高考试题(天津卷)
2020年高考真题：数学(天津卷)【含答案及解析】
2020年普通高等学校招生全国统一考试（天津卷）(数学)第I 卷参考公式：如果事件A 与事件B 互斥，那么()()()È=+P A B P A P B ．如果事件A 与事件B 相互独立，那么()()()P AB P A P B =．球的表面积公式24S R p =，其中R 表示球的半径．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.设全集{3,2,1,0,1,2,3}U =---，集合{1,0,1,2},{3,0,2,3}A B =-=-，则()U A B =I ð（）A.{3,3}- B.{0,2}C.{1,1}-D.{3,2,1,1,3}---2.设a ÎR ，则“1a >”是“2a a >”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3.函数241xy x =+的图象大致为（）A.B.C. D.4.从一批零件中抽取80个，测量其直径（单位：mm ），将所得数据分为9组：[5.31,5.33),[5.33,5.35),,[5.45,5.47],[5.47,5.49]L ，并整理得到如下频率分布直方图，则在被抽取的零件中，直径落在区间[5.43,5.47)内的个数为（）A. 10B. 18C. 20D. 365.若棱长为）A.12pB.24pC.36pD.144p6.设0.80.70.713,,log 0.83a b c -æö===ç÷èø，则,,a b c 的大小关系为（）A.a b c<< B.b a c<< C.b c a<< D.c a b<<7.设双曲线C 的方程为22221(0,0)x y a b a b-=>>，过抛物线24y x =的焦点和点(0,)b 的直线为l ．若C 的一条渐近线与l 平行，另一条渐近线与l 垂直，则双曲线C 的方程为（）A.22144x y -= B.2214y x -= C.2214x y -= D.221x y -=8.已知函数()sin 3f x x p æö=+ç÷èø．给出下列结论：①()f x 的最小正周期为2p ；②2f p æöç÷èø是()f x 的最大值；③把函数sin y x =的图象上所有点向左平移3p个单位长度，可得到函数()y f x =的图象．其中所有正确结论的序号是A.① B.①③C.②③D.①②③9.已知函数3,0,(),0.x x f x x x ì=í-<î…若函数2()()2()g x f x kx xk =--ÎR 恰有4个零点，则k 的取值范围是（）A.1,)2æö-¥-+¥ç÷èøUB.1,(0,2æö-¥-ç÷èøUC.(,0)(0,-¥UD.(,0))-¥+¥U2020年普通高等学校招生全国统一考试（天津卷）(数学)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．2．本卷共11小题，共105分．二、填空题：本大题共6小题，每小题5分，共30分．试题中包含两个空的，答对1个的给3分，全部答对的给5分．10.i 是虚数单位，复数82ii-=+_________．11.在522x x æö+ç÷èø的展开式中，2x 的系数是_________．12.已知直线80x +=和圆222(0)x y r r +=>相交于,A B 两点．若||6AB =，则r 的值为_________．13.已知甲、乙两球落入盒子的概率分别为12和13．假定两球是否落入盒子互不影响，则甲、乙两球都落入盒子的概率为_________；甲、乙两球至少有一个落入盒子的概率为_________．14.已知0,0a b >>，且1ab =，则11822a b a b+++的最小值为_________．15.如图，在四边形ABCD 中，60,3B AB °Ð==，6BC =，且3,2AD BC AD AB l =×=-u u u r u u u r u u u r u u u r ，则实数l 的值为_________，若,M N 是线段BC 上的动点，且||1MN =u uu u r ，则DM DN ×u u u u r u u u r的最小值为_________．三、解答题：本大题共5小题，共75分．解答应写出文字说明，证明过程或演算步骤．16.在ABC V 中，角,,A B C 所对的边分别为,,a b c ．已知5,a b c ===．（Ⅰ）求角C 的大小；（Ⅱ）求sin A 的值；（Ⅲ）求sin 24A p æö+ç÷èø的值．17.如图，在三棱柱111ABC A B C -中，1CC ^平面,,2ABC AC BC AC BC ^==，13CC =，点,D E 分别在棱1AA 和棱1CC 上，且12,AD CE M ==为棱11A B 的中点．（Ⅰ）求证：11C M B D ^；（Ⅱ）求二面角1B B E D --的正弦值；（Ⅲ）求直线AB 与平面1DB E 所成角的正弦值．18.已知椭圆22221(0)x y a b a b+=>>的一个顶点为(0,3)A -，右焦点为F ，且||||OA OF =，其中O 为原点．（Ⅰ）求椭圆的方程；（Ⅱ）已知点C 满足3OC OF =u u u r u u u r，点B 在椭圆上（B 异于椭圆的顶点），直线AB 与以C 为圆心的圆相切于点P ，且P 为线段AB 的中点．求直线AB 的方程．19.已知{}n a 为等差数列，{}n b 为等比数列，()()115435431,5,4a b a a a b b b ===-=-．（Ⅰ）求{}n a 和{}n b 的通项公式；（Ⅱ）记{}n a 的前n 项和为n S ，求证：()2*21n n n S S S n ++<ÎN；（Ⅲ）对任意的正整数n ，设()21132,,,.n nn n n n n a b n a a c a n b +-+ì-ïï=íïïî为奇数为偶数求数列{}n c 的前2n 项和．20.已知函数3()ln ()f x x k x k R =+Î，()f x ¢为()f x 的导函数．（Ⅰ）当6k =时，（i ）求曲线()y f x =在点(1,(1))f 处的切线方程；（ii ）求函数9()()()g x f x f x x¢=-+的单调区间和极值；（Ⅱ）当3k -…时，求证：对任意的12,[1,)x x Î+¥，且12x x >，有()()()()1212122f x f x f x f x x x ¢¢+->-．答案及解析第I 卷一、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.设全集{3,2,1,0,1,2,3}U =---，集合{1,0,1,2},{3,0,2,3}A B =-=-，则()U A B =I ð（）A.{3,3}- B.{0,2}C.{1,1}-D.{3,2,1,1,3}---【答案】C 【解析】【分析】首先进行补集运算，然后进行交集运算即可求得集合的运算结果.【详解】由题意结合补集的定义可知：{}U 2,1,1B =--ð，则(){}U 1,1A B =-I ð.故选：C.【点睛】本题主要考查补集运算，交集运算，属于基础题.2.设a ÎR ，则“1a >”是“2a a >”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】A 【解析】【分析】首先求解二次不等式，然后结合不等式的解集即可确定充分性和必要性是否成立即可.【详解】求解二次不等式2a a >可得：1a >或0a <，据此可知：1a >是2a a >的充分不必要条件.故选：A.【点睛】本题主要考查二次不等式的解法，充分性和必要性的判定，属于基础题.3.函数241xy x =+的图象大致为（）A.B.C. D.【答案】A 【解析】【分析】由题意首先确定函数的奇偶性，然后考查函数在特殊点的函数值排除错误选项即可确定函数的图象.【详解】由函数的解析式可得：()()241xf x f x x --==-+，则函数()f x 为奇函数，其图象关于坐标原点对称，选项CD 错误；当1x =时，42011y ==>+，选项B 错误.故选：A.【点睛】函数图象的识辨可从以下方面入手：(1)从函数的定义域，判断图象的左右位置；从函数的值域，判断图象的上下位置．(2)从函数的单调性，判断图象的变化趋势．(3)从函数的奇偶性，判断图象的对称性．(4)从函数的特征点，排除不合要求的图象．利用上述方法排除、筛选选项．4.从一批零件中抽取80个，测量其直径（单位：mm ），将所得数据分为9组：[5.31,5.33),[5.33,5.35),,[5.45,5.47],[5.47,5.49]L ，并整理得到如下频率分布直方图，则在被抽取的零件中，直径落在区间[5.43,5.47)内的个数为（）A. 10B. 18C. 20D. 36【答案】B 【解析】【分析】根据直方图确定直径落在区间[)5.43,5.47之间的零件频率，然后结合样本总数计算其个数即可.【详解】根据直方图，直径落在区间[)5.43,5.47之间的零件频率为：()6.25 5.000.020.225+´=，则区间[)5.43,5.47内零件的个数为：800.22518´=.故选：B.【点睛】本题主要考查频率分布直方图的计算与实际应用，属于中等题.5.若棱长为）A.12pB.24pC.36pD.144p【答案】C 【解析】【分析】求出正方体的体对角线的一半，即为球的半径，利用球的表面积公式，即可得解.【详解】这个球是正方体的外接球，其半径等于正方体的体对角线的一半，即3R ==，所以，这个球的表面积为2244336S R p p p ==´=.故选：C.【点睛】本题考查正方体的外接球的表面积的求法，求出外接球的半径是本题的解题关键，属于基础题.求多面体的外接球的面积和体积问题，常用方法有：（1）三条棱两两互相垂直时，可恢复为长方体，利用长方体的体对角线为外接球的直径，求出球的半径；（2）直棱柱的外接球可利用棱柱的上下底面平行，借助球的对称性，球心为上下底面外接圆的圆心连线的中点，再根据勾股定理求球的半径；（3）如果设计几何体有两个面相交，可过两个面的外心分别作两个面的垂线，垂线的交点为几何体的球心.6.