ?n c?n1 的通项公式 :为a2nc 2n 19 4n 12n 2n2n2n(ii )a &a i a C i 1a ia c 2i1i 1i 1i 1i 12n 2n 1n2 n4-39 412i 14 1 4n3 ?2 n5 2n 19n1 427 _2n•1J 112N*25 2n n【点睛】本题主要考查等差数列？等比数列的通项公式及其前 n 项和公式等基础知识.考查化归与转化思想和数列求和的基本方法以及运算求解能力.【详解】(I )设等差数列a n 的公db n 的公比为q .依题意得6q6q 2（2019上海）18•已知数列｛a n ｝ , a 1 3，前n 项和为S n •（1）若｛an ｝为等差数列,且 a 4 15, 求S n ；（2）若｛a n ｝为等比数列,且 lim n S n 12，求公比 q 的取值范围【解答】解：（1） Q a 4 a 3d 3 3d 15 ,d 4 ,n(n 1),S n 3n4 2n 2 n；2lim S n 存在，nlim 3(^ 2 ,n1 q 1 q3 4公比q 的取值范围为（1 , 0）（0 , 3）.42综上，d -或者d3Hm S n存在， lim S n n （2019上海）21.已知等差数列｛务｝的公差d (0，］，数列｛b n ｝满足 b n sin (a n )，集合 S x|xb n ,n2 、（1 ）若a 1 0,d 一，求集合 30,d —,3｛乜，0, △.2 2根据三角函数线，①等差数列｛a n ｝的终边落在y 轴的正负半轴上时，集合S 恰好有两个元素，此时此时d —,3（2）若a 1，求d 使得集合 2 S 恰好有两个（3）若集合S 恰好有三个元素: b n T b n , T 是不超过7的正整数，求 T 的所有可能的值.【解答】解：（1） Q 等差数列｛a n ｝的公差d (0，］，数列｛b n ｝满足 b n sin (a n )，集合 S x|xb n ,n当a 1集合S (2) Q，数列｛b n ｝满足 b n sin (a .),2集合S x|x N *恰好有两个元素，如图:②a 1终边落在OA 上，要使得集合 S 恰好有两个元素，可以使 a 2, a 3的终边关于y 轴对称，如图OB , OC ,(3)①当T 3 时，b n 3 b n，集合S {bl，b2, b3},符合题意.②当T 4 时，b n 4 b n ，sin(a n 4d) sina. a n 4d a n 2k ，或者a n 4d 2k a n ，4d a n 2k，又k 1，2当k1时满足条件，此时S {，1, 1}.③当T 5时，b n 5b n，si n(a n5d)sina n，故k1，2.当k1时，S{sin—，1，sin}满足题意1010④当T 6时，b n 6b n，sin (an6d)sina n，a na n等差数列{a n}的公差d (0，］，故a n5d a n 2k ，或者a n 5d 2k a n，因为 d (0 ,所以6d a n 2k 或者a n 6d 2k a n，d (0，1 , 2, 3.1时，S {-^O, —3}，满足题意.2 2⑤当T 7 时，b n 7 b n，si n(a n 7d) si na n si na n，所以a n 7d a n 2k ，或者a n 7d 2k a n，d (0，故k 1 , 2, 31时，因为b i ~b7对应着3个正弦值，故必有一个正弦值对应着3个点，必然有a m a n 2 ，d m 7，不符合条件.k 2时，因为b i~b7对应着3个正弦值，故必有一个正弦值对应着3个点，必然有a m a n 2 ，d n不是整数，不符合条件.k 3时，因为bi ~ b7对应着3 个正弦值，故必有一个正弦值对应着3个点，必然有a m a n—，或者d7—，此时，m n均不是整数，不符合题意.7综上，T3，4，5，6.(2019江苏)8.已知数列{a n}( n N*)是等差数列，S n是其前n项和若a2^ 兎0,S9 27 ,则Q的值是 _____________________ 【答案】16【解析】【分析】由题意首先求得首项和公差，然后求解前8项和即可.a 2a 5CBa 1 d a-i 4d7d 0【详解】由题意可得：9 8S99a 1 9 8d227解得： a 1 51 ，则 S 8 8a 1 8 7d40 28 216.d 22【点睛】等差数列、等比数列的基本计算问题，是高考必考内容，解题过程中要注意应用函数方程思想，灵活应用通项公式、求和公式等，构建方程(组)，如本题，从已知出发，构建a 1, d 的方程组.(2019江苏)20.定义首项为1且公比为正数的等比数列为“M—数列”.(1)已知等比数列｛a n ｝满足：a ?a 4 a 5,a 3 4a ? 4印 0 ,求证：数列｛a n ｝为“M—数列”；u . 1 2 2(2)已知数列｛b n ｝满足：b 1 1,S b b ，其中S 为数列｛b n ｝的前n 项和.S n b n b n 1① 求数列｛b n ｝的通项公式；② 设m 为正整数，若存在 “M—数列” ｛｝ (n € N *)，对任意正整数k ，当k 呦时，都有C k b k q 1成立，求m 的最大值.【答案】(1)见解析； (2［① b n = n n N * :② 5. 【解析】【分析】(1 )由题意分别求得数列的首项和公比即可证得题中的结论； (2)①由题意利用递推关系式讨论可得数列｛b n ｝是等差数列，据此即可确定其通项公式；②由①确定b k 的值，将原问题进行等价转化，构造函数，结合导函数研究函数的性质即可求得【详解】(1)设等比数列｛a n ｝的公比为q ,所以a 1^0, q 丰0.因此数列｛a n ｝为M —数列”1 22 (2) ①因S n—，所以b nb nbn11 2 2由b| 1,S 1th 得1 1 ，则 b 22.1由2 2 得 S nb n b n 1m 的最大值.a 2&4 a s由a 3 4a ： 4ci|。

