2003年高考试题上海卷(L)

1、2003年普通高等学校招生全国统一考试（上海卷）数学（理工农医类）考生注意： 1答卷前，考生务必将姓名、高考准考证号、校验码等填写清楚 2本试卷共有22道试题，满分150分考试时间120分钟请考生用钢笔或圆珠笔将答案直接写在试卷上 3华东师大二附中、大同中学、格致中学考生请注意试卷最后的符号说明一填空题（本大题满分48分）本大题共有12题，只要求直接填写结果，每个空格填对得4分，否则一律得零分（1）函数的最小正周期t= .（2）若是方程2cos（xa）1的解，其中a（0，2p），则a .（3）在等差数列an中，a53，a62，则a4a5a10 .（4）在极坐标系中，定点，点b在直线rcosqr

2、sinq0上运动，当线段ab最短时，点b的坐标是 .（5）在正四棱锥pabcd中，若侧面与底面所成二面角的大小为60，则异面直线pa与bc所成的大小等于 .（结果用反三角函数值表示）（6）设集合a=x| |x|4，b=x| x24x30和a2x2b2xc20的解集分别为集合m和n，那么是“mn”的 （a）充分非必要条件（b）必要非充分条件（c）充要条件（d）既非充分又非必要条件（16）f（x）是定义在区间c，c上的奇函数，其图象如图所示.令g（x）=af（x）+b，则下列关于函数g（x）的叙述正确的是 （a）若a0，则函数g（x）的图象关于原点对称.（b）若a1，0b0且a1）的图象y=x与的

3、图象有公共点，证明：f（x）= axm;（3）若函数f（x）sink xm，求实数k的取值范围符号意义本试卷所有符号等同于实验教材符号正切、余切tg、ctgtan、cot2003年普通高等学校招生全国统一考试（上海卷）数学参考答案（理工农医类）说明 1本解答列出试题的一种或几种解法，如果考生的解法与所列解法不同，可参照解答中评分标准的精进行评分。2评阅试卷，应坚持每题评阅到底，不要因为考生的解答中出现错误而中断对该题的评阅，当考生的解答在某一步出现错误，影响了后继部分，但该步以后的解答未改变这一的内容和难度时，可视影响程度决定后面部分的给分，这时原则上不应超过后面部分应给分数之半，如果有较严重

4、的概念性错误，就不给分。一、（第1题到第12题）（1）p （2）（3）49（4） （5）arctg2（6）1，3（7）（8）（a10，0q0，得 v8，故 （2）由得b（10，5），于是直线ob方程：由条件可知圆的标准方程为：（x3）2（y1）210，得圆心（3，1），半径为设圆心（3，1）关于直线ob的对称点为（x，y），则得故所求圆的方程为（x1）2（y3）210（3）设p（x1，y1），q（x2，y2）为抛物线上关于直线ob对称的两点，则得即x1、x2为方程的两个相异实根，于是由得故当时，抛物线y =ax21上总有关于直线ob对称的两点（22）解（1）对于非零常数t，f （x+t）=x+

5、t，tf （x）tx 因为对任意xr，x+t =tx不能恒成立， 所以f （x）x m （2）因为函数f （x）=ax （a0且a1）的图象与函数y=x的图象有公共点，所以方程组： 有解，消去y得ax=x，显然x=0不是方程的ax=x解，所以存在非零常数t，使at=t于是对于f （x）=ax ，有f （xt）=ax+t = atax=tax =t f （x），故f （x）=axm（3）当k=0时，f （x）=0，显然f （x）=0m当k0时，因为f （x）=sinkxm，所以存在非零常数t，对任意xr，有f （xt）= t f （x）成立，即sin（kxkt）= t sinkx因为k0时，且x

6、r，所以kxr，kxktr，于是sinkx1，1，sin（kxkt） 1，1，故要使sin（kxkt） = tsinkx成立，只有t=1当t1时，sin（kxk）= sinkx成立，则k=2mp，mz当t1时，sin（kxk）= sinkx成立，即sin（kxkp） = sinkx成立，则kp =2mp，mz，即k= （2m1） p，mz综合得，实数k的取值范围是k | k= mp，mz qcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiaw

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