设0.80.70.713,,log 0.83a b c -æö===ç÷èø，则,,a b c 的大小关系为（）A.a b c <<B.b a c<< C.b c a<< D.c a b<<【答案】D 【解析】【分析】利用指数函数与对数函数的性质，即可得出,,a b c 的大小关系.【详解】因为0.731a =>，0.80.80.71333b a -æö==>=ç÷èø，0.70.7log 0.8log 0.71c =<=，所以1c a b <<<.故选：D.【点睛】本题考查的是有关指数幂和对数值的比较大小问题，在解题的过程中，注意应用指数函数和对数函数的单调性，确定其对应值的范围.比较指对幂形式的数的大小关系，常用方法：（1）利用指数函数的单调性：x y a =，当1a >时，函数递增；当01a <<时，函数递减；（2）利用对数函数的单调性：log a y x =，当1a >时，函数递增；当01a <<时，函数递减；（3）借助于中间值，例如：0或1等.7.设双曲线C 的方程为22221(0,0)x y a b a b-=>>，过抛物线24y x =的焦点和点(0,)b 的直线为l ．若C 的一条渐近线与l 平行，另一条渐近线与l 垂直，则双曲线C 的方程为（）A.22144x y -= B.2214y x -= C.2214x y -= D.221x y -=【答案】D 【解析】【分析】由抛物线的焦点()1,0可求得直线l 的方程为1yx b+=，即得直线的斜率为b -，再根据双曲线的渐近线的方程为b y x a =±，可得b b a -=-，1bb a-´=-即可求出,a b ，得到双曲线的方程．【详解】由题可知，抛物线的焦点为()1,0，所以直线l 的方程为1yx b+=，即直线的斜率为b -，又双曲线的渐近线的方程为b y x a =±，所以b b a -=-，1bb a-´=-，因为0,0a b >>，解得1,1a b ==．故选：D ．【点睛】本题主要考查抛物线的简单几何性质，双曲线的几何性质，以及直线与直线的位置关系的应用，属于基础题．8.已知函数()sin 3f x x p æö=+ç÷èø．给出下列结论：①()f x 的最小正周期为2p ；②2f p æöç÷èø是()f x 的最大值；③把函数sin y x =的图象上所有点向左平移3p个单位长度，可得到函数()y f x =的图象．其中所有正确结论的序号是A.① B.①③ C.②③D.①②③【答案】B 【解析】【分析】对所给选项结合正弦型函数的性质逐一判断即可.【详解】因为()sin(3f x x p =+，所以周期22T pp w ==，故①正确；51()sin()sin 122362f p p pp =+==¹，故②不正确；将函数sin y x =的图象上所有点向左平移3p个单位长度，得到sin()3y x p =+的图象，故③正确.故选：B.【点晴】本题主要考查正弦型函数的性质及图象的平移，考查学生的数学运算能力，逻辑分析那能力，是一道容易题.9.已知函数3,0,(),0.x x f x x x ì=í-<î…若函数2()()2()g x f x kx xk =--ÎR 恰有4个零点，则k 的取值范围是（）A.1,)2æö-¥-+¥ç÷èøUB.1,(0,2æö-¥-ç÷èøUC.(,0)(0,-¥UD.(,0))-¥+¥U 【答案】D 【解析】【分析】由(0)0g =，结合已知，将问题转化为|2|y kx =-与()()||f x h x x =有3个不同交点，分0,0,0k k k =<>三种情况，数形结合讨论即可得到答案.【详解】注意到(0)0g =，所以要使()g x 恰有4个零点，只需方程()|2|||f x kx x -=恰有3个实根即可，令()h x =()||f x x ，即|2|y kx =-与()()||f x h x x =的图象有3个不同交点.因为2,0()()1,0x x f x h x x x ì>==í<î，当0k =时，此时2y =，如图1，2y =与()()||f x h x x =有2个不同交点，不满足题意；当k 0<时，如图2，此时|2|y kx =-与()()||f x h x x =恒有3个不同交点，满足题意；当0k >时，如图3，当2y kx =-与2y x =相切时，联立方程得220x kx -+=，令0D =得280k -=，解得k =，所以k >综上，k的取值范围为(,0))-¥+¥U .故选：D.【点晴】本题主要考查函数与方程的应用，考查数形结合思想，转化与化归思想，是一道中档题.2020年普通高等学校招生全国统一考试（天津卷）数学第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．2．本卷共11小题，共105分．二、填空题：本大题共6小题，每小题5分，共30分．试题中包含两个空的，答对1个的给3分，全部答对的给5分．10.i是虚数单位，复数82ii-=+_________．【答案】32i-【解析】【分析】将分子分母同乘以分母的共轭复数，然后利用运算化简可得结果.【详解】()()()()828151032 2225i ii iii i i----===-++-.故答案为：32i -.【点睛】本题考查复数的四则运算，属于基础题.11.在522x x æö+ç÷èø的展开式中，2x 的系数是_________．【答案】10【解析】【分析】写出二项展开式的通项公式，整理后令x 的指数为2，即可求出．【详解】因为522x x æö+ç÷èø的展开式的通项公式为()5531552220,1,2,3,4,5rr r rr r r T C x C x r x --+æö==××=ç÷èø，令532r -=，解得1r =．所以2x 的系数为15210C ´=．故答案为：10．【点睛】本题主要考查二项展开式的通项公式的应用，属于基础题．12.已知直线80x +=和圆222(0)x y r r +=>相交于,A B 两点．若||6AB =，则r 的值为_________．【答案】5【解析】【分析】根据圆的方程得到圆心坐标和半径，由点到直线的距离公式可求出圆心到直线的距离d ，进而利用弦长公式||AB =，即可求得r ．【详解】因为圆心()0,0到直线80x +=的距离4d ==，由||AB =可得6==5r ．故答案为：5．【点睛】本题主要考查圆的弦长问题，涉及圆的标准方程和点到直线的距离公式，属于基础题．13.已知甲、乙两球落入盒子的概率分别为12和13．假定两球是否落入盒子互不影响，则甲、乙两球都落入盒子的概率为_________；甲、乙两球至少有一个落入盒子的概率为_________．【答案】 (1).16(2).23【解析】【分析】根据相互独立事件同时发生的概率关系，即可求出两球都落入盒子的概率；同理可求两球都不落入盒子的概率，进而求出至少一球落入盒子的概率.【详解】甲、乙两球落入盒子的概率分别为11,23，且两球是否落入盒子互不影响，所以甲、乙都落入盒子的概率为111236´=，甲、乙两球都不落入盒子的概率为111(1)(1)233-´-=，所以甲、乙两球至少有一个落入盒子的概率为23.故答案为：16；23.【点睛】本题主要考查独立事件同时发生的概率，以及利用对立事件求概率，属于基础题.14.已知0,0a b >>，且1ab =，则11822a b a b+++的最小值为_________．【答案】4【解析】【分析】根据已知条件，将所求的式子化为82a b a b+++，利用基本不等式即可求解.【详解】0,0,0a b a b >>\+>Q ，1ab =,11882222ab ab a b a b a b a b\++=++++4==，当且仅当a b +=4时取等号，结合1ab =,解得22a b =-=+，或22a b =+=.故答案为：4【点睛】本题考查应用基本不等式求最值，“1”的合理变换是解题的关键，属于基础题.15.如图，在四边形ABCD 中，60,3B AB °Ð==，6BC =，且3,2AD BC AD AB l =×=-u u u r u u u r u u u r u u u r ，则实数l 的值为_________，若,M N 是线段BC 上的动点，且||1MN =u uu u r ，则DM DN ×u u u u r u u u r的最小值为_________．【答案】 (1).16 (2).132【解析】【分析】可得120BAD Ð=o ，利用平面向量数量积的定义求得l 的值，然后以点B 为坐标原点，BC 所在直线为x 轴建立平面直角坐标系，设点(),0M x ，则点()1,0N x +（其中05x ££），得出DM DN ×u u u u r u u u r关于x 的函数表达式，利用二次函数的基本性质求得DM DN ×u u u u r u u u r的最小值.【详解】AD BC l =u u u r u u u rQ ，//AD BC \，180120BAD B \Ð=-Ð=o o ，cos120AB AD BC AB BC AB l l ×=×=×ouu u r uu u r uu u r u uu r u uu r u uu r1363922l l æö=´´´-=-=-ç÷èø，解得16l =，以点B 为坐标原点，BC 所在直线为x 轴建立如下图所示的平面直角坐标系xBy，()66,0BC C =\Q ，,∵3,60AB ABC =Ð=°，∴A的坐标为3,22A æöç÷ç÷èø,∵又∵16AD BC =uu u r u u u r ,则5,22D æöç÷ç÷èø，设(),0M x ，则()1,0N x +（其中05x ££），5,22DM x æö=--ç÷èøu u u u r，3,22DN x æö=--ç÷èøu u u r，()2225321134222222DM DN x x x x x æöæöæö×=--+=-+=-+ç÷ç÷ç÷ç÷èøèøèøu u u u r u u u r ，所以，当2x =时，DM DN ×u u u u r u u u r 取得最小值132.