2019年3月天津高考试题及答案

2019年3月天津高考试题第一卷第一部分：第一节：1.–Everybody is helping out with dinner. Would you clean the fish?--______. I’d rather make the salad.A. A piece of cakeB. No problemC. Couldn’t be betterD. Anything but that2. You should respect the views of others, and at the same time, ______what you think is right.A. care forB. look atC. insist onD. meet with3. –Edward, do you mind giving me a ride to the railway station?--______. I’d be glad to.A. Yes. I doB. Of course notC. Never mindD. Go ahead4. The captain of the ship was advised to turn back ______ a sudden heavy storm.A. due toB. by means ofC. in addition toD. instead of5. Watch out for injuries while exercising. Always stop ______ you begin to feel any pain.A. in order thatB. even ifC. ever sinceD. as soon as6. ______ volcanoes for many years, I am still amazed at their beauty as well as their potential to cause great damage.A. To studyB. StudyingC. Having studiedD. Studied7. Even though we live in a high-tech age, it’s still impossible to predict he weather ______.A. activelyB. accuratelyC. cautiouslyD. originally8. The opinion ______ learning is a lifelong process has been expressed by education experts throughout the years.A. whichB. thatC. whatD. how9. I took notes during the meeting, but I ______ a chance to write them into a report so far.A. didn’t haveB. haven’t hadC. hadn’t hadD. won’t have10. The sign on the wall of the library says, “No magazine is allowed ______ out of the reading room”.A. being takenB. to takeC. to be takenD. taking11. People believe writing poems provides a ______ through which they can express their feelings.A. shelterB. sourceC. channelD. background12. We are determined that our training should ______ the current development in education.A. keep pace withB. take possession ofC. make room forD. give birth to13. Paul did a great job in the speech contest. He ______ many times last week.A. need have practicedB. might practiceC. must have practicedD. could practice14. The course normally attracts about 100 students per year, ______ up to half will be from abroad.A. whoB. of whichC. whenD. of whom15. Mary’s description of the party was so vivid that I felt as if I ______ there.A. would beB. might have beenC. would have beenD. had been第二节：In the summer before I entered middle school, I read the book they Cage the Animals at Night.It’s a story about Jennings, a boy living in various orphanages（孤儿院）with only his stuffed animal（毛绒玩具）, Doggie, for companionship. It’s a fascinating book, but little did I know it would __16__ my life and the lives of others.One day, as I looked across my room at the pile of stuffed animals, a(n) __17__ came to me.I would __18__ stuffed animals for children like Jennings. __19__, I contacted agencies that support those children and they said the stuffed animals would certainly __20__ children’s spirits.I named my project “Cuddle Buddies”. I wrote articles for the local and school newspapers, __21__ donations of the “buddies”- stuffed animals. My phone __22__ ringing; schools, families, and toy factories all wanted to help. Much to my __23__, by the second week, my living room looked like a zoo. Whenever Mom and I __24__ the “buddies” to the agencies, the kids would be waiting there with their eyes down, too __25__ to look but shaking with excitement.Six years after its launch, Cuddle Buddies continues to __26__. Now over 25,000stuffed animals have been __27__ to agencies worldwide. Simon, a seven-year-old boy in an African Children’s Home, couldn’t sleep at night after she lost his parents. When asked to choose his “buddies”, Simon __28__a panda and soon after was sleeping through the night. I never dreamed Cuddle Buddies would __29__ this way.Upon graduation from high school, I designed a website, , to __30__ the project. The response was __31__. More and more people joined me and two dozen Cuddle Buddies __32__ were established across the nation.This has been a great experience. I’ve learned how to follow through on an idea and how to __33__ impact a life. I will go to college with these __34__ in mind and continue my work with Cuddle Buddies, even when I __35__ a whole new set of exciting academic and nonacademic pursuits.16. A. touch B. risk C. lead D. save17. A. chance B. idea C. explanation D. word18. A. display B. repair C. purchase D. collect19. A. Previously B. Occasionally C. Immediately D. Gradually20. A. capture B. reflect C. lift D. record21. A. putting aside B. turning down C. asking for D. getting back22. A. delayed B. kept C. stopped D. meant23. A. knowledge B. taste C. regret D. delight24. A. delivered B. applied C. rented D. returned25. A. shy B. content C. weak D. eager26. A. reform B. expand C. dominate D. divide27. A. produced B. adopted C. sold D. donated28. A. picked out B. came across C. showed off D. brought up29. A. decline B. work C. increase D. react30. A. promote B. start C. evaluate D. examine31. A. confusing B. amusing C. automatic D. enthusiastic32. A. companies B. committees C. branches D. institutes33. A. financially B. positively C. socially D. physically34. A. plans B. images C. lessons D. motives35. A. work out B. keep off C. rely on D. engage in第二部分：AWelcome to University of Waikato. Our mission is to create a welcoming environment in which residents feel comfortable, safe and a sense of belonging.Student Village and Waikato ApartmentsWaikato Accommodation Services offers two types of residential accommodation: Students Village and Waikato Apartments.Student Village consists of fifteen independent units. Each contain five bedrooms, toilet facilities, and a shared kitchen with a fridge and cupboard. There is also a games room and a TV room.Waikato Apartments are situated in a three-level apartment building. Each level has four apartments containing six bedrooms, toilet facilities, and a kitchen. Each level has a shared area with a TV, a study desk and easy chairs.Security and SafetyCampus security is available to assist 24 hours, 7 days a week. If you see anything suspicious, please call Campus Security on 0800863800.Please ensure that you have your own insurance for your personal belongings as Waikato Accommodation Services takes no responsibility for any theft or damage to personal belongings. Housekeeping and MaintenanceHousekeeping staff service all shared areas. Residents are responsible for keeping their rooms, kitchens and shower rooms clean and tidy.Maintenance staff ensure necessary repairs to facilities are made. If your room or other shared area needs maintenance, please inform the Accommodation Office. Maintenance staff may be required to enter your room to tend to a maintenance request – this will be prearranged. Termination（终止）of Residence and Checking OutYou are required to pay accommodation fees for the full period of your residency contract. At the end of your contract you must arrange a time with the office staff to check out. A room inspection will take place and you will be responsible for payment for any damage or missing items. If you fail to check out in this manner, the staff will hold an inspection in your absence and assess whether any payment is required.36. What facilities are available in both Student Village and Waikato Apartments?A. Kitchen and TV.B. Study desk and fridge.C. Cupboard and study room.D. Games room and toilet facilities.37. What should you do when you suspect that someone is destroying school facilities?A. Report to the insurance company.B. Contact Campus Security.C. Inform the assistant teacher.D. Call Waikato Accommodation Services.38. The responsibility of the maintenance staff includes ______.A. making plans for Accommodation OfficeB. monitoring the use of kitchen facilitiesC. fixing a broken windowD. cleaning shower rooms39. If you can’t check out in the required way, the staff will ______.A. ask you to pay higher accommodation feesB. assess whether an inspection is necessaryC. still carry out the room inspectionD. wait until you can arrange a time40. Which of the following is the passage probably taken from?A. A technical instruction book.B. A university newspaper.C. A residency contract.D. A student guide.BAn unpleasant smell floated into my nose as the bowl of fried stinky tofu （臭豆腐）settled onto my grandmother’s dinner table. I immediately sat back. The smell ruined my appetite, and I pushed the dish away.“It’s supposed to stink. Just give it a try!” my grandmother said.Holding my breath, I unwillingly took a bite. To my surprise, the tofu beneath the fried skin was warm, soft, and mouthwatering. Since then, whenever I visited my grandmother’s house, I would rush to the kitchen for the stinky tofu with excitement. Thus, stinky tofu became more than a household tradition. It became my favorite dish.Along with the stinky tofu, I would meet my Uncle Chan on every visit. As an immature boy, I never understood my uncle’s condition of Down syndrome（唐氏综合症）.He was an unfriendly and terrible creature in my eyes. He constantly walked around the house, talking to himself. Whenever he was within view, I would run away in fear.However, one day, my view of my uncle suddenly changed. I was just getting out of my bed when I heard soft steps approaching. It was my uncle. My muscles froze. He slowly sat next to me and touched my face gently, as a mother would stroke（抚摸）a baby. A wide smile emerged from his blank expression. How beautiful his smile was! I could not help but smile back. My fear disappeared gradually, and my muscles relaxed. For the first time, I could see softness and warmth in his eyes.My uncle very much resembles stinky tofu. Stinky tofu smells unpleasant on the outside, yet feels warm and soft on the inside. Like stinky tofu, my uncle’s Down syndrome made me keep my distance at first. Yet, I learned that deep inside, he is harmless and has a loving nature.Some people tend to look down upon disabled people and regard them as unfit for a regular life. As I was unable to see through the outer surface of the tofu, these people are unable to see through the surface of prejudice（偏见）. As a result, they judge the disabled with one glance.The outside and inside of the stinky tofu exist as two extremely different worlds. Perhaps that is what makes it such a tasty dish. Not only is it delicious, stinky tofu offers a valuable lesson: never judge people at first glance, for true beauty comes from the inside.41. Why did stinky tofu become the author’s favorite dish?A. It was related to family tradition.B. It turned out to be quite delicious.C. Grandma encouraged him to try it.D. His appetite for food had increased.42. What did the author think of his uncle at first?A. He was a quiet person.B. He was an energetic person.C. He was an immature person.D. He was a strange creature.43. The author’s attitude towards Uncle Chan changed when ______.A. Chan’s footsteps woke up the authorB. Chan showed his friendliness to the authorC. Chan displayed his love towards the authorD. Chan’s blank expression made the author smile44. The comparison between stinky tofu and the author’s uncle serves to ______.A. highlight the nature of Uncle ChanB. analyze the family relationshipC. describe a childhood memoryD. introduce a traditional Chinese food45. What does the author mainly want to tell us in the passage?A. We shouldn’t judge by appearances.B. A person’s taste changes over time.C. Blood is thicker than water.D. You are what you eat.CEssentially, everyone has two ages: a chronological（按时间计算的）age, how old the calendar says you are, and a biological age, basically the age at which your body functions as it compares to average fitness or health levels.