故答案为：16；132.【点睛】本题考查平面向量数量积的计算，考查平面向量数量积的定义与坐标运算，考查计算能力，属于中等题.三、解答题：本大题共5小题，共75分．解答应写出文字说明，证明过程或演算步骤．16.在ABC V 中，角,,A B C 所对的边分别为,,a b c．已知5,a b c ===．（Ⅰ）求角C 的大小；（Ⅱ）求sin A 的值；（Ⅲ）求sin 24A p æö+ç÷èø的值．【答案】（Ⅰ）4C p =；（Ⅱ）sin 13A =；（Ⅲ）sin 2426A p æö+=ç÷èø.【解析】【分析】（Ⅰ）直接利用余弦定理运算即可；（Ⅱ）由（Ⅰ）及正弦定理即可得到答案；（Ⅲ）先计算出sin ,cos ,A A 进一步求出sin 2,cos 2A A ，再利用两角和的正弦公式计算即可.【详解】（Ⅰ）在ABC V中，由5,a b c ===及余弦定理得222cos 22a b c C ab +-===，又因为(0,)C p Î，所以4C p=；（Ⅱ）在ABC V 中，由4C p =，a c ==及正弦定理，可得sin sin a C A c ´===13；（Ⅲ）由a c <知角A为锐角，由sin 13A =，可得cos A ==13，进而2125sin 22sin cos ,cos 22cos 11313A A A A A ===-=，所以125sin(2)sin2cos cos2sin 444132132A A A p p p +=+=´+´=26.【点晴】本题主要考查正、余弦定理解三角形，以及三角恒等变换在解三角形中的应用，考查学生的数学运算能力，是一道容易题.17.如图，在三棱柱111ABC A B C -中，1CC ^平面,,2ABC AC BC AC BC ^==，13CC =，点,D E 分别在棱1AA 和棱1CC 上，且12,AD CE M ==为棱11A B 的中点．（Ⅰ）求证：11C M B D ^；（Ⅱ）求二面角1B B E D --的正弦值；（Ⅲ）求直线AB 与平面1DB E 所成角的正弦值．【答案】（Ⅰ）证明见解析；（Ⅱ）6；（Ⅲ）3.【解析】【分析】以C 为原点，分别以1,,CA CB CC u u u r u u u r u u u u r的方向为x 轴，y 轴，z 轴的正方向建立空间直角坐标系.（Ⅰ）计算出向量1C M u u u u r 和1B D u u u u r 的坐标，得出110C M B D ×=uu u u r u uu u r，即可证明出11C M B D ^；（Ⅱ）可知平面1BB E 的一个法向量为CA uu u r ，计算出平面1B ED 的一个法向量为n r，利用空间向量法计算出二面角1B B E D --的余弦值，利用同角三角函数的基本关系可求解结果；（Ⅲ）利用空间向量法可求得直线AB 与平面1DB E 所成角的正弦值.【详解】依题意，以C 为原点，分别以CA uu u r 、CB u u u r、1CC u u u u r 的方向为x 轴、y 轴、z 轴的正方向建立空间直角坐标系（如图），可得()0,0,0C 、()2,0,0A 、()0,2,0B 、()10,0,3C 、()12,0,3A 、()10,2,3B 、()2,0,1D 、()0,0,2E 、()1,1,3M .（Ⅰ）依题意，()11,1,0C M =uu uu r ，()12,2,2B D =--uu u u r，从而112200C M B D ×=-+=uu u u r u u u u r，所以11C M B D ^；（Ⅱ）依题意，()2,0,0CA =uu u r是平面1BB E 的一个法向量，()10,2,1EB =u uu r ，()2,0,1ED =-u u u r．设(),,n x y z =r为平面1DB E 的法向量，则100n EB n ED ì×=ïí×=ïîr u u u r r u u u r ，即2020y z x z +=ìí-=î，不妨设1x =，可得()1,1,2n =-r．cos ,6C CA n A C n A n ×<>===×u u u r ru u u r u r u u r r，sin ,6CA n \<>==u u u r r ．所以，二面角1B B E D --的正弦值为6；（Ⅲ）依题意，()2,2,0AB =-u u u r．由（Ⅱ）知()1,1,2n =-r 为平面1DB E的一个法向量，于是cos ,3AB n AB n AB n ×<>===-×u u u r ru u u r r u u u r r ．所以，直线AB 与平面1DB E所成角的正弦值为3.【点睛】本题考查利用空间向量法证明线线垂直，求二面角和线面角的正弦值，考查推理能力与计算能力，属于中档题.18.已知椭圆22221(0)x y a b a b+=>>的一个顶点为(0,3)A -，右焦点为F ，且||||OA OF =，其中O 为原点．（Ⅰ）求椭圆的方程；（Ⅱ）已知点C 满足3OC OF =u u u r u u u r，点B 在椭圆上（B 异于椭圆的顶点），直线AB 与以C 为圆心的圆相切于点P ，且P 为线段AB 的中点．求直线AB 的方程．【答案】（Ⅰ）221189x y +=；（Ⅱ）132y x =-，或3y x =-．【解析】【分析】（Ⅰ）根据题意，并借助222a b c =+，即可求出椭圆的方程；（Ⅱ）利用直线与圆相切，得到CP AB ^，设出直线AB 的方程，并与椭圆方程联立，求出B 点坐标，进而求出P 点坐标，再根据CP AB ^，求出直线AB 的斜率，从而得解.【详解】（Ⅰ）Q 椭圆()222210x ya b a b+=>>的一个顶点为()0,3A -，\3b =，由OA OF =，得3c b ==，又由222a b c =+，得2228313a =+=，所以，椭圆的方程为221189x y +=；（Ⅱ）Q 直线AB 与以C 为圆心的圆相切于点P ，所以CP AB ^，根据题意可知，直线AB 和直线CP 的斜率均存在，设直线AB 的斜率为k ，则直线AB 的方程为3y kx +=，即3y kx =-，2231189y kx x y =-ìïí+=ïî，消去y ，可得()2221120k x kx +-=，解得0x =或21221k x k =+.将21221k x k =+代入3y kx =-，得222126321213k y k k k k =×--=++，所以，点B 的坐标为2221263,2121k k k k æö-ç÷++èø，因为P 为线段AB 的中点，点A 的坐标为()0,3-，所以点P 的坐标为2263,2121kk k -æöç÷++èø，由3OC OF =u u u r u u u r，得点C 的坐标为()1,0，所以，直线CP 的斜率为222303216261121CPk k k k k k --+=-+-+=，又因为CP AB ^，所以231261k k k ×=--+，整理得22310k k -+=，解得12k =或1k =.所以，直线AB 的方程为132y x =-或3y x =-.【点睛】本题考查了椭圆标准方程的求解、直线与椭圆的位置关系、直线与圆的位置关系、中点坐标公式以及直线垂直关系的应用，考查学生的运算求解能力，属于中档题.当看到题目中出现直线与圆锥曲线位置关系的问题时，要想到联立直线与圆锥曲线的方程.19.已知{}n a 为等差数列，{}n b 为等比数列，()()115435431,5,4a b a a a b b b ===-=-．（Ⅰ）求{}n a 和{}n b 的通项公式；（Ⅱ）记{}n a 的前n 项和为n S ，求证：()2*21n n n S S S n ++<ÎN；（Ⅲ）对任意的正整数n ，设()21132,,,.n nn n n n n a b n a a c a n b +-+ì-ïï=íïïî为奇数为偶数求数列{}n c 的前2n 项和．【答案】（Ⅰ）n a n =，12n n b -=；（Ⅱ）证明见解析；（Ⅲ）465421949n nn n +--+´.【解析】【分析】(Ⅰ)由题意分别求得数列的公差、公比，然后利用等差、等比数列的通项公式得到结果；(Ⅱ)利用(Ⅰ)的结论首先求得数列{}n a 前n 项和，然后利用作差法证明即可；(Ⅲ)分类讨论n 为奇数和偶数时数列的通项公式，然后分别利用指数型裂项求和和错位相减求和计算211nk k c-=å和21nkk c=å的值，据此进一步计算数列{}n c 的前2n 项和即可.【详解】(Ⅰ)设等差数列{}n a 的公差为d ，等比数列{}n b 的公比为q .由11a =，()5435a a a =-，可得d =1.从而{}n a 的通项公式为n a n =.由()15431,4b b b b ==-，又q ≠0，可得2440q q -+=，解得q =2，从而{}n b 的通项公式为12n n b -=.(Ⅱ)证明：由(Ⅰ)可得(1)2n n n S +=，故21(1)(2)(3)4n n S S n n n n +=+++，()()22211124n S n n +=++，从而2211(1)(2)02n n n S S S n n ++-=-++<，所以221n n n S S S ++<.(Ⅲ)当n 为奇数时，()111232(32)222(2)2n n n n n nn n a b n c a a n n n n-+-+--===-++，当n 为偶数时，1112n n n n a n c b -+-==，对任意的正整数n ，有222221112221212121k k nnnk k k c k k n --==æö=-=-ç÷+-+èøåå，和223111211352321444444nnk k n n k k k n n c -==---==+++++ååL ①由①得22314111352321444444n k nn k n n c +=--=+++++åL ②由①②得22111211312221121441444444414n n k n n n k n n c ++=æö-ç÷--èø=+++-=---åL ，由于11211121221121156544144334444123414n n n n n n n n ++æö-ç÷--+èø--=-´--´=-´-，从而得：21565994nk nk n c =+=-´å.因此，2212111465421949n nnnk k k nk k k n c c c n -===+=+=--+´ååå.所以，数列{}n c 的前2n 项和为465421949n nn n +--+´.【点睛】本题主要考查数列通项公式的求解，分组求和法，指数型裂项求和，错位相减求和等，属于中等题.20.已知函数3()ln ()f x x k x k R =+Î，()f x ¢为()f x 的导函数．（Ⅰ）当6k =时，（i ）求曲线()y f x =在点(1,(1))f 处的切线方程；（ii ）求函数9()()()g x f x f x x¢=-+的单调区间和极值；（Ⅱ）当3k -…时，求证：对任意的12,[1,)x x Î+¥，且12x x >，有()()()()1212122f x f x f x f x x x ¢¢+->-．【答案】（Ⅰ）（i ）98y x =-；（ii ）()g x 的极小值为(1)1g =，无极大值；（Ⅱ）证明见解析.【解析】【分析】(Ⅰ) (i)首先求得导函数的解析式，然后结合导数的几何意义求解切线方程即可；(ii)首先求得()g x ¢的解析式，然后利用导函数与原函数的关系讨论函数的单调性和函数的极值即可；（Ⅱ）首先确定导函数的解析式，然后令12x t x =，将原问题转化为与t 有关的函数，然后构造新函数，利用新函数的性质即可证得题中的结论.【详解】(Ⅰ) (i)当k =6时，()36ln f x x x =+，()26'3f x x x=+.可得()11f =，()'19f =，所以曲线()y f x =在点()()1,1f 处的切线方程为()191y x -=-，即98y x =-.(ii)依题意，()()32336ln ,0,g x x x x x x=-++Î+¥.从而可得()2263'36g x x x x x=-+-，整理可得：323(1)(1)()x x g x x¢-+=，令()'0g x =，解得1x =.当x 变化时，()()',g x g x 的变化情况如下表：x ()0,11x =()1,+?()'g x -0+()g x 单调递减极小值单调递增所以，函数g (x )的单调递减区间为(0，1)，单调递增区间为(1，+∞)；g (x )的极小值为g (1)=1，无极大值.（Ⅱ）证明：由3()ln f x x k x =+，得2()3k f x x x¢=+.对任意的12,[1,)x x Î+¥，且12x x >，令12(1)x t t x =>，则()()()()()()()1212122x x f x f x f x f x ¢¢-+--()22331121212122332ln x k k x x x x x x k x x x æöæö=-+++--+ç÷ç÷èøèø3322121121212212332ln x x x x x x x x x k k x x x æö=--++--ç÷èø()332213312ln x t t t k t t t æö=-+-+--ç÷èø.①令1()2ln ,[1,)h x x x x x=--Î+¥.当x >1时，22121()110h x x x x ¢æö=+-=->ç÷èø，由此可得()h x 在[)1,+¥单调递增，所以当t >1时，()()1h t h >，即12ln 0t t t-->.因为21x ³，323331(1)0t t t t -+-=->，3k ³-，所以()()332322113312ln 33132ln x t t t k t t t t t t t ttæöæö-+-+-------ç÷+ç÷èøèø (323)36ln 1t t t t=-++-.②由(Ⅰ)(ii)可知，当1t >时，()()1g t g >，即32336ln 1t t t t-++>，故32336ln 10t t t t-++->③由①②③可得()()()()()()()12121220x x fx f x f x f x ¢¢-+-->.所以，当3k ³-时，任意的[)12,1,x x Î+¥，且12x x >，有()()()()1212122f x f x f x f x x x ¢¢+->-.【点睛】导数是研究函数的单调性、极值(最值)最有效的工具，而函数是高中数学中重要的知识点，对导数的应用的考查主要从以下几个角度进行：(1)考查导数的几何意义，往往与解析几何、微积分相联系．(2)利用导数求函数的单调区间，判断单调性；已知单调性，求参数．(3)利用导数求函数的最值(极值)，解决生活中的优化问题．(4)考查数形结合思想的应用．。
2024年天津高考语文试卷
高考语文试题一、语言文字基础1．对下列句子所用修辞手法的分析，正确的一项是（）①汗水在他那络腮胡根上聚成了一粒粒晶亮的露珠。
②军队驻扎一个月，没有动过群众的一针一线。
③他的日历上是工作，工作，工作，从来没有节假日。
2.下列各项中，作家、作品、人物的对应关系错误的一项是（）A．曹禺——《雷雨》——周冲B．海明威——《老人与海》——桑地亚哥C．狄更斯——《大卫·科波菲尔》——米考伯D．施耐庵——《水浒传》——冷子兴3.仿照示例,运用所给三组材料仿写三个句子,要求逻辑严密,语意连贯,信息完整,句式一致,并与所给示例构成一组排比句。
示例:卧冰求鲤,叨陪鲤对,敬养母聆父训,中国人向来都不缺乏孝顺意识。
第一组:诚信建功抗争第二组:诛暴秦御外侮轻生死重然诺持金戈破巨浪第三组:立木取信闻鸡起舞军民抗倭尾生抱柱破釜沉舟击楫中流A．①借代②夸张③排比 B．①比喻②夸张③排比C．①借代②比喻③反复 D．①比喻②借代③反复4.把下列句子组成语意连贯的语段，排序最恰当的一项是( )①以及乐曲与你之间的故事，如《童年》让你想起无忧无虑的童年生活。
②不管你什么年龄，无论你走到哪里，只要你重新听到那熟悉的旋律，就会触动你那颗敏感的心，引起你久久的怀念。
③这些故事包括乐曲本身的故事，如《月光曲》与贝多芬、《二泉映月》与阿炳。
④感人的乐曲留给人的记忆是长久的。
⑤想一想，哪一首最让你怀念，哪一支曲子最让你浮想联翩，由此你联想起怎样的故事。
A．②④①⑤③ B．②④⑤①③C．④②⑤③① D．④②①③⑤5.下列对材料有关内容的概述,不正确的一项是( ）A.李陵擅长骑马射箭,受命在酒泉、张掖训练军队。
李广利出击匈奴时,李陵请求以少击众，率五千步卒深入单于王庭，王夫之认为这是“自炫其勇”。
B.李陵遭遇单于三万人马,奋力作战,单于震恐,召八万人围攻李陵,李陵率军杀敌数千。
对于李陵的战功,司马迁称赞他虽败犹荣,而王夫之则未置一词。
C.李陵投降后,武帝大怒,司马迁竭力替李陵辩白。
2024年高考语文作文试题及范文(天津卷)
2024年高考语文作文试题及范文（天津卷）阅读下面的材料，根据要求写作。
（60分）在缤纷的世界中，无论是个人、群体还是国家，都会面对别人对我们的定义。
我们要认真对待“被定义”，明辨是非，去芜存真，为自己的提升助力；也要勇于通过“自定义”来塑造自我，彰显风华，用自己的方式前进。
以上材料能引发你怎样的联想与思考？请结合你的体验和感悟，写一篇文章。
要求：①自选角度，自拟标题；②文体不限（诗歌除外），文体特征明显；③不少于800字；④不得抄袭，不得套作。
范文一：定义自己的人生蒲公英选择四海为家，于是它们纷纷挣开妈妈的怀抱，随风万里安家；小雏鹰选择畅游天地，于是它们奋力练习翱翔，以期一朝乘风破浪；长尾蜉蝣选择化蝶翩飞，于是它们用尽生平一丝气力，即便朝生暮死也要在一刹那绽放……自然界的生灵们每时每刻都在用自己的智慧和勇气定义着自己独特的人生。
作为万物灵长的人类，更应如此。
面对他人的定义，我们需要保持清醒，明辨是非。
这世界上总有人喜欢给别人贴标签、下定义，或出于好心，或别有用心。
如果我们轻易地被他人的定义所左右，就可能会迷失自我，失去前进的方向。
就像那只被贴上“不会飞”标签的鹰，如果它从此认定自己真的不会飞，那它就永远无法翱翔蓝天。
然而，真正的智者会在他人的定义中保持冷静，客观分析，取其精华，去其糟粕，让那些合理的定义成为自己成长的助力，而不是束缚。
当然，我们不能仅仅满足于被动地接受他人的定义，更要勇于通过“自定义”来塑造自我。
每个人都是独一无二的，都有自己的兴趣、爱好、天赋和潜力。
我们应该根据自己的特点和优势，为自己设定明确的目标和方向，用自己的方式去追求梦想，实现人生价值。
“杂交水稻之父”袁隆平，他没有满足于别人定义的“农民”身份，而是凭借自己对农业的热爱和执着，在田间地头辛勤耕耘，不断探索创新，最终为解决全球粮食问题做出了巨大贡献，也为自己赢得了“当代神农”的赞誉。
马云，一个曾经多次被人否定的创业者，他没有因为别人说他不行就放弃，而是坚定地走自己的路，用自己独特的商业模式打造了一个庞大的商业帝国，成为无数人心目中的创业英雄。
2020年高考数学天津卷附答案解析版
（Ⅰ）当 k 6 时，
（i）求曲线 y f x 在点1，f1处的切线方程；
（ii）求函数 g x f x f x 9 的单调区间和极值；
x
（ Ⅱ ） 当 k≥ 3 时 ， 求 证 ： 对 任 意 的 x1，x21， ， 且 x1＞x2 ， 有
f x1 f x2 ＞f x1 f x2 .
r 1 .所以
x 2的 系 数 为
C51 2 10 .故答案为：10. 【考点】二项展开式的通项公式的应用
2
x1 x2
数学试卷 第 6 页（共 6 页）
2020年普通高等学校招生全国统一考试（天津卷）
数学答案解析
第Ⅰ卷
一、选择题 1. 【答案】C 【解析】首先进行补集运算，然后进行交集运算即可求得集合的运算结果.由题意结合补集的定义可知：
UB 2，1，1，则 A U B 1，1.故选：C.