“Chronological age isn’t how old we really are. It’s merely a number,” said Professor David Sinclair at Harvard University. “It is biological age that determines our health and ultimately our lifespan. We all age biologically at different rates according to our genes, what we eat, how much we exercise, and what environment we live in. Biological age is the number of candles we really should be blowing out. In the future, with advances in our ability to control biological age, we may have even fewer candles on our birthday cake than the previous one.To calculate biological age, Professor Levine at Yale University identified nine biomarkers（生物标志）that seemed to be the most influential on lifespan by a simple blood test. The numbers of those markers, such as blood sugar and immune（免疫的）measures, can be put into the computer, and the algorithm（算式，算法）does the rest.Perhaps what’s most important here is that these measures can be changed. Doctors can take this information and help patients make changes to lifestyle, and hopefully take steps to improve their biological conditions. “I think the most exciting thing about this research is that these things aren’t set in stone,” Levine said. “People can be given the information earlier and take steps to improve their health before it’s too late.”Levine even entered her own numbers into the algorithm. She was surprised by the results. “ I always considered myself a very healthy person. I’m physically active; I eat what I consider a fairly healthy diet. But I did not find my results to be as good as I had hoped they would be. It was a wake-up call,” she said.Levine is working with a group to provide access to the algorithm online so that anyone can calculate their biological age, identify potential risks and take steps to improve their health in the long run. “No one wants to live an extremely long life with a lot of chronic（慢性的）diseases,” Levine said. “By delaying the development of mental and physical functioning problems, people can still be engaged in society in their senior years. That is the ideal we should be pursuing.”46. Biological age depends on ______.A. whether we can adapt ourselves to the environmentB. how well our body works compared with our peers’C. when we start to take outdoor exerciseD. what the calendar says about our age47. By saying “ we may have even fewer candles on our birthday cake than the previous one” in Para. 2, the author means ______.A. we don’t have to celebrate our birthday every yearB. we are chronologically older than last yearC. we might be less happy than the previous yearD. we may be biologically younger than the year before48. According to Professor Levine, what is the most exciting thing about her research?A. The measures are not fixed.B. Health services can be set up.C. The problems are not that serious.D. The research results are written in stone.49. What does the author want to tell us by Levine’s example in Para. 5?A. It is necessary to change our diet regularly.B. The test results may give us wrong information.C. Waking up early in the morning is good for our fitness.D. The algorithm can reveal our potential health problems.50. The ultimate goal of Levine’s research is to ______.A. free people from chronic diseasesB. work out a solution to genetic problemsC. keep people socially active even in old ageD. provide people with access to scientific theoryDWhen I graduated from high school, I wanted to major in comparative literature. But, once I found out my friends were going into “real”majors, like marketing, nursing, and engineering, I figured I needed to do the same to ensure a good career. So I changed my mind and choose Business Management as my major. I forced myself to believe that I would enjoy it and succeed in the future, but eventually I exhausted myself understanding the economy and trying new marketing techniques. Coming out of this experience, I realize it is ok to be different from others and to study things like classic art history and other majors offered in the College of Humanities（人文学科）.The worries most people have about a Humanities degree and finding a career afterwards are that the majors are too abstract, and one will not obtain any applicable skills. Actually ideas discussed in Humanities classes which are occasionally different from what people commonly believe, offer a broad perspective（视角）. How could one effectively be an unbiased（无偏见的）writer with only a knowledge of the popular opinion of society? How could one speak persuasively with closed minds? Only seeing the world through a single perspective leads to missed learning, missed friendships, and missed growth! With a broader perspective, we will be more free, more open-minded, and less limited in what we can become.STEM(science, technology, engineering and mathematics) graduates learn actual skills in their studies while humanities majors learn “soft skills” like communicating effectively through writing critically and speaking persuasively, synthesizing（综合）ideas through gathering and interpretinginformation, and developing cultural awareness. Do those soft skills sound useless and inapplicable to you? Think of it. How often do you communicate with others? Produce ideas? Encounter people from other cultures? Every day. Every SECOND of every day. So why not master these skills?If you choose a major in the College of Humanities, you will be needed. The job market is quietly creating thousands of openings a week for people who can bring a humanist’s grace to our rapidly evolving high-tech future. Your skills will be valuable to any workplace you hope to be in. Chase after your dream major with all your energy, no matter what other people think.51. Why did the author choose Business Management as his major at first?A. He was tired of learning comparative literature.B. He came to enjoy learning marketing techniques.C. He wanted to go to the same university as his friends.D. He believed Business Management was more practical.52. According to the author, what is a possible way to gain a broad perspective?A. Making more friends and learning from them.B. Exchanging ideas in a philosophy seminar.C. Opening your mind to future possibilities.D. Getting to know the popular opinion of society.53. In which of the following situations do soft skills play the most important role?A. Writing a software program.B. Performing a heart operation.C. Playing a musical instrument.D. Negotiating with a business partner.54. The author believes the job market for Humanities graduates is ______.A. recoveringB. depressingC. promisingD. challenging55. What could be the best title for this passage?A. Stick to Your Desired MajorB. Broaden Your PerspectiveC. How to Acquire Soft SkillsD. Humanities vs. STEM第二卷第三部分：写作第一节：My teacher Peter Bullock proved that science wasn’t boring and made physics great fun. In every lesson he would produce one of those worksheets in purple ink that smelled fresh from the machine. Every sheet contained a poem or a joke and we looked forward to them being handed out. I’ve kept them all to show to my children. There was one poem in particular that’s stayed in my memory, which went:In early days Man measured TimeBy sunrise and sunset sublime;As well as being able to turn physics into poetry, Peter had energy and passion and conveyed the sense that science was part of the world around you and not a dull boring academic discipline. I’m sure he was sticking pretty much to the standard topics – waves and electricity and so on – but he managed to convert everything into a joke or a story and did so in a clever and imaginative way.Peter Bullock was not only an exciting teacher, but also the man who founded the debate clubin my school. He encouraged me to take part and express myself in a forceful way that I’ve been doing ever since – and getting paid for it. That made me what I am today – a commercial lawyer.Peter Bullock taught me throughout my secondary schooling and I took physics, chemistry and maths at A-level. Everyone in his class enjoyed learning. He was inspiring. He turned me on to the idea that the things you are interested in might seem dull on the surface, but you can still excite others. For example, when I tell people I’m interested in law, they may think: “How boring.” I guess being a physics teacher is much the same. Peter showed me that what matters is how you convey your own passion so that you excite other people.56. Why did the students expect to get the worksheets in Peter’s class?(no more than 10 words)57. What is the meaning of the underlined word in Para. 2?(1 word)58. What does Para. 3 mainly tell us?(no more than 15 words)59. How did Peter make his students interested in the things that seem dull?(no more than 10 words)60. What do you think is the most important quality of a good teacher? Please explain. (no more than 20 words)第二节：假设你是晨光中学的李津。

2019年高考理科数学天津卷(附参考答案和详解)

之
前
到
校
的
概率
均
为
$ +
!假
定
甲
/乙
两
位
同
学
到
校
情
况
互
不影响 #且任一同学每天到校情况相互独立 !
$!%用 8 表示甲同学上学期间的三天中 7,+# 之前到校的
天数#求随机变量 8 的分布列和数学期望-
$$%设 " 为事件&上学期间的三天中#甲同学在 7,+# 之前
*!槡$
,!槡+
-!$
.!槡"
&!已知''123"$#('123#!"#!$#)'#!"#!$#则'#(#)的大小关系
为
$! ! %
*!'#)#(
,!'#(#)
-!(#)#'
.!)#'#(
7!已知函数 *$#%'+9/:$#0%$+)##)##""#%是奇函数#将 &'*$#%的图象上所有点的横坐标伸长到原来的 $
'!!!!!
三解答题解答应写出文字说明证明过程或演算步骤
!"!$本小题满分!+分%在'+0. 中#内角 +#0#. 所对的边分别为'#(#)!已知(0)'$'#+)9/:0')'9/:.! $!%求5290 的值-

2019年全国Ⅰ卷高考理科数学试题及答案详细解析

（一）必考题：共60分。
17. 的内角A，B，C的对边分别为a，b，c，设．
（1）求A；
（2）若，求sinC．
解：（1）
即：
由正弦定理可得：
（2），由正弦定理得：
又，
整理可得：
解得：或
因为所以，故 .
（2）法二：，由正弦定理得：
又，
整理可得：，即
或
且
考点：正弦定理、余弦定理解三角形的问题，涉及到两角和差正弦公式、同角三角函数关系的应用，解题关键是能够利用正弦定理对边角关系式进行化简，得到余弦定理的形式或角之间的关系.
解：
由知是的中点，，又是的中点，所以为中位线且，所以，因此，又根据两渐近线对称，，所以， .
考点：，双曲线及其渐近线的对称性.
三、解答题：共70分。解答应写出文字说明、证明过程或演算步骤。第17~21题为必考题，每个试题考生都必须作答。第22、23题为选考题，考生根据要求作答。
A. B.
C. D.
解：由，得是奇函数，其图象关于原点对称．又．故选D．
考点：本题考查函数的性质与图象，利用函数奇偶性和特殊点即可解决这类问题．
6.我国古代典籍《周易》用“卦”描述万物的变化．每一“重卦”由从下到上排列的6个爻组成，爻分为阳爻“——”和阴爻“——”，如图就是一重卦．在所有重卦中随机取一重卦，则该重卦恰有3个阳爻的概率是
2019年普通高等学校招生全国统一考试
理科数学
一、选择题：本题共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.已知集合，则 =
A. B. C. D.
解：，．故选C．
考点：一元二次不等式解法，集合的交集.

【数学】2019年高考真题——天津卷(理)(精校版)