【考点】补集运算，交集运算 2. 【答案】A 【解析】首先求解二次不等式，然后结合不等式的解集即可确定充分性和必要性是否成立即可.求解二次不 等式a2＞a 可得： a＞1或 a＜0，据此可知： a＞1是 a2＞a 的充分不必要条件.故选：A.
【考点】函数图象的识辨 4. 【答案】B
【解析】根据直方图确定直径落在区间5.43，5.47 之间的零件频率，然后结合样本总数计算其个数即可.根 据直方图，直径落在区间5.43，5.47 之间的零件频率为：6.25 5.00 0.02 0.225 ，则区间5.43，5.47内
零件的个数为： 80 0.225 18 .故选：B.
A.12
B. 24
C. 36
6.设
a
30.7
，
b
1 3
，0.8
(完整版)天津高考英语试题及答案
2019 年 3 月一般高等学校招生全国一致考试（天津卷）英语笔试本试卷分为第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，共 130 分，考试用时 100 分钟。
第Ⅰ卷 1 至 10 页，第Ⅱ卷 11 至 12 页。
答卷前，考生务势必自己的姓名、准考号填写在答题卡上，并在规定地点粘贴考试用条形码。
答卷时，考生务势必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。
如需变动，用橡皮擦洁净后，再选涂其余答案标号。
2.本卷共 55 小题，共 95 分。
第一节 :单项填空(共15小题;每题1分，满分15分)从 A 、B、C、D 四个选项中，选出能够填入空白处的最正确选项。
例 :Stand over there you’ll be able to see it better.D. while答案是 B 。
1.—Everybody is helping out with dinner. Would you clean the fish?—______. I ’d rather make the salad.A. A piece of cake.B. 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 beginto 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 aswell as theirpotential to cause great damage.A.To study C.Having studied7. Even though we live in a high- tech age, it ’s still impossible to predict the weather ______.8.The opinion ______ learning is a lifelong process has been expressedby educationexperts throughout the years.9.I took notes during the meeting, but I ______ a chance to writethem into a report so far.A. didn ’t have’t had’t had’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 taken11. People believe writing poems provides a ______ through which they can express their feelings.12.We are determined that our training should ______ the currentdevelopment ineducation.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 lastweek.A.need have practisedB.might practiseC.must have practisedD.could practise14.The course normally attracts about 100 students per year, ______ upto half will be from abroad.B.of which D.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：完型填空（共20 小题；每题 1.5 分，满分 30 分）阅读下边短文，掌握其粗心，而后从 16~35 各题所给的 A、B、C、D四个选项中，选出最正确选项。
高考物理真题(天津卷)(试题+答案解析)
一般高等学校招生全国统一考试（天津卷）理科综合物理部分第Ⅰ卷一、单项选择题（每题6分，共30分。
每题给出旳四个选项中，只有一种选项是对旳旳）1．质点做直线运动旳速度—时间图象如图所示，该质点（）A．在第1秒末速度方向发生了变化B．在第2秒末加速度方向发生了变化C．在前2秒内发生旳位移为零D．第3秒末和第5秒末旳位置相似2．如图所示，电路中R1、R2均为可变电阻，电源内阻不能忽视，平行板电容器C旳极板水平放置。
闭合开关S，电路到达稳定期，带电油滴悬浮在两板之间静止不动。
假如仅变化下列某一种条件，油滴仍能静止不动旳是（）A．增大R1旳阻值B．增大R2旳阻值C．增大两板间旳距离D．断开开关S3．研究表明，地球自转在逐渐变慢，3亿年前地球自转旳周期约为22小时。
假设这种趋势会持续下去，地球旳其他条件都不变，未来人类发射旳地球同步卫星与目前旳相比（）A．距地面旳高度变大B．向心加速度变大C．线速度变大D．角速度变大4．如图所示，平行金属板A、B水平正对放置，分别带等量异号电荷。
一带电微粒水平射入板间，在重力和电场力共同作用下运动，轨迹如图中虚线所示，那么（）A．若微粒带正电荷，则A板一定带正电荷B．微粒从M点运动到N点电势能一定增长C．微粒从M点运动到N点动能一定增长D．微粒从M点运动到N点机械能一定增长5．平衡位置处在坐标原点旳波源S在y轴上振动，产生频率为50Hz旳简谐横波向x轴正、负两个方向传播，波速均为100m/s。
平衡位置在x轴上旳P、Q两个质点随波源振动着，P、Q旳x轴坐标分别为x P＝3.5m、x Q＝－3 m。
当S位移为负且向－y方向运动时，P、Q两质点旳（）A．位移方向相似、速度方向相反B．位移方向相似、速度方向相似C．位移方向相反、速度方向相反D．位移方向相反、速度方向相似二、不定项选择题（每题6分，共18分。
每题给出旳四个选项中，均有多种选项是对旳旳。
所有选对旳得6分，选对但不全旳得3分，选错或不答旳得0分）6．下列说法对旳旳是（）A．玻尔对氢原子光谱旳研究导致原子旳核式构造模型旳建立B．可运用某些物质在紫外线照射下发出荧光来设计防伪措施C．天然放射现象中产生旳射线都能在电场或磁场中发生偏转D．观测者与波源互相远离时接受到波旳频率与波源频率不一样7．如图1所示，在匀强磁场中，一矩形金属线圈两次分别以不一样旳转速，绕与磁感线垂直旳轴匀速转动，产生旳交变电动势图象如图2中曲线a、b所示，则（）A．两次t＝0时刻线圈平面均与中性面重叠B．曲线a、b对应旳线圈转速之比为2∶3C．曲线a表达旳交变电动势频率为25 HzD．曲线b表达旳交变电动势有效值为10 V8．一束由两种频率不一样旳单色光构成旳复色光从空气射入玻璃三棱镜后，出射光提成a、b两束，如图所示，则a、b两束光（）A．垂直穿过同一块平板玻璃，a光所用旳时间比b光长B．从同种介质射入真空发生全反射时，a光顾界角比b光旳小C．分别通过同一双缝干涉装置，b光形成旳相邻亮条纹间距小D．若照射同一金属都能发生光电效应，b光照射时逸出旳光电子最大初动能大第Ⅱ卷注意事项：本卷共4题，共72分。
(2022年高考真题)2022年全国普通高等学校招生统一考试英语试卷 天津卷(含解析)
2022年全国普通高等学校招生统一考试天津卷英语试卷养成良好的答题习惯，是决定成败的决定性因素之一。
做题前，要认真阅读题目要求、题干和选项，并对答案内容作出合理预测;答题时，切忌跟着感觉走，最好按照题目序号来做，不会的或存在疑问的，要做好标记，要善于发现，找到题目的题眼所在，规范答题，书写工整;答题完毕时，要认真检查，查漏补缺，纠正错误。
一、单选题1、—I worked on your car the whole night. How is it running?— It is running great! _____. You were such a big help!A. It's a pityB.I couldn't agree moreC. Forget itD.I can hardly thank you enough2、Food and medical supplies _____ to all the residents after the hurricane last Sunday.A. distributeB. distributedC. are distributedD. were distributed3、_____ we achieve great success in our work, we should not be too proud.A. Ever sinceB. Even ifC. In caseD. As though4、_____gardening may be hard physical work, those who love it find it very relaxing mentally.A. AlthoughB. OnceC. SinceD. Unless5、Critical reasoning, together with problem-solving, _____ teenagers to make better decisions.A. prepareB. preparesC. is preparingD. are preparing6、_____ his restless students occupied with an indoor sport on rainy days, James Naismith created basketball.A. To be keptB. KeptC. To keepD. Keeping7、The children failed to hide their disappointment when they found out the school _____ the party.A. cancelsB. will cancelC. has cancelledD. had cancelled8、When people are depressed, some experience a loss of _____ while others can't stop eating.A. appetiteB. powerC. memoryD. sight9、The city temperatures have returned from record low to normal, _____ the citizens to enjoy the outdoors again.A. allowingB. being allowedC. having allowedD. having been allowed10、The experienced climber was _____ the potential danger in such extreme weather and decided to wait until the following day.A. completely blind toB. totally lost inC. pretty keen onD. well aware of11、Mental health involves _____ you procees things such as stress and anxiety.A. howB. whatC. whyD. which12、I'm far _____ and I'll never get this report done by Friday.A. below surfaceB. beyond controlC. behind scheduleD. above average13、If we continue to _____ environmental problems, we will regret it sooner or later.A. highlightB. identifyC. ignoreD. prevent14、—Angela just doesn't like me. She won't even say hello.—_____. Actually, she's very shy.A.I have no ideaB. Don't jump to conclusionsC. Don't mention itD. There is no doubt about it15、Guide books are prepared to suit the convenience of the traveler, _____ routes round a city or a site are often suggested.A. for whichB. with whichC. for whomD. with whom二、完形填空（20空）When I was in sixth grade, I joined the band pro-gram to learn to play the clarinet(单簧管). The beginning of the year had gone 16 . But as most students progressed, I seemed to fall behind. One day, when my teacher told us to play in front of the other students, I was filled with fear. I knew I would 17 . When I began to play, my rhythms(节奏) were good, but my tone was another 18 . “Did you practice your les-son?” the teacher barked at me. I felt so 19 and my world came 20 down in an instant.From then on, I hated playing the clarinet and I kept getting 21 . With the day of the new performance approaching, I grew increasingly upset. In a moment of 22 , I asked for sick leave. It was so relieving and such a(n) 23 way out.The avoidance of my lessons continued until my mum asked me about it. “I want to quit.” My tears started 24 . “If you really want to quit, why are you crying?” asked mum. She25 and I realized I wanted to stay in band and, by not facing my fears, I had created a black hole that would be difficult to 26 out of. I made a 27 not to hide from my fears and to stand up to even the worst of them, so a 28 could be achieved.The next day I met with my other band teacher and told her I was having a problem and couldn't 29 why. She asked me gently to play for her. I tried, but only an unpleasant sound came out. She didn’t30 at me and handed me a new reed(簧片). I put it in place and tried again. To my great 31 I could play well. My problem was solved and my fear 32 improved a lot that year.33 I'm glad that I overcame my fear. Fear can 34 everything in a person's life. Hiding from those very fears only cligs a hole, which makes a person stay 35 inside. After facing up to a fear, one may find life easier and much more enjoyable.16、A. badly B. endlessly C. randomly D. smoothly17、A. mess up B. move on C. set out D. take off18、A. impression B. essay C. story D. factor19、A. ashamed B. starved C. excited D. relaxed20、A、crashing B. moving C. selling D. bending21、A. stricter B. worse C. happier D. smarter22、A. joy B. panic C. doubt D. sympathy23、A. funny B. important C. easy D. traditional24、A. drying B. disappearing C. flowing D. separating25、A. had a point B. made a change C. reached a level D. took a break26、A. send B. bring C. pick D. climb27、A. request B. resolution C. presentation D. proposal28、A. balance B. degree C. position D. solution29、A. figure out B. give away C. think over D. make up30、A、aim B. smile C. wave D. shout31、A. anger B. sorrow C. disappointment D. surprise32、A. felt B. shown C. removed D. voiced33、A. Carrying on B. Looking back C. Stepping aside D. Turning around34、A. consume B. examine C. reflect D. rescue35、A. unknown B. unpunished C. interested D. trapped三、阅读理解AGetting into college is a big step for high school graduates, and it comes with a lot of changes. For most students, it’s the first time they’re living away from home and mana ging their own life. Not surprisingly, adapting to this new lifestyle can be challenging. The following four tips will make high school graduates better prepared for college life.Goal settingWhen setting goals, whether they're academic, career, or personal, re-member they should be attainable but not too easy, so that you really have to push yourself to achieve them, and feel rewarded when you do. Writing down your goals and breaking down each huge, long-term goal into smaller more practical ones can help make it feel more real, and writing out a plan for achieving it can give you a roadmap to success.Interpersonal skillsAt college, you will interact with fellow students, professors, librarians, and many others. Strong interpersonal skills will help you build relationships during this time, and get more out of them. If you feel that your interpersonal skills need some work, practice asking thoughtful questions and listening closely, develop your understanding by putting yourself in someone else's shoes, and enhance your self-confidence.StudyingWith fewer in-class hours and more on-your-own learning, you're required to really digest learning material rather than simply memorize facts. To be successful in college you’ll need to learn how to int egrate large amounts of information obtained through reading, do research, and write pa-pers. Organization is the key, so if you are not someone who is naturally organized, set up your study schedule.BudgetingManaging money is a critical life skill, and for many, it is at college hat they develop it for the first time. Start by estimating your financial ballince. Then give high priority to the expenses on basic needs and determine low much money to set aside every month to cover those costs. Don't for get about savings…and the fun stuff(movies, dinners out), too.36、Who is this passage most probably written for?A. College teachers.B. Universily graduates.C. High school teachers.D. Would-be college students.37、What is the author's suggestion for reaching a huge goal?A. Divide it into smaller, more achievable ones.B. Reward oneself for each goal one has set.C. Purchase a clear, updated roadmap.D. Push oneself to an upper level.38、One of the suggested ways to enhance your interpersonal skills is to _____.A. prepare complicated questionsB. try on someone else's shoesC. listen to others carefullyD. take advantage of others39、What is the key to successful college study according to the author?A. Being well-organized.B. Being well-informed.C. Effective reading skills.D. Reliable research methods.40、To learn how to manage money, the first thing to do is _____.A. save money for financial investmentB. estimate one's income and expensesC. set aside money for fun activitiesD. open a personal bank accountBI'm an 18-year-old pre-medical student, tall and good-looking, with two short story books and quite a number of essays my credit. Why am I singing such praises of myself? Just to explain that he attainment of self-pride comes from a great deal of self-love, and to attain it, one must first learn to accept oneself as one is. That was where my struggle began.Born and raised in Africa，I had always taken my African origin as burden. My self-dislike was further fueled when my family had to relocate to Norway, where I attended a high school. Compared to all the white girls around me, with their golden hair and delicate lips, I, a black girl, had curly hair and full, red lips. My nose often had a thin sheet of sweat on it, whatever the weathe r was. I just wanted to bury myself in my shell crying “I’m so different!”What also contributed to my self-dislike was my occasional stuttering (口吃), which had weakened myself-confidence. It always stood between me and any fine opportunity. I'd taken it as an excuse to avoid any public speaking sessions, and unknowingly let it rule over me.Fortunately, as I grew older, there came a turning point. One day a white girl caught my eye on the school buswhen she suddenly turned back. To my astonishment, she had a thin sheet of sweat on her nose too, and it was in November! “Wow,” I whispered to myself, “this isn't a genetic(遗传的) disorder after all. It's perfectly normal.” Days later, my life took an-other twist(转折). Searching the internet for stuttering cures, I accidentally learned that such famous people as Isaac Newton and Winston Churchill also stuttered. I was greatly relieved and then an idea suddenly hit me—if I'm smart, I shouldn't allow my stuttering to stand between me and my success.Another boost to my self-confidence came days later as I was watching the news about Oprah Winfrey, the famous talkshow host and writer—she's black too! Whenever I think of her story and my former dislike of my color, I'm practically filled with shame.Today, I've grown to accept what I am with pride; it simply gives me feeling of uniqueness. The idea ofself-love has taken on a whole new meaning for me: there's always something fantastic about us, and what we need to do is learn to appreciate it.41、What affected the author's adjustment to her school life in Norway!A. Her appearanceB. Social discrimɪnation.C. Her changing emotions.D. The climate in Norway.42、What did the author's occasional stuttering bring about according on Paragraph 3?A. Her lack of self-confidence.B. Her loss of interest in school.C. Her unwillingness to greet her classmates.D. Her desire for chances to improve herself.43、How did the author feel on noticing the similarity between her and ne girl on the bus?A. Blessed and proud.B. Confused and afraid.C. Amazed and relieved.D. Shocked and ashamed.44、What lesson did the author learn from the cases of Newton and Churchill?A. Great minds speak alike.B. Stuttering is no barrier to success.C. Wisdom counts more than hard work.D. Famous people can't live with their weaknesses.45、What can best summarize the message contained in the passage?A. Pride comes before a fall.B. Where there is a will, there is a way.C. Self-acceptance is based on the love for oneself.D. Self-love is key to the attainment of self-pride.CIs it true that our brain