2019年普通高等学校招生全国统一考试（天津卷）理科数学第Ⅰ卷一、选择题1．设集合A ＝{－1,1,2,3,5}，B ＝{2,3,4}，C ＝{x ∈R |1≤x ＜3}，则(A ∩C )∪B 等于( ) A ．{2} B ．{2,3} C ．{－1,2,3} D ．{1,2,3,4}答案 D解析由条件可得A ∩C ＝{1,2}，故(A ∩C )∪B ＝{1,2,3,4}．2．设变量x ，y 满足约束条件＋－，－＋，－，－，则目标函数z ＝－4x ＋y 的最大值为( )A ．2B ．3C ．5D ．6 答案 C解析画出可行域如图中阴影部分(含边界)所示，作出直线－4x ＋y ＝0，并平移，可知当直线过点A 时，z 取得最大值．由＝－，－＋＝，可得＝－，＝，所以点A 的坐标为(－1,1)，故z max ＝－4×(－1)＋1＝5.3．设x ∈R ，则“x 2－5x ＜0”是“|x －1|＜1”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答案 B解析由x 2－5x ＜0可得0＜x ＜5.由|x －1|＜1可得0＜x ＜2.由于区间(0,2)是(0,5)的真子集，故“x 2－5x ＜0”是“|x －1|＜1”的必要不充分条件．4．阅读如图所示的程序框图，运行相应的程序，输出S 的值为()A．5 B．8 C．24 D．29答案 B解析执行程序框图，S＝1，i＝2，j＝1，S＝1＋4＝5，i＝3，S＝8，i＝4，满足i≥4，输出的S＝8.5．已知抛物线y2＝4x的焦点为F，准线为l.若l与双曲线－＝1(a＞0，b＞0)的两条渐近线分别交于点A和点B，且|AB|＝4|OF|(O为原点)，则双曲线的离心率为()A. B.C．2 D.答案 D解析由题意，可得F(1,0)，直线l的方程为x＝－1，双曲线的渐近线方程为y＝±x.将x ＝－1代入y＝±x，得y＝±，所以点A，B的纵坐标的绝对值均为.由|AB|＝4|OF|可得＝4，即b＝2a，b2＝4a2，故双曲线的离心率e＝＝＝.6．已知a＝log52，b＝log0.50.2，c＝0.50.2，则a，b，c的大小关系为()A．a＜c＜b B．a＜b＜cC．b＜c＜a D．c＜a＜b答案 A解析a＝log52＜log5＝.而c＝0.50.2＞0.51＝，故a＜c；b＝log0.50.2＞log0.50.25＝2，而c ＝0.50.2＜0.50＝1，故c＜b.所以a＜c＜b.7．已知函数f(x)＝A sin(ωx＋φ)(A＞0，ω＞0，|φ|＜π)是奇函数，将y＝f(x)的图象上所有点的横坐标伸长到原来的2倍(纵坐标不变)，所得图象对应的函数为g(x)．若g(x)的最小正周期为2π，且g＝，则f 等于()A．－2 B．－ C.D．2答案 C解析由f(x)为奇函数可得φ＝kπ(k∈Z)，又|φ|＜π，所以φ＝0，所以g(x)＝A sin .由g(x)的最小正周期为2π，可得＝2π，故ω＝2，g(x)＝A sin x，g＝A sin ＝，所以A＝2，所以f(x)＝2sin 2x，故f＝2sin ＝.8．已知a∈R.设函数f(x)＝－＋，，－，＞若关于x的不等式f(x)≥0在R上恒成立，则a的取值范围为()A．[0,1]B．[0,2]C．[0，e]D．[1，e]答案 C解析方法一当a＝0时，不等式f(x)≥0恒成立，排除D；当a＝e时，f(x)＝－＋，，－，＞，当x≤1时，f(x)＝x2－2e x＋2e的最小值为f(1)＝1＞0，满足f(x)≥0；当x＞1时，由f(x)＝x－eln x可得f′(x)＝1－＝，易得f(x)在x＝e处取得极小值(也是最小值)f(e)＝0，满足f(x)≥0恒成立，排除A，B.故选C.方法二若x≤1，f(x)＝x2－2ax＋2a＝(x－a)2－a2＋2a.当a≤1时，可得f(x)的最小值为f(a)＝－a2＋2a，令f(a)≥0，解得0≤a≤2，故0≤a≤1；当a＞1时，可得f(x)的最小值为f(1)＝1≥0，满足条件．所以a≥0.若x＞1，由f(x)＝x－a ln x可得f′(x)＝1－＝，当a≤1时，f′(x)＞0，则f(x)在(1，＋∞)上单调递增，故只需1－a ln 1≥0，显然成立；当a＞1时，由f′(x)＝0可得x＝a，易得f(x)的最小值为f(a)＝a－a ln a，令f(a)≥0，解得a≤e，故1＜a≤e，所以a≤e.综上，a的取值范围是[0，e]．第Ⅱ卷二、填空题9．i是虚数单位，则的值为________．答案解析方法一＝＝＝2－3i，故＝＝.方法二＝＝＝＝.10.8的展开式中的常数项为________．答案28解析二项展开式的通项T r＋1＝(2x)8－r r＝r·28－r x8－4r，令8－4r＝0可得r＝2，故常数项为2×26×＝28.11．已知四棱锥的底面是边长为的正方形，侧棱长均为，若圆柱的一个底面的圆周经过四棱锥四条侧棱的中点，另一个底面的圆心为四棱锥底面的中心，则该圆柱的体积为________．答案解析由题意可得，四棱锥底面对角线的长为2，则圆柱底面的半径为，易知四棱锥的高为＝2，故圆柱的高为1，所以圆柱的体积为π×2×1＝.12．设a∈R，直线ax－y＋2＝0和圆＝＋，＝＋(θ为参数)相切，则a的值为________．答案解析由已知条件可得圆的直角坐标方程为(x－2)2＋(y－1)2＝4，其圆心为(2,1)，半径为2，由直线和圆相切可得＝2，解得a＝.13．设x＞0，y＞0，x＋2y＝5，则的最小值为________．答案4解析＝＝＝2＋.由x＋2y＝5得5≥2，即≤，即xy≤，当且仅当x＝2y＝时等号成立．所以2＋≥2＝4，当且仅当2＝，即xy＝3时取等号，结合xy≤可知，xy可以取到3，故的最小值为4.14．在四边形ABCD中，AD∥BC，AB＝2，AD＝5，∠A＝30°，点E在线段CB的延长线上，且AE＝BE，则·＝________.答案－1解析方法一在等腰△ABE中，易得∠BAE＝∠ABE＝30°，故BE＝2，则·＝(－)·(＋)＝·＋·－2－·＝5×2×cos 30°＋5×2×cos 180°－12－2×2×cos 150°＝15－10－12＋6＝－1.方法二在△ABD中，由余弦定理可得BD＝＝，所以cos∠ABD＝＝－，则sin ∠ABD＝.设与的夹角为θ，则cos θ＝cos(180°－∠ABD＋30°)＝－cos(∠ABD－30°)＝－cos∠ABD·cos 30°－sin∠ABD·sin 30°＝－，在△ABE中，易得AE＝BE＝2，故·＝×2×＝－1.三、解答题15．在△ABC中，内角A，B，C所对的边分别为a，b，c，已知b＋c＝2a,3c sin B＝4a sin C.(1)求cos B的值；(2)求sin的值．解(1)在△ABC中，由正弦定理＝，得b sin C＝c sin B，又由3c sin B＝4a sin C，得3b sin C＝4a sin C，又sin C≠0，所以3b＝4a.又因为b＋c＝2a，所以b＝a，c＝a，由余弦定理可得cos B＝＝＝－.(2)由(1)可得sin B＝＝，从而sin 2B＝2sin B cos B＝－，cos 2B＝cos2B－sin2B＝－，故sin＝sin 2B cos ＋cos 2B sin ＝－×－×＝－.16．设甲、乙两位同学上学期间，每天7：30之前到校的概率均为，假定甲、乙两位同学到校情况互不影响，且任一同学每天到校情况相互独立．(1)用X表示甲同学上学期间的三天中7：30之前到校的天数，求随机变量X的分布列和数学期望；(2)设M为事件“上学期间的三天中，甲同学在7：30之前到校的天数比乙同学在7：30之前到校的天数恰好多2”，求事件M发生的概率．解(1)因为甲同学上学期间的三天中到校情况相互独立，且每天7：30之前到校的概率均为，故X～B，从而P(X＝k)＝k3－k，k＝0,1,2,3.所以，随机变量X的分布列为随机变量X的数学期望E(X)＝3×＝2.(2)设乙同学上学期间的三天中7：30之前到校的天数为Y，则Y～B，且M＝{X＝3，Y ＝1}∪{X＝2，Y＝0}．由题意知事件{X＝3，Y＝1}与{X＝2，Y＝0}互斥，且事件{X＝3}与{Y＝1}，事件{X＝2}与{Y＝0}均相互独立，从而由(1)知P(M)＝P({X＝3，Y＝1}∪{X＝2，Y＝0})＝P({X＝3，Y＝1})＋P({X＝2，Y＝0})＝P({X＝3})P({Y＝1})＋P({X＝2})P({Y＝0})＝×＋×＝.17．如图，AE⊥平面ABCD，CF∥AE，AD∥BC，AD⊥AB，AB＝AD＝1，AE＝BC＝2.(1)求证：BF∥平面ADE；(2)求直线CE与平面BDE所成角的正弦值；(3)若二面角E－BD－F的余弦值为，求线段CF的长．(1)证明以A为坐标原点，分别以，，的方向为x轴，y轴，z轴正方向建立空间直角坐标系(如图所示)，可得A(0,0,0)，B(1,0,0)，C(1,2,0)，D(0,1,0)，E(0,0,2)．设CF＝h(h ＞0)，则F(1,2，h)．由题意知，＝(1,0,0)是平面ADE的法向量，又＝(0,2，h)，所以·＝0，又因为直线BF⊄平面ADE，所以BF∥平面ADE.(2)解由(1)知，＝(－1,1,0)，＝(－1,0,2)，＝(－1，－2,2)．设n＝(x，y，z)为平面BDE的法向量，则即－＋＝，－＋＝，不妨令z＝1，可得n＝(2,2,1)．所以cos〈，n〉＝＝－.所以直线CE与平面BDE所成角的正弦值为.(3)解设m＝(x′，y′，z′)为平面BDF的法向量，＝(－1,1,0)，＝(0,2，h)．则即－＋＝，＋＝，不妨令y′＝1，可得m＝.由题意，有|cos〈m，n〉|＝＝＝，解得h＝，经检验，符合题意．所以线段CF的长为.18．设椭圆＋＝1(a＞b＞0)的左焦点为F，上顶点为B.已知椭圆的短轴长为4，离心率为.(1)求椭圆的方程；(2)设点P在椭圆上，且异于椭圆的上、下顶点，点M为直线PB与x轴的交点，点N在y 轴的负半轴上．若|ON|＝|OF|(O为原点)，且OP⊥MN，求直线PB的斜率．解(1)设椭圆的半焦距为c，依题意，2b＝4，＝，又a2＝b2＋c2，可得a＝，b＝2，c ＝1.所以椭圆的方程为＋＝1.(2)由题意，设P(x P，y P)(x P≠0)，M(x M,0)，直线PB的斜率为k(k≠0)，又B(0,2)，则直线PB 的方程为y＝kx＋2，与椭圆方程联立得整理得(4＋5k2)x2＋20kx＝0，可得x P＝－，代入y＝kx＋2得y P＝.所以直线OP的斜率为＝.在y＝kx＋2中，令y＝0，得x M＝－.由题意得N(0，－1)，所以直线MN的斜率为－.由OP⊥MN，得·＝－1，化简得k2＝，从解得k＝±.所以直线PB的斜率为或－.19．设{a n}是等差数列，{b n}是等比数列．已知a1＝4，b1＝6，b2＝2a2－2，b3＝2a3＋4.(1)求{a n}和{b n}的通项公式；(2)设数列{c n}满足c1＝1，c n＝其中k∈N*.(ⅰ)求数列{a2n(c2n－1)}的通项公式；(ⅱ)求(n∈N*)．解(1)设等差数列{a n}的公差为d，等比数列{b n}的公比为q.依题意得＝＋，＝＋，解得＝，＝，所以a n＝a1＋(n－1)d＝4＋(n－1)×3＝3n＋1，b n＝b1·q n－1＝6×2n－1＝3×2n.所以{a n }的通项公式为a n ＝3n ＋1，{b n }的通项公式为b n ＝3×2n . (2)(ⅰ)a 2n (c 2n －1)＝a 2n (b n －1)＝(3×2n ＋1)(3×2n －1)＝9×4n －1. 所以数列{a 2n (c 2n －1)}的通项公式为a 2n (c 2n －1)＝9×4n －1. (ⅱ)a i c i ＝[a i ＋a i (c i －1)] ＝a i ＋a 2i (c 2i －1)＝＋ (9×4i －1)＝(3×22n －1＋5×2n －1)＋9×－n＝27×22n －1＋5×2n －1－n －12(n ∈N *)． 20．设函数f (x )＝e x cos x ，g (x )为f (x )的导函数． (1)求f (x )的单调区间；(2)当x ∈时，证明：f (x )＋g (x )≥0；(3)设x n 为函数u (x )＝f (x )－1在区间内的零点，其中n ∈N ，证明：2n π＋－x n ＜.(1)解由已知f (x )的定义域为R ，f ′(x )＝e x (cos x －sin x )．当x ∈(k ∈Z )时，有sin x ＞cos x ，则f ′(x )＜0，f (x )单调递减；当x ∈(k ∈Z )时，有sin x ＜cos x ，则f ′(x )＞0，f (x )单调递增．所以f (x )的单调递增区间为(k ∈Z )，f (x )的单调递减区间为(k ∈Z )．(2)证明记h (x )＝f (x )＋g (x )，由题意及(1)得，g (x )＝e x (cos x －sin x )，所以g ′(x )＝－2e x sin x ．当x ∈时，g ′(x )＜0，故h ′(x )＝f ′(x )＋g ′(x )·＋g (x )·(－1)＝g ′(x )＜0.所以h (x )在区间上单调递减，所以h (x )≥h ＝f＝0.所以当x ∈⎣⎡⎦⎤π4，π2时，f (x )＋g (x )⎝⎛⎭⎫π2－x ≥0. (3)证明由题意得，u (x n )＝f (x n )－1＝0，即 cos x n ＝1.记y n ＝x n －2n π，则y n ∈，且f (y n )＝ cos y n ＝－2n πcos(x n －2n π)＝e －2n π(n ∈N )．由f (y n )＝e－2n π≤1＝f (y 0)及(1)，得y n ≥y 0.由(2)知，当x∈时，g′(x)＜0，所以g(x)在上为减函数，所以g(y n)≤g(y0)＜g＝0.又由(2)知，f(y n)＋g(y n)≥0，故－y n≤－＝－≤－＝＜. 所以2nπ＋－x n＜.。