alone is responsible for human cognition(认知)? What about our body? Is it possible for thought and behavior to originate from somewhere other than our brain? Psychologists who study Embodied Cognition(EC) ask similar questions. The EC theory suggests our body is also responsible for thinking or problem-solving. More precisely, the mind shapes the body and the body shapes the mind in equal measure.If you think about it for a moment, it makes total sense. When you smell something good or hear amusing sounds, certain emotions are awakened. Think about how newborns use their senses to understand the world around them. They don't have emotions so much as needs—they don't feel sad, they're just hungry and need food. Even unborn babies can feel their mothers’ heartbeats and this has a calming effect. In the real world，they cry when they're cold and then get hugged. That way, they start to as—sociate being warm with being loved.Understandably, theorists have been arguing for years and still dis-agree on whether the brain is the nerve centre that operates the rest of the body. Older Western philosophers and mainstream language researchers believe this is fact, while EC theorises that the brain and body are working together as an organic supercomputer, processing everything and forming your reactions.Further studies have backed up the mind-body interaction. In one experiment, test subjects(实验对象) were asked to judge people after being handed a hot or a cold drink.They all made warm evaluations when their fingertips perceived warmth rather than coolness. And it works the other way too; in another study, subjects’ fingertip temperatures were measured after being included in or “rejected” from a group task. Those who were included felt physically warmer.For further proof, we can look at the metaphors(比喻说法) that we use without even thinking. A kind and sympathetic person is frequently referred to as one with a soft heart and someone who is very strong and calm in difficult situations is often described as solid as a rock. And this kind of metaphorical use is common across languages.Now that you have the knowledge of mind-body interaction, why not use it? If you're having a bad day, a warm cup of tea will give you a flash of pleasure. If you know you're physically cold, warm up before making any interpersonal decisions.46、According to the author, the significance of the EC theory lies in _____.A. facilitating our understanding of the origin of psychologyB. revealing the major role of the mind in human cognitionC. offering a clearer picture of the shape of human brainD. bringing us closer to the truth in human cognition47、Where does the new borns’ understanding of their surroundings start from?A. Their personal looks.B. Their mental needs.C. Their inner emotions.D. Their physical feelings.48、The experiments mentioned in Paragraph 4 further prove _____.A. environment impacts how we judge othersB. how body temperature is related to healthC. the mind and the body influence each otherD. how humans interact with their surroundings49、What does the author intend to prove by citing the metaphors in Paragraph 5?A. Human speech is alive with metaphors.B. Human senses have effects on thinking.C. Human language is shaped by visual images.D. Human emotions are often compared to natural materials.50、What is the author's purpose in writing the last paragraph?A. To share with the reader ways to release their emotions.B. To guide the reader onto the path to career success.C. To encourage the reader to put EC into practice.D. To deepen the reader's understanding of EC.DRalph Emerson once said that the purpose of life is not to be happy, but to be useful, to be loving, to make some difference in the world. While we appreciate such words of wisdom, we rarely try to follow them in our lives.Most people prefer to live a good life themselves, ignoring their responsibilities for the world. This narrow perception of a good life may provide short-term benefits, but is sure to lead to long-term harm and suffering. A good life based on comfort and luxury may eventually lead to more pain because we spoil our health and even our character, principles, ideals, and relationships.What then, is the secret of a good life? A good life is a process, not a state of being a direction, not a destination. We have to earn a good life by first serving others without any expectation in return because theirhappiness is the very source of our own happiness. More importantly, we must know ourselves inside out. Only when we examine ourselves deeply can we discover our abilities and recognize our limitations, and then work accordingly to create a better world.The first requirement for a good life is having a loving heart. When we do certain right things merely as a duty, we find our job so tiresome that we’ll soon burn out. However, when we do that same job out of love, we not only enjoy what we do, but also do it with an effortless feeling.However, love alone is insufficient to lead a good life. Love sometimes blinds us to the reality. Consequently, our good intentions may not lead to good results. To achieve desired outcome, those who want to do good to others also need to equip themselves with accurate world knowledge. False knowledge is more dangerous than ignorance. If love is the engine of a car knowledge is the steering wheel(方向盘). If the engine lacks power, the car can't move; if the driver loses control of the steering, a road accident probably occurs. Only with love in heart and the right knowledge in mind can we lead a good life.With love and knowledge, we go all out to create a better world by doing good to others. When we see the impact of our good work on the world we give meaning to our life and earn lasting joy and happiness.51、What effect does the narrow perception of a good life have on us?A. Making us simple-mindedB. Making us short-signted.C. Leading us onto a busy road.D. Keeping us from comfort and luxury.52、According to the author, how can one gain true happiness?A. Through maintaining good health.B. By going through pain and suffering.C. By recognizing one's abilities and limitations.D. Through offering help much needed by others.53、According to Paragraph 4, doing certain right things with a loving heart makes one_____.A. less selfishB. less annoyingC. more motivatedD. more responsible54、In what case may good intentions fail to lead to desired results?A. When we have wrong knowledge of the world.B. When our love for the world is insufficient.C. When we are insensitive to dangers in life.D. When we stay blind to the reality.55、According to Paragraph 5, life can be made truly good when _____.A. inspired by love and guided by knowledgeB. directed by love and pushed by knowledgeC. purified by love and enriched by knowledgeD. promoted by love and defined by knowledge四、阅读表达It was a dark and stormy night. The ferocious wind shook the windows wildly, as though someone outside were beating on the glass. It was also New Year's Eve. We were having our annual party and had a house full of people just starting to celebrate.Suddenly, we heard loud explosions. Looking outside and up into the hills, we saw sparks(火花) flying from electrical transformers(变压器). One area after another went dark up in those hills. Then there was the loudest explosion of them all and our house went dark too. I tried to find every candle we had and lit them. The candles made