2019年天津高考生物试卷及答案

2019年普通高等学校招生全国统一考试（天津卷）理科综合生物部分第I卷1.用3H标记胸腺嘧啶后合成脱氧核苷酸，注入真核细胞，可用于研究A. DNA复制的场所B. mRNA与核糖体的结合C. 分泌蛋白的运输D. 细胞膜脂质的流动【答案】A【解析】【分析】本题主要考查DNA的复制、蛋白质的合成及运输及细胞膜的功能等有关知识。

DNA复制需要脱氧核苷酸作为原料；mRNA与核糖体结合，翻译形成蛋白质；分泌蛋白的运输需要内质网和高尔基体形成囊泡运输；细胞膜脂质的流动与物质跨膜运输有关。

【详解】DNA复制需要DNA模板、原料脱氧核苷酸、能量ATP和DNA聚合酶，A正确；mRNA与核糖体的结合，开始翻译mRNA上的密码子，需要tRNA运输氨基酸，不需要脱氧核苷酸，B错误；分泌蛋白的需要内质网的加工，形成囊泡运到高尔基体，加工、分类和包装，形成分泌小泡，运到细胞膜，胞吐出去，与脱氧核苷酸无关，C错误；细胞膜脂质的流动与物质跨膜运输有关，无需脱氧核苷酸，D错误。

因此，本题答案选A。

【点睛】解答本题关键要熟悉细胞中不同的生理活动的具体过程，来判断是否需要脱氧核苷酸作为原料。

2.下列过程需ATP水解提供能量的是A. 唾液淀粉酶水解淀粉B. 生长素的极性运输C. 光反应阶段中水在光下分解D. 乳酸菌无氧呼吸的第二阶段【答案】B【解析】【分析】本题主要考查细胞中生理活动的相关知识。