everything look lovely. But we had problems. We had fifteen people standing around and we still had to cook dinner. How would we do that without electricity?The barbecue! Why not cook on the barbecue? We men went outside, some holding flashlights and others cooking. We did a wonderful job. The women stayed inside and got the salads ready. Everything was delicious. There were still a few hours to go before the beginning of the new year, so we all sat around the dining room table and sang up until a few minutes be-fore midnight. We couldn't watch the ball drop in Times Square on television but that wouldn't stop us from celebrating. I stood on a chair and, with the help of someone's watch to tell us the time, we all counted down and I dropped a tennis ball! We all screamed Happy New Year. We didn’t need elec tricity for that!Nowadays, we still get together with the same group to celebrate the New Year and we still talk about that special night. I don't think we have ever laughed so much as we did on that New Year’s Eve.56、What does the underlined word mean in Paragraph 1? (1 word)___________________________________________________________57、What made dinner preparation difficult according to Paragraph 2? (no more than 6 words)___________________________________________________________58、How did the people celebrate on New Year's Eve according to the passage? (no more than 10 words)___________________________________________________________59、How does the author feel about that particular New Year’s Eve? (no more than 8 words)___________________________________________________________60、What do you think is the most necessary quality when dealing with an unexpected difficult situation? Please explain why. (no more than 25words)___________________________________________________________五、写作61、假设你是晨光中学英语口语社的成员李津。
【2024年高考真题】天津市历史试题(有答案)
【2024年高考真题】天津市历史试题（有答案）一、选择题：本大题共15小题，每小题3分，共45分。
单项选择题：1—13题，每题3分，共39分。
每题只有一个正确选项。
1．周人认为，商之所以灭亡是因为“敛怨以为德”，为上天所弃；为此，周奉行“惟德是依”“敬天保民”。
这主要体现了周人A．能够借鉴历史经验B．推崇神权至上思想C．沿袭前朝价值观念D．坚持礼法并重传统2．唐中宗下令修改《氏族志》，至唐玄宗初年撰成《姓氏录》；书中对相关家族按照道德、功勋、血缘、籍贯，“等而次之”，对有官爵的“夷番酋长”也授予相应等级。
此举A．维护了门阀制度B．破除了门第界限C．改革了选官制度D．削弱了士族集团3．在编撰《资治通鉴》过程中，面对多政权并存时期如何确定编年标准的问题，司马光认为，政权无论华夷大小强弱，须同等对待，不能独尊一国为正统，“而其余皆为僭伪”，从中可以看出司马光A．坚持务实治史的追求B．推进华夏认同的志向C．强调民族对立的意图D．独尊中原王朝的立场4．下表为天津两所学堂的基本情况及主要课程设置，表中内容的联系与变化反映了天津武备学堂（1885年，军事专科学堂）天津中西学堂（1895年，综合性学堂）课程分为学、术两科：中国经史、测绘、算学、战法兵器等，马、步、炮队操演阵式等通学课程包括：高级应用英文课程，西学基础课程如几何学、微分学等，应用性课程如驾驶、万国公法、理财富国学等A．百日维新的成效B．统一学制的建立C．社会制度的演进D．洋务人士的反思5．1919年4月底，虽经中国代表的据理力争，巴黎和会上西方列强仍然决定将德国在山东的权益转让给日本，列强的这一行径，引发了国人“积压已久的不满与愤怒”。
从形成原因上，对“积压已久”理解全面的是A．五四爱国运动的激发B．群众爱国意识的觉醒C．近代历史发展的结果D．列强压迫中国的反弹6．下图表现了劳动人民边生产边学习的情景，它反映出A．统一战线受到了广泛拥护B．土地革命赢得了农民支持C．抗日民主根据地巩固发展D．敌后战场开始成为主战场7．1954年，中国与英国建立代办级外交关系；1964年，中法两国建立大使级外交关系；1972年，中美发表《联合公报》，两国关系开始走向正常化。
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天津理综答卷前，考生务必将自己的姓名、准考证号填写在答题卡上，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，答在试卷上无效，考试结束后，将本试卷哈答疑卡一并交回。
祝各位考生考生顺利！第Ⅰ卷注意事项：1.每题选出答案后，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，再选图其他答案标号。
2.本卷共8题，每题6分，共48分。
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.下列能揭示原子具有核式结构的实验是A. 光电效应实验B. 伦琴射线的发现C. α粒子散射实验D. 氢原子光谱的发现2.如图所示，A 、B 两物块叠放在一起，在粗糙的水平面上保持相对静止地向右做匀减速直线运动，运动过程中B 受到的摩擦力A. 方向向左，大小不变B. 方向向左，逐渐减小C. 方向向右，大小不变D. 方向向右，逐渐减小3. 质点做直线运动的位移x 与时间t 的关系为25x t t =+（各物理量均采用国际单位），则改质点A. 第1s 内的位移是5mB. 前2s 内的平均速度是6m/sC. 任意相邻的1s 内位移差都是1mD. 任意1s 内的速度增量都是2m/s4. 在匀强磁场中，一矩形金属线框绕与磁感线垂直的转轴匀速转动，如图1所示。
产生的交变电动势的图像如图2所示，则A. t =0.005s 时线框的磁通量变化率为零B. t =0.01s 时线框平面与中性面重合C. 线框产生的交变电动势有效值为311VD. 线框产生的交变电动势频率为100HZ5．板间距为d 的平等板电容器所带电荷量为Q 时，两极板间电势差为1U ，板间场强为1E 现将电容器所带电荷量变为2Q ，板间距变为12d ，其他条件不变，这时两极板间电势差2U ，板间场强为2E ，下列说法正确的是A. 2121,U U E E ==B. 21212,4U U E E ==C. 2121,2U U E E ==D. 21212,2U U E E ==6.甲、乙两单色光分别通过一双缝干涉装置得到各自的干涉图样，设相邻两个亮条纹的中心距离为0x ∆>，若∆x 甲> ∆x 乙，则下列说法正确的是A. 甲光能发生偏振现象，乙光则不能发生B. 真空中甲光的波长一定大于乙光的波长C. 甲光的光子能量一定大于乙光的光子能量D. 在同一均匀介质甲光的传播速度大于乙光7.位于坐标原点处的波源A 沿y 轴做简谐运动。
A 刚好完成一次全振动时，在介质中形成简谐横波的波形如图所示。
B 是沿波传播方向上介质的一个质点，则A ．波源A 开始振动时的运动方向沿y 轴负方向。
B. 此后的14周期内回复力对波源A 一直做负功。
C.经半个周期时间质点B 将向右迁移半个波长D.在一个周期时间内A 所收回复力的冲量为零8.质量为m 的探月航天器在接近月球表面的轨道上飞行，其运动视为匀速圆周运动。
已知月球质量为M ，月球半径为R ，月球表面重力加速度为g ，引力常量为G ，不考虑月球自转的影响，则航天器的A.线速度GM v R =B.角速度w gR =C.运行周期2R T gπ= D.向心加速度2Gm a R = 9.（18）（1）某同学利用测力计研究在竖直方向运行的电梯运动状态，他在地面上用测力计测量砝码的重力，实数是G ，他在电梯中用测力计仍测量同一砝码的重力，则测力计的实数小于G ，由此判断此时电梯的运动状态能是 。
（2）用螺旋测微器测量某金属丝直径的结果如图所示。
该金属丝的直径是 mm 。
（3）某同学用大头针、三角板、量角器等器材测量玻璃砖的折射率，开始玻璃砖位置如图中实线所示，使大头针1P 、2P 圆心O 在同一直线上，该直线垂直于玻璃砖的直径边，然后使玻璃砖绕圆心 缓缓转动，同时在玻璃砖的直径边一侧观察1P 、2P 的像，且2P 的像挡住1P 的像，如此只需测量出 ，即可计算出玻璃砖的折射率，请用你的方法表示出折射率n =（4）某同学测量阻值约为25k Ω的电阻R X ,现备有下列洗菜：A. 电流表（量程122μA, 内阻约2k Ω）；B. 电流表（量程500μA, 内阻约300Ω）；C. 电流表（量程15V, 内阻约100k Ω）；D. 电流表（量程50, 内阻约500k Ω）；E. 直流电源（20V, 允许最大电流1A ）；F. 滑动变阻器（最大阻值1k Ω，额定功率1W ）；G. 电键和导线若干。
电流表应选_________.电压表应_________.(填字母代号)该同学正确选择仪器后连接了以下电路，为保证实验顺利进行，并使测量误差尽量减小，实验前请你检查该电路，指出电路在接线上存在的问题：①_________________________________________.②_________________________________________.注意事项：1. 用黑色墨水的钢笔或签字笔将答案卸载答题卡上。
2. 本卷共4题，共64分7.（14分）图中X 、Y 、Z 为单质，其他为化学物，它们之间存在如下转化关系（部分产物已略去）。
其中，A 俗称磁性氧化铁：E 是不溶于水的酸性氧化物，能与氢氟酸反应。
回答下列问题：（1）组成单质Y 的元素在周期表中的位置是 ；M 中存在的化学键类型为 ；R 的化学式是 。
（2）一定条件下，Z 与2H 反应生成4ZH ,4ZH 的电子式为 。
（3）已知A与1molAl反应转化为X时（所有物质均为固体）。
放出akl热量。
写出该反应的热化学方程式：。
（4）写出A和D的稀溶液反应生成G的离子方程式：（5）问含4mol D 的稀溶液中，逐渐加入X3粉末至过量。
假设生成的气体只有一种，请在坐标系中画出n(X2-)随n（X）变化的示意图，并标出n(X2-)的最大值。
8.（18分）已知：1.冠心平F是降血脂、降胆固醇的药物，它的一条合成路线如下：（1）A为一元羧酸，8.8g A与足量NaHCO3溶液反应生成2.24L CO2（标准状况），A的分子式为_____________________________________________。
（2）写出符合A分子式的所有甲酸酯的结构简式：_________________________________________________________。
（3）B是氯代羧酸，其核磁共振氢谱有两个峰，写出B→C的反应方程式：__________________________________________________________。
（4）C+E→F的反应类型为________________________。
（5）写出A和F的结构简式：A______________________； F__________________________。
（6）D 的苯环上有两种氢，它所含官能团的名称为___________________；写出a 、b 所代表的试剂：a ______________; b___________。
Ⅱ.按如下路线，由C 可合成高聚物H ：（7）C →G 的反应类型为_____________________.（8）写出G →H 的反应方程式：_______________________。
9.（18分）某研究性学习小组为合成1-丁醇，查阅资料得知一条合成路线：请填写下列空白：（1）实验室现有锌粒、稀硝酸、稀盐酸、浓硫酸、2-丙醇，从中选择合适的试剂制备氦气，丙烯。
写出化学方程式： ， 。
（2）若用以上装置制备干燥纯净的CO,装置中a 和b 的作用分别是 ， ；C 和d 中承装的试剂分别是 ， 。
若用以上装置制备H 2， 气体发生装置中必需的玻璃仪器名称是 ；在虚线框内画出收集H 2干燥的装置图。
（3）制丙烯时，还产生少量2SO ,2CO 及水蒸气，该小组用以下试剂检验这四种气体，混合气体通过试剂的顺序是________________（填序号）①饱和23Na SO 溶液 ②酸性4KMnO 溶液 ③石灰水④无水4CuSO ⑤品红溶液（4）合成正丁醛的反应为正向放热的可逆反应，为增大反应速率和提高原料气的转化率，你认为应该采用的适宜反应条件是______________。
a. 低温、高压、催化剂b. 适当的温度、高压、催化剂c. 常温、常压、催化剂d. 适当的温度、常压、催化剂（5）正丁醛经催化剂加氢得到含少量正丁醛的1-丁醇粗品，为纯化10-丁醇，该小组查阅文献得知：①3R CHO NaHSO -+ （饱和）→()3RCH OH SO Na ↓；②沸点：乙醚34°C ，1-丁醇118°C ，并设计出如下提纯路线：试剂1为_________，操作1为________，操作2为_______，操作3为_______。
10．（14分）工业废水中常含有一定量的，它们会对人类及生态系统产生很大的伤害，必须进行处理。
常用的处理方法有两种。
方法1：还原沉淀法该法的工艺流程为其中第①步存在平衡：（1）若平衡体系的pH=2，则溶液显 色.（2）能说明第①步反应达平衡状态的是 。
a ．Cr 2O -27和CrO -24的浓度相同b ．2v (Cr 2O -27) =v (CrO -24)c.溶液的颜色不变（3）第②步中，还原1mol Cr 2O -27离子，需要 mol 的FeSO 4·7H 2O 。
（4）第③步生成的Cr(OH)3在溶液中存在以下沉淀溶解平衡：常温下，Cr(OH)的溶度积，要使降至3105 mol/L，溶液的pH应调至。
方法2：电解法该法用Fe做电极电解含的酸性废水，随着电解进行，在阴极附近溶液pH升高，沉淀。
产生Cr(OH)3（5）用Fe做电极的原因为。
（6）在阴极附近溶液pH升高的原因是（用电极反应解释）。
溶液中同时生成的沉淀还有。
4.玉米花药培养的单倍体幼苗，经秋水仙素处理后形成倍体植株，下图是该过程中某时段细胞核DNA含量变化示意图。
下列叙述错误的是：A.a-b过程中细胞不会发生基因变化B.c-d过程中细胞内发生了染色体数加倍C.c点后细胞内各染色体组的基因组成相同D.f-g过程中同源染色体分离，染色体数减半5.土壤农杆菌能将自身Ti的质粒的T-DNA整合到植物染色体DNA上，诱发植物形成肿瘤。
T-DNA中含有植物生长素合成酶基因（S）和细胞分裂素合成酶基因（R）,它们的表达与否能影响相应植物激素的含量，进而调节肿瘤组织的生长与分化。
据图分析，下列叙述错误的是A．当细胞分裂素与生长素的比值升高时，诱发肿瘤生芽B．清楚肿瘤中的土壤农杆菌后，肿瘤不再生长与分化C．图中肿瘤组织可在不含细胞分裂与生长的培养基izhong生长D．基因通过控制酶的合成控制代谢，进而控制肿瘤组织生长与分化6、某致病基因h位于X染色体上，该基因和正常基因H中的某一特定序列BclI酶切后，可产生大小不同的片段（如图1，bp表示碱基对），据此可进行基因诊断。