细胞中蛋白质、淀粉和多糖在酶的作用下水解，无需消耗能量；生长素在幼嫩组织中从形态学上端向形态学下端运输属于极性运输，其方式为主动运输，需要消耗能量；光合作用的光反应阶段，水在光下分解，不需要消耗ATP的能量；无氧呼吸的第二阶段是丙酮酸分解成酒精和CO2或乳酸，都不消耗能量。

【详解】唾液淀粉酶水解淀粉，形成麦芽糖，不消耗能量，A错误；.生长素的极性运输是以主动运输的方式，在幼嫩组织中从形态学上端运到形态学下端，需要ATP提供能量，B正确；光反应阶段中水在光下分解，需要光能，不需要ATP功能，C错误；乳酸菌无氧呼吸的第二阶段是丙酮酸变成乳酸，不需要ATP供能，D错误。

【2019高考真题天津卷】2019年天津卷理数高考试题(有答案)

绝密★启用前2019年普通高等学校招生全国统一考试（天津卷）数学（理工类）本试卷分为第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，共150分，考试用时120分钟。

第Ⅰ卷1至2页，第Ⅱ卷3至5页。

答卷前，考生务必将自己的姓名、准考号填写在答题卡上，并在规定位置粘贴考试用条形码。

答卷时，考生务必将答案涂写在答题卡上，答在试卷上的无效。

考试结束后，将本试卷和答题卡一并交回。

祝各位考生考试顺利！第Ⅰ卷注意事项：1．每小题选出答案后，用铅笔将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。

2．本卷共8小题，每小题5分，共40分。

参考公式：·如果事件A 、B 互斥，那么()()()P AB P A P B =+．·如果事件A 、B 相互独立，那么()()()P AB P A P B =．·圆柱的体积公式V Sh =，其中S 表示圆柱的底面面积，h 表示圆柱的高． ·棱锥的体积公式13V Sh =，其中S 表示棱锥的底面面积，h 表示棱锥的高．一、选择题：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．设集合{1,1,2,3,5},{2,3,4},{|13}A B C x x =-==∈≤<R ，则()A CB =A ．{}2B ．{}2,3C ．{}1,2,3-D ．{}1,2,3,42．设变量,x y 满足约束条件20,20,1,1,x y x y x y +-≤⎧⎪-+≥⎪⎨≥-⎪⎪≥-⎩则目标函数4z x y =-+的最大值为A ．2B ．3C ．5D ．63．设x ∈R ，则“250x x -<”是“|1|1x -<”的A ．充分而不必要条件B ．必要而不充分条件C ．充要条件D ．既不充分也不必要条件4．阅读下边的程序框图，运行相应的程序，输出S 的值为A ．5B ．8C ．24D ．295．已知抛物线24y x =的焦点为F ，准线为l ，若l 与双曲线22221(0,0)x y a b a b-=>>的两条渐近线分别交于点A 和点B ，且||4||AB OF =（O 为原点），则双曲线的离心率为A BC ．2D 6．已知5log 2a =，0.5og 2.l 0b =，0.20.5c =，则,,a b c 的大小关系为 A ．a c b <<B ．a b c <<C ．b c a <<D ．c a b <<7．已知函数()sin()(0,0,||)f x A x A ωϕωϕ=+>><π是奇函数，将()y f x =的图象上所有点的横坐标伸长到原来的2倍（纵坐标不变），所得图象对应的函数为()g x ．若()g x 的最小正周期为2π，且4g π⎛⎫= ⎪⎝⎭38f π⎛⎫= ⎪⎝⎭A ．2-B ．C D ．28．已知a ∈R ，设函数222,1,()ln ,1.x ax a x f x x a x x ⎧-+≤=⎨->⎩若关于x 的不等式()0f x ≥在R 上恒成立，则a 的取值范围为 A ．[]0,1B ．[]0,2C ．[]0,e D ．[]1,e第Ⅱ卷注意事项：1．用黑色墨水的钢笔或签字笔将答案写在答题卡上。

2019年高考理综天津卷(附答案与解析)

绝密★启用前天津市2019年高等学校招生全国统一考试理科综合理科综合共300分,考试用时150分钟.物理部分物理试卷分为第I 卷（选择题）和第Ⅱ卷两部分，共120分。

第Ⅰ卷一、单项选择题（每小题6分，共30分。

每小题给出的四个选项中，只有一个选项是正确的）1.2018年12月8日，肩负着亿万中华儿女探月飞天梦想的嫦娥四号探测器成功发射，“实现人类航天器首次在月球背面巡视探测，率先在月背刻上了中国足迹”。

已知月球的质量为M 、半径为R ，探测器的质量为m ，引力常量为G ，嫦娥四号探测器围绕月球做半径为r 的匀速圆周运动时，探测器的（）A .周期为234r GMB .动能为2GMmRC .角速度为3GmrD .向心加速度为2GMR2.2018年10月23日，港珠澳跨海大桥正式通车。

为保持以往船行习惯，在航道处建造了单面索（所有钢索均处在同一竖直面内）斜拉桥，其索塔与钢索如图所示。

下列说法正确的是（）A .增加钢索的数量可减小索塔受到的向下的压力B .为了减小钢索承受的拉力，可以适当降低索塔的高度C .索塔两侧钢索对称且拉力大小相同时，钢索对索塔的合力竖直向下D .为了使索塔受到钢索的合力竖直向下，索塔两侧的钢索必须对称分布3.如图所示，在水平向右的匀强电场中，质量为m 的带电小球，以初速度v 从M 点竖直向上运动，通过N 点时，速度大小为2v ，方向与电场方向相反，则小球从M 运动到N 的过程（）A .动能增加212mv B .机械能增加22mv C .重力势能增加232mvD .电势能增加22mv4.笔记本电脑机身和显示屏对应部位分别有磁体和霍尔元件。

当显示屏开启时磁体远离霍尔元件，电脑正常工作；当显示屏闭合时磁体靠近霍尔元件，屏幕熄灭，电脑进入休眠状态。

如图所示，一块宽为a 、长为c 的矩形半导体霍尔元件，元件内的导电粒子是电荷量为e 的自由电子，通入方向向右的电流时，电子的定向移动速度为0v 当显示屏闭合时元件处于垂直于上表面、方向向下的匀强磁场中，于是元件的前、后表面间出现电压U ，以此控制屏幕